MathsCBSE Class 10 · Chapter 2

Polynomials Class 10: Zeros and Their Relationship with Coefficients

Q: Is the division algorithm for polynomials in the Class 10 syllabus for 2026-27?

No. The division algorithm for polynomials and the relationship between the zeros and coefficients of a cubic polynomial are not in the 2026-27 syllabus. Only the quadratic relationship is assessed.

Q: Do I have to draw graphs of polynomials in the board exam?

Usually you read a given or described graph rather than plotting one. You need to count the points where it meets the x-axis and read off the zeros. Sometimes you also form the polynomial from those points and one extra point.

Q: Why is the answer to 'find a quadratic polynomial with these zeros' not unique?

Multiplying a polynomial by any non-zero number does not change its zeros, so there are infinitely many correct answers. Write the answer with a constant k, then pick a convenient value, usually one that clears the fractions. Any correct choice earns full marks.

Q: What if I cannot split the middle term to find the zeros?

First check whether the polynomial is a difference of squares, like x squared minus 3, which works even with surds. You can also complete the square. The quadratic formula from the Quadratic Equations chapter also gives the zeros, because the zeros of p(x) are the roots of p(x) = 0.

Q: Is it zeros or zeroes?

Both spellings are correct. The NCERT textbook uses zeroes and the syllabus uses zeros. Examiners accept either.

Q: How can I quickly check my zeros in the exam?

Add your two zeros and compare the result with minus b over a, then multiply them and compare with c over a. If both match, your zeros are almost certainly right. For extra certainty, substitute one zero back into the polynomial and confirm that you get 0.

Updated 13 October 2026 · Fact-checked

Polynomials in Class 10 CBSE covers the zeros of a polynomial. You read them from a graph as the points where it meets the x-axis, and you link the zeros of a quadratic to its coefficients. Board questions ask you to count zeros from graphs, find and verify zeros, form a quadratic from its zeros, and find unknown coefficients.

What Polynomials is about

This chapter is about one idea: the zeros of a polynomial. A zero is the value of $x$ that makes the polynomial equal to $0$ . You will find zeros in two ways. Graphically, they are where the graph meets the x-axis. Algebraically, you factorise. For a quadratic $a x^{2} + b x + c$ , the zeros are tied to the coefficients by two neat rules: the sum of zeros is $- \frac{b}{a}$ and the product is $\frac{c}{a}$ .

The chapter builds on Class 9 polynomials: degree, the value $p (k)$ , and factorisation by splitting the middle term. It feeds straight into Quadratic Equations, because the zeros of $p (x)$ are exactly the roots of $p (x) = 0$ . Reading points off a graph also links it to Coordinate Geometry.

Board questions take a few clear shapes:

MCQs where you count zeros from a described graph.
Short answers where you find the zeros and verify the sum and product rules.
Forming a quadratic from given zeros, or from a given sum and product.
Finding an unknown $k$ from a condition on the zeros.
Evaluating expressions like $α^{2} + β^{2}$ without finding $α$ and $β$ .
Case studies built on parabolic shapes such as arches and the path of a ball.

Weightage: This chapter is part of Unit II, Algebra, which carries 20 of the 80 theory marks together with Pair of Linear Equations in Two Variables, Quadratic Equations and Arithmetic Progressions.Source: CBSE Mathematics syllabus 2026-27, Unit II Algebra

In real board papers: Standard, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 3.8 marks per paper; Basic, asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.2 marks per paper. See the questions

Key concepts

Polynomials and their degree

NCERT section 2.1 Introduction

A polynomial in $x$ is a sum of terms of the form $a x^{n}$ , where each power $n$ is a whole number and each coefficient is a real number. The degree is the highest power of $x$ whose coefficient is not zero.

Linear: $a x + b$ , with $a \neq = 0$ .
Quadratic: $a x^{2} + b x + c$ , with $a \neq = 0$ .
Cubic: $a x^{3} + b x^{2} + c x + d$ , with $a \neq = 0$ .

Expressions like $x + 1$ or $x + \frac{1}{x}$ are not polynomials, because the powers $\frac{1}{2}$ and $- 1$ are not whole numbers.

Why does degree matter? It decides the shape of the graph and the greatest number of zeros the polynomial can have. Be careful when a coefficient contains an unknown such as $k$ . If that coefficient becomes $0$ , the term disappears and the degree drops. So 'non-zero coefficient' is part of the definition, not a side detail.

Worked example

For what value of $k$ is $p (x) = (k - 2) x^{3} + 3 x^{2} - k x + 5$ a quadratic polynomial? For this value of $k$ , find $p (- 1)$ .

Try it first, then show the solutionSolution

For $p (x)$ to be quadratic, the coefficient of $x^{3}$ must be $0$ and the coefficient of $x^{2}$ must be non-zero.
$k - 2 = 0 \Rightarrow k = 2$ .
The coefficient of $x^{2}$ is $3 \neq = 0$ , so the degree is exactly 2.
With $k = 2$ : $p (x) = 3 x^{2} - 2 x + 5$ .
$p (- 1) = 3 (- 1)^{2} - 2 (- 1) + 5 = 3 + 2 + 5 = 10$ .

Answer: $k = 2$ , and then $p (- 1) = 10$ .

Zeros of a polynomial

NCERT section 2.1 Introduction

A real number $k$ is a zero of $p (x)$ if $p (k) = 0$ . To test a number, substitute it. If you get $0$ , it is a zero.

A linear polynomial $a x + b$ has exactly one zero. Solving $a x + b = 0$ gives $x = - \frac{b}{a}$ . Watch the sign: the zero of $2 x + 3$ is $- \frac{3}{2}$ , not $\frac{3}{2}$ .

The zeros of $p (x)$ are the same as the roots of the equation $p (x) = 0$ . If $k$ is a zero, then $(x - k)$ is a factor of $p (x)$ . This is the factor theorem from Class 9. So when a question gives you one zero and an unknown coefficient, take two steps:

Substitute the zero to find the coefficient.
Factorise to get the remaining zero.

This 'unknown first, then the rest' pattern is very common in short-answer questions.

Worked example

If $x = 2$ is a zero of $p (x) = x^{2} + k x - 10$ , find the value of $k$ . Hence find the other zero.

Try it first, then show the solutionSolution

Since 2 is a zero, $p (2) = 0$ .
$(2)^{2} + k (2) - 10 = 0 \Rightarrow 4 + 2 k - 10 = 0$ .
$2 k = 6 \Rightarrow k = 3$ .
So $p (x) = x^{2} + 3 x - 10$ .
Split the middle term: $x^{2} + 5 x - 2 x - 10 = x (x + 5) - 2 (x + 5) = (x - 2) (x + 5)$ .
$x + 5 = 0 \Rightarrow x = - 5$ .
Check: the product of zeros is $2 \times (- 5) = - 10 = \frac{c}{a}$ . Correct.

Answer: $k = 3$ , and the other zero is $- 5$ .

Geometrical meaning of zeros

NCERT section 2.2 Geometrical Meaning of the Zeroes of a Polynomial

Draw the graph of $y = p (x)$ . At a zero $k$ we have $y = p (k) = 0$ , so the point $(k, 0)$ lies on the x-axis. So:

**The zeros of $p (x)$ are the x-coordinates of the points where the graph of $y = p (x)$ meets the x-axis**, whether it cuts the axis or just touches it.

To count zeros from a graph, count those meeting points.

A linear polynomial $a x + b$ has a straight-line graph. It cuts the x-axis exactly once, at $(- \frac{b}{a}, 0)$ . That is why it has exactly one zero.

The point where the graph crosses the y-axis is $(0, p (0))$ . This gives the constant term. It is not a zero, so never count it.

A quadratic graph is symmetric, so its axis of symmetry lies exactly midway between its two zeros.

Worked example

The front edge of a garden arch follows the graph of a quadratic polynomial $y = p (x)$ . The curve opens downward. It cuts the x-axis at $(- 1, 0)$ and $(3, 0)$ , and its highest point is $(1, 4)$ .
(i) How many zeros does $p (x)$ have, and what are they?
(ii) Find $p (x)$ .
(iii) At which point does the arch cross the y-axis?

Try it first, then show the solutionSolution

(i) The graph meets the x-axis at two points, so $p (x)$ has two zeros: $- 1$ and $3$ .
(ii) Since $- 1$ and $3$ are zeros, $(x + 1)$ and $(x - 3)$ are factors, so $p (x) = a (x + 1) (x - 3)$ .
The point $(1, 4)$ lies on the graph: $4 = a (1 + 1) (1 - 3) = - 4 a$ .
So $a = - 1$ .
$p (x) = - (x + 1) (x - 3) = - (x^{2} - 2 x - 3) = - x^{2} + 2 x + 3$ .
Check: $a = - 1 < 0$ , so the graph opens downward, as given. The sum of zeros is $- 1 + 3 = 2$ , and $- \frac{b}{a} = - \frac{2}{- 1} = 2$ . Correct.
(iii) $p (0) = 3$ , so the arch crosses the y-axis at $(0, 3)$ .

Answer: (i) Two zeros: $- 1$ and $3$ . (ii) $p (x) = - x^{2} + 2 x + 3$ . (iii) At $(0, 3)$ .

Shape of a quadratic graph and the number of zeros

NCERT section 2.2 Geometrical Meaning of the Zeroes of a Polynomial

The graph of $y = a x^{2} + b x + c$ is a curve called a parabola.

It opens upward (like a cup) if $a > 0$ .
It opens downward (like a cap) if $a < 0$ .

There are three possible cases:

The graph cuts the x-axis at two distinct points: two zeros.
The graph touches the x-axis at one point: the two zeros are equal, so there is one zero value.
The graph does not meet the x-axis: no real zeros.

So a quadratic has at most 2 zeros. In general, a polynomial of degree $n$ has at most $n$ zeros. For example, a cubic graph can meet the x-axis at most 3 times.

To decide without drawing, rewrite the quadratic as a perfect square plus a constant. A square is never negative. So if the constant is positive and $a > 0$ , the value can never be $0$ .

Worked example

The number of zeros of the polynomial $p (x) = x^{2} + 4 x + 5$ is
(a) 0
(b) 1
(c) 2
(d) 3

Try it first, then show the solutionSolution

Complete the square: $x^{2} + 4 x + 5 = (x^{2} + 4 x + 4) + 1 = (x + 2)^{2} + 1$ .
$(x + 2)^{2} \geq 0$ for every real $x$ , so $p (x) \geq 1$ .
So $p (x)$ is never $0$ .
The graph opens upward with its lowest point at $(- 2, 1)$ , above the x-axis, so it never meets the x-axis.

Answer: (a) 0

Finding zeros algebraically by factorisation

NCERT section 2.3 Relationship between Zeroes and Coefficients of a Polynomial

To find the zeros of a quadratic, set $p (x) = 0$ and factorise.

To split the middle term of $a x^{2} + b x + c$ , find two numbers whose **product is $a c$ and whose sum is $b$ **. Rewrite $b x$ using them, then group the terms in pairs.

Why does this give the zeros? A product is $0$ only when one of its factors is $0$ . So each factor set equal to $0$ gives one zero.

Some useful cases:

For forms like $x^{2} - 3$ , use $a^{2} - b^{2} = (a - b) (a + b)$ even with surds: $(x - 3) (x + 3)$ .
When the coefficients contain surds, $a c$ often turns out rational, which makes the split easy to spot.

Your final answer must state the zeros (values of $x$ ), not just the factors.

Worked example

Find the zeros of $p (x) = 43 x^{2} + 5 x - 23$ .

Try it first, then show the solutionSolution

Here $a c = 43 \times (- 23) = - 24$ .
We need two numbers with product $- 24$ and sum $5$ : these are $8$ and $- 3$ .
$p (x) = 43 x^{2} + 8 x - 3 x - 23$ .
Write $3 x = 3 \cdot 3 x$ and $23 = 3 \cdot 2$ : $p (x) = 4 x (3 x + 2) - 3 (3 x + 2)$ .
$p (x) = (4 x - 3) (3 x + 2)$ .
$4 x - 3 = 0 \Rightarrow x = \frac{3}{4}$ .
$3 x + 2 = 0 \Rightarrow x = - \frac{2}{3} = - \frac{2 3}{3}$ .
Check: the product of zeros is $\frac{3}{4} \times (- \frac{2}{3}) = - \frac{1}{2}$ , and $\frac{c}{a} = \frac{- 2 3}{4 3} = - \frac{1}{2}$ . Correct.

Answer: The zeros are $\frac{3}{4}$ and $- \frac{2 3}{3}$ .

Relationship between zeros and coefficients of a quadratic

NCERT section 2.3 Relationship between Zeroes and Coefficients of a Polynomial

Let $α$ and $β$ be the zeros of $a x^{2} + b x + c$ . Then $(x - α)$ and $(x - β)$ are factors, so

$a x^{2} + b x + c = a (x - α) (x - β) = a x^{2} - a (α + β) x + a α β$ .

Now compare the coefficients on both sides:

$b = - a (α + β)$ , so $α + β = - \frac{b}{a}$ .
$c = a α β$ , so $α β = \frac{c}{a}$ .

This is exactly why the minus sign appears in the sum but not in the product.

Use these results to:

verify zeros you have found;
check your factorisation;
find unknown coefficients;
learn about the zeros without finding them.

For example, if $\frac{c}{a} < 0$ , the product of the zeros is negative, so the zeros have opposite signs.

Worked example

Find the zeros of $p (x) = 6 x^{2} - 7 x - 3$ and verify the relationship between the zeros and the coefficients.

Try it first, then show the solutionSolution

$a c = 6 \times (- 3) = - 18$ . We need two numbers with product $- 18$ and sum $- 7$ : these are $- 9$ and $2$ .
$6 x^{2} - 9 x + 2 x - 3 = 3 x (2 x - 3) + 1 (2 x - 3) = (3 x + 1) (2 x - 3)$ .
Zeros: $x = - \frac{1}{3}$ and $x = \frac{3}{2}$ .
Sum of zeros $= \frac{3}{2} - \frac{1}{3} = \frac{9 - 2}{6} = \frac{7}{6}$ .
$- \frac{b}{a} = - \frac{- 7}{6} = \frac{7}{6}$ . The sum is verified.
Product of zeros $= \frac{3}{2} \times (- \frac{1}{3}) = - \frac{1}{2}$ .
$\frac{c}{a} = \frac{- 3}{6} = - \frac{1}{2}$ . The product is verified.

Answer: The zeros are $\frac{3}{2}$ and $- \frac{1}{3}$ . Sum $= \frac{7}{6} = - \frac{b}{a}$ and product $= - \frac{1}{2} = \frac{c}{a}$ , so the relationship is verified.

Forming a quadratic polynomial from its zeros

NCERT section 2.3 Relationship between Zeroes and Coefficients of a Polynomial

Run the relationship backwards. If the zeros are $α$ and $β$ , every quadratic with these zeros has the form

$k [x^{2} - (α + β) x + α β]$ , where $k$ is any non-zero real number.

In words: $x^{2}$ minus (sum) $x$ plus (product).

Why the $k$ ? Multiplying a polynomial by a non-zero constant does not change where it equals $0$ . So infinitely many polynomials share the same zeros. The board accepts any correct one. Usually you choose $k$ to clear the fractions.

If the question gives the sum $S$ and product $P$ directly, write $k (x^{2} - S x + P)$ .

When the zeros are a conjugate surd pair such as $p + q$ and $p - q$ , the surds cancel in both the sum and the product. So the polynomial has rational coefficients.

If an extra condition is given, such as the value at $x = 0$ , use it to fix $k$ .

Worked example

Find a quadratic polynomial whose zeros are $\frac{3 + 5}{2}$ and $\frac{3 - 5}{2}$ . Also find the polynomial with these zeros whose value at $x = 0$ is $4$ .

Try it first, then show the solutionSolution

Sum $= \frac{3 + 5}{2} + \frac{3 - 5}{2} = \frac{6}{2} = 3$ .
Product $= \frac{( 3 + 5 ) ( 3 - 5 )}{4} = \frac{9 - 5}{4} = 1$ .
The required polynomial is $k (x^{2} - 3 x + 1)$ . Taking $k = 1$ gives $x^{2} - 3 x + 1$ .
Check with $α = \frac{3 + 5}{2}$ : $α^{2} = \frac{14 + 6 5}{4} = \frac{7 + 3 5}{2}$ and $3 α = \frac{9 + 3 5}{2}$ .
So $α^{2} - 3 α + 1 = \frac{7 + 3 5 - 9 - 3 5}{2} + 1 = - 1 + 1 = 0$ . Correct.
For the second part: $p (0) = k (0 - 0 + 1) = k$ . We need $k = 4$ .
So $p (x) = 4 x^{2} - 12 x + 4$ .

Answer: $x^{2} - 3 x + 1$ (or any non-zero multiple of it). The one with value $4$ at $x = 0$ is $4 x^{2} - 12 x + 4$ .

Expressions in the zeros without finding the zeros

NCERT section 2.3 Relationship between Zeroes and Coefficients of a Polynomial

Often the zeros are irrational, or the question asks for something like $α^{2} + β^{2}$ or $\frac{1}{α} + \frac{1}{β}$ . Finding $α$ and $β$ first would be slow and messy.

Instead, rewrite the expression using only $α + β$ and $α β$ . You already know both from the coefficients. This works for any expression that stays the same when you swap $α$ and $β$ .

The key rewrites are:

$α^{2} + β^{2} = (α + β)^{2} - 2 α β$
$(α - β)^{2} = (α + β)^{2} - 4 α β$
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$
$α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)$

The same trick lets you build a new polynomial whose zeros are related to the old ones, such as $\frac{1}{α}$ and $\frac{1}{β}$ . Find the new sum and the new product, then use the forming rule.

Worked example

If $α$ and $β$ are the zeros of $p (x) = x^{2} - 6 x + 4$ , find (i) $α^{2} + β^{2}$ and (ii) a quadratic polynomial whose zeros are $\frac{1}{α}$ and $\frac{1}{β}$ .

Try it first, then show the solutionSolution

Here $a = 1$ , $b = - 6$ , $c = 4$ , so $α + β = - \frac{- 6}{1} = 6$ and $α β = \frac{4}{1} = 4$ .
(i) $α^{2} + β^{2} = (α + β)^{2} - 2 α β = 36 - 8 = 28$ .
(ii) New sum $= \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{6}{4} = \frac{3}{2}$ .
New product $= \frac{1}{α} \cdot \frac{1}{β} = \frac{1}{α β} = \frac{1}{4}$ .
The required polynomial is $k (x^{2} - \frac{3}{2} x + \frac{1}{4})$ . Taking $k = 4$ gives $4 x^{2} - 6 x + 1$ .
Check: the zeros of $x^{2} - 6 x + 4$ are $3 \pm 5$ . Then $\frac{1}{3 + 5} + \frac{1}{3 - 5} = \frac{6}{9 - 5} = \frac{3}{2}$ . Correct.

Answer: (i) $α^{2} + β^{2} = 28$ . (ii) $4 x^{2} - 6 x + 1$ (or any non-zero multiple of it).

Finding an unknown coefficient from a condition on the zeros

NCERT section 2.3 Relationship between Zeroes and Coefficients of a Polynomial

Many longer questions give a polynomial with an unknown such as $k$ , plus a condition on its zeros. The method has three steps:

Write the sum and the product of zeros in terms of $k$ .
Translate the condition into an equation.
Solve it.

Common conditions and how to handle them:

**One zero is $m$ times the other:** call the zeros $α$ and $m α$ , and use the sum to find $α$ first.
Zeros are reciprocals: the product is $1$ , so $c = a$ .
Zeros are negatives of each other: the sum is $0$ , so $b = 0$ .
**One zero is $0$ :** then $c = 0$ .
Difference of zeros given: use $(α - β)^{2} = (α + β)^{2} - 4 α β$ , or solve the sum and the difference together.

Always finish by checking. Put $k$ back in, factorise, and confirm the condition holds.

Worked example

One zero of the polynomial $p (x) = 2 x^{2} - 9 x + k$ is twice the other. Find the value of $k$ and both zeros.

Try it first, then show the solutionSolution

Let the zeros be $α$ and $2 α$ .
Sum of zeros: $α + 2 α = - \frac{- 9}{2} = \frac{9}{2}$ .
$3 α = \frac{9}{2} \Rightarrow α = \frac{3}{2}$ , so the zeros are $\frac{3}{2}$ and $3$ .
Product of zeros: $\frac{3}{2} \times 3 = \frac{k}{2}$ .
$\frac{9}{2} = \frac{k}{2} \Rightarrow k = 9$ .
Check: $2 x^{2} - 9 x + 9 = 2 x^{2} - 6 x - 3 x + 9 = 2 x (x - 3) - 3 (x - 3) = (2 x - 3) (x - 3)$ .
The zeros are $\frac{3}{2}$ and $3$ , and $3$ is twice $\frac{3}{2}$ . Correct.

Answer: $k = 9$ . The zeros are $\frac{3}{2}$ and $3$ .

Maths in real life: Polynomials

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Zeros of a polynomial

Suppose your class runs a snack stall at the school fete. After studying last year's sales, your teacher models the stall's profit in rupees on selling $x$ items as $p (x) = - x^{2} + 60 x - 800$ . Selling too few items does not cover the costs, and pushing too many means cutting the price too much.

Find the zeros of $p (x)$ and explain what they mean for the stall. Then use the sum of the zeros to find how many items should be sold for the greatest profit, and find that greatest profit.

Try it first, then show the solutionSolution

Set $p (x) = 0$ : $- x^{2} + 60 x - 800 = 0$ , that is, $x^{2} - 60 x + 800 = 0$ .
We need two numbers with product $800$ and sum $- 60$ : these are $- 20$ and $- 40$ .
$x^{2} - 20 x - 40 x + 800 = x (x - 20) - 40 (x - 20) = (x - 20) (x - 40)$ .
So the zeros are $20$ and $40$ .
Check: $p (20) = - 400 + 1200 - 800 = 0$ and $p (40) = - 1600 + 2400 - 800 = 0$ . At 20 items and at 40 items the stall just breaks even.
Sum of zeros $= - \frac{b}{a} = - \frac{60}{- 1} = 60$ , which agrees with $20 + 40$ .
The coefficient of $x^{2}$ is negative, so the graph opens downward, and its highest point lies midway between the zeros: $x = \frac{60}{2} = 30$ .
Greatest profit $= p (30) = - 900 + 1800 - 800 = 100$ .

Answer: The zeros are 20 and 40, the break-even numbers of items. The greatest profit comes from selling 30 items, and it is Rs 100.

Memory hook: Break-even points are the zeros, and the best sale sits halfway between them at half the sum of zeros.

Geometrical meaning of zeros

Imagine that on sports day a student kicks a football from the ground. Its height above the ground, $y$ metres, when it has travelled $x$ metres horizontally is modelled by $y = - \frac{1}{10} x^{2} + 2 x$ .

Find the zeros of $p (x) = - \frac{1}{10} x^{2} + 2 x$ and explain what each one means on the graph and on the field. Then use the symmetry of the graph to find how far from the kick the ball reaches its greatest height, and what that height is.

Try it first, then show the solutionSolution

Take out the common factor: $p (x) = - \frac{1}{10} x (x - 20)$ .
Set $p (x) = 0$ : either $x = 0$ or $x - 20 = 0$ . So the zeros are $0$ and $20$ .
The graph of $y = p (x)$ meets the x-axis at $(0, 0)$ and $(20, 0)$ . At these two points the height of the ball is zero.
$x = 0$ is where the ball is kicked, and $x = 20$ is where it lands, 20 m away.
The coefficient of $x^{2}$ is $- \frac{1}{10} < 0$ , so the parabola opens downward and has a highest point.
The axis of symmetry lies midway between the zeros: $x = \frac{0 + 20}{2} = 10$ .
The greatest height is $p (10) = - \frac{1}{10} (100) + 2 (10) = - 10 + 20 = 10$ .

Answer: The zeros are $0$ and $20$ . The ball is kicked at $0$ m and lands 20 m away. It is highest 10 m from the kick, at a height of 10 m.

Memory hook: Where the ball touches the ground, the graph touches the x-axis: those x-values are the zeros.

Finding zeros algebraically by factorisation

Suppose your family wants a rectangular vegetable patch on the terrace. Its length must be 7 m more than its breadth, and the area must be 60 square metres. If the breadth is $x$ metres, the area condition becomes $x (x + 7) = 60$ .

Form the polynomial $p (x) = x^{2} + 7 x - 60$ and find its zeros by splitting the middle term. Decide which zero gives the breadth, and hence find the length and the perimeter of the patch.

Try it first, then show the solutionSolution

From $x (x + 7) = 60$ we get $x^{2} + 7 x - 60 = 0$ , so $p (x) = x^{2} + 7 x - 60$ .
We need two numbers whose product is $a c = - 60$ and whose sum is $b = 7$ . These are $12$ and $- 5$ .
Split the middle term: $x^{2} + 12 x - 5 x - 60$ .
Group the terms: $x (x + 12) - 5 (x + 12) = (x + 12) (x - 5)$ .
Set each factor equal to $0$ : the zeros are $x = - 12$ and $x = 5$ .
A breadth cannot be negative, so the breadth is $x = 5$ m.
Length $= 5 + 7 = 12$ m. Check: $5 \times 12 = 60$ .
Perimeter $= 2 (12 + 5) = 34$ m.

Answer: The zeros are $5$ and $- 12$ . The breadth is 5 m, the length is 12 m and the perimeter is 34 m.

Memory hook: Split the middle term to find the zeros, then keep only the zero that makes sense for a real garden.

Relationship between zeros and coefficients of a quadratic

Imagine your class is making a rectangular rangoli for Diwali. The design sheet says its length and breadth, in metres, are the zeros of $p (x) = 2 x^{2} - 7 x + 6$ . You must buy a coloured border to go all round it.

Without finding the zeros, find the length of border needed (the perimeter) and the area of the rangoli. Then find the zeros by factorisation and verify the relationship between the zeros and the coefficients.

Try it first, then show the solutionSolution

Here $a = 2$ , $b = - 7$ , $c = 6$ . Let the length and breadth be $α$ and $β$ .
Sum of zeros: $α + β = - \frac{b}{a} = \frac{7}{2}$ .
Perimeter $= 2 (α + β) = 2 \times \frac{7}{2} = 7$ m.
Product of zeros: $α β = \frac{c}{a} = \frac{6}{2} = 3$ , so the area is $3$ square metres.
To factorise, split $- 7 x$ using two numbers with product $a c = 12$ and sum $- 7$ : $- 4$ and $- 3$ .
$2 x^{2} - 4 x - 3 x + 6 = 2 x (x - 2) - 3 (x - 2) = (x - 2) (2 x - 3)$ , so the zeros are $2$ and $\frac{3}{2}$ .
Verify the sum: $2 + \frac{3}{2} = \frac{7}{2} = - \frac{b}{a}$ .
Verify the product: $2 \times \frac{3}{2} = 3 = \frac{c}{a}$ .

Answer: The border needed is 7 m and the area is 3 square metres. The sides are 2 m and 1.5 m, and both relations are verified.

Memory hook: Border is the sum, floor is the product: $- \frac{b}{a}$ gives half the perimeter and $\frac{c}{a}$ gives the area.

Forming a quadratic polynomial from its zeros

Suppose your school is fitting 1 m by 1 m floor tiles in a rectangular activity room. The caretaker knows only that the perimeter of the room is 26 m and that it needs exactly 40 tiles to cover the floor.

Form a quadratic polynomial whose zeros are the length and the breadth of the room. Use it to find the dimensions, taking the length as the longer side.

Try it first, then show the solutionSolution

Let the length and breadth be $α$ and $β$ metres.
Perimeter: $2 (α + β) = 26$ , so $α + β = 13$ .
Area: 40 tiles of $1 m^{2}$ each, so $α β = 40$ .
The polynomial is $k [x^{2} - (α + β) x + α β] = k (x^{2} - 13 x + 40)$ . Taking $k = 1$ gives $x^{2} - 13 x + 40$ .
Factorise: two numbers with product $40$ and sum $- 13$ are $- 8$ and $- 5$ , so $x^{2} - 13 x + 40 = (x - 8) (x - 5)$ .
The zeros are $8$ and $5$ .
Check: $8 + 5 = 13$ and $8 \times 5 = 40$ .

Answer: One such polynomial is $x^{2} - 13 x + 40$ ; any $k (x^{2} - 13 x + 40)$ with $k \neq = 0$ also works. The room is 8 m long and 5 m wide.

Memory hook: Know the sum and the product? Write $x^{2}$ minus sum $x$ plus product, and the room's sides fall out.

Expressions in the zeros without finding the zeros

Imagine your class is building a rectangular photo-booth frame for the school festival. A wooden strip must be fixed along one diagonal to keep the frame rigid. The plan gives the length and breadth of the frame, in feet, as the zeros of $p (x) = x^{2} - 17 x + 60$ .

Without finding the length and breadth, find the length of the diagonal strip. Then find the total length of wood needed for the four sides and the diagonal.

Try it first, then show the solutionSolution

Let the sides be $α$ and $β$ . Here $a = 1$ , $b = - 17$ , $c = 60$ .
Sum: $α + β = - \frac{b}{a} = 17$ . Product: $α β = \frac{c}{a} = 60$ .
By Pythagoras, $diagonal^{2} = α^{2} + β^{2}$ .
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 289 - 120 = 169$ .
So the diagonal $= 169 = 13$ feet.
Four sides: $2 (α + β) = 34$ feet.
Total wood $= 34 + 13 = 47$ feet.
Check: $x^{2} - 17 x + 60 = (x - 5) (x - 12)$ , so the sides are 5 and 12, and $5^{2} + 1 2^{2} = 169$ .

Answer: The diagonal strip is 13 feet long, and the total wood needed is 47 feet.

Memory hook: Diagonal from the coefficients: $α^{2} + β^{2} = (sum)^{2} - 2 (product)$ , with no need to find the sides.

Topics in this chapter

Polynomials formula sheet

Zero of a linear polynomial: $p (x) = a x + b, a \neq = 0 \Rightarrow zero = - \frac{b}{a}$; A linear polynomial has exactly one zero. Its graph is a line cutting the x-axis at this value.
Definition of a zero: $k is a zero of p (x) ⟺ p (k) = 0$; Use this to test a value, or to find an unknown coefficient when one zero is given.
Standard form of a quadratic polynomial: $p (x) = a x^{2} + b x + c, a \neq = 0$; Always rearrange into this order before reading off a, b and c.
Standard form of a cubic polynomial: $p (x) = a x^{3} + b x^{2} + c x + d, a \neq = 0$; You only need its degree and how to read its zeros from a graph. Zero-coefficient relations for cubics are not in the syllabus.
Maximum number of zeros: $degree n \Rightarrow at most n zeros$; A quadratic has at most 2 zeros and a cubic at most 3. Use this to reject impossible graph options.
Geometrical meaning of zeros: $zeros of p (x) = x -coordinates of points where y = p (x) meets the x -axis$; Count the meeting points (cuts or touches) with the x-axis only. The y-axis crossing is never a zero.
Shape of a quadratic graph: $a > 0 : opens upward; a < 0 : opens downward$; Use this to match a polynomial to its graph, or to decide the sign of a from a described curve.
y-intercept of the graph: $p (0) = c$; The graph crosses the y-axis at (0, c). This is useful in case studies and when fixing k.
Sum of zeros of a quadratic: $α + β = - \frac{b}{a} = - \frac{coefficient of x}{coefficient of x ^{2}}$; Use this to verify zeros, find unknowns, or get the sum without solving. Note the minus sign.
Product of zeros of a quadratic: $α β = \frac{c}{a} = \frac{constant term}{coefficient of x ^{2}}$; There is no minus sign here. A negative product means the zeros have opposite signs.
Factored form: $a x^{2} + b x + c = a (x - α) (x - β)$; This is where the two relations come from. It is also useful when a graph gives you the zeros and one more point.
Quadratic from its zeros: $k [x^{2} - (α + β) x + α β], k \neq = 0$; Use this when the zeros, or their sum and product, are given. Choose k to clear fractions.
Sum of squares of zeros: $α^{2} + β^{2} = (α + β)^{2} - 2 α β$; This is the most used rewrite. Never write it as just the square of the sum.
Square of the difference of zeros: $(α - β)^{2} = (α + β)^{2} - 4 α β$; Use this when the difference of the zeros is given or asked for.
Sum of reciprocals of zeros: $\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}$; Use this to build a polynomial whose zeros are the reciprocals of the given zeros.
Ratio sum of zeros: $\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β}$; Find the sum of squares first, then divide by the product.
Sum of reciprocal squares: $\frac{1}{α ^{2}} + \frac{1}{β ^{2}} = \frac{α ^{2} + β ^{2}}{( α β ) ^{2}}$; This is the same idea as the sum of reciprocals, applied to the squares.
Sum of cubes of zeros: $α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)$; This appears in harder 3-mark and 5-mark questions on expressions in the zeros.
Special conditions on zeros: $reciprocal zeros : c = a; zeros α, - α : b = 0; one zero is 0 : c = 0$; These are quick ways to turn a condition into an equation in k.

How the board asks Polynomials

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Count the zeros from a described graph	An MCQ describes the graph of y = p(x). It tells you where the graph cuts or touches the x-axis and sometimes where it crosses the y-axis. You are asked how many zeros p(x) has.	Count only the points on the x-axis, including any point where the graph just touches it. Ignore the y-axis crossing. Remember that a quadratic can have at most 2 zeros and a cubic at most 3.
Find the zeros and verify the relationship	You are given a quadratic such as $6 x^{2} - 7 x - 3$ , $4 u^{2} + 8 u$ or $t^{2} - 15$ . You must find its zeros and verify the sum and product relations. This is usually worth 2 or 3 marks.	First rearrange into standard form. Then factorise by splitting the middle term, by taking out a common factor, or with $a^{2} - b^{2}$ . State the zeros. Then show the sum against $- \frac{b}{a}$ and the product against $\frac{c}{a}$ on separate lines.
Form a quadratic from its zeros or from a sum and product	Find a quadratic polynomial whose zeros are, for example, $2 + 3$ and $2 - 3$ , or whose sum and product of zeros are given numbers.	Find the sum S and the product P. Write $k (x^{2} - S x + P)$ and choose $k$ to clear fractions. If the question gives an extra condition, such as the value at $x = 0$ , use it to fix $k$ .
Evaluate an expression in the zeros	If $α$ and $β$ are the zeros of a given quadratic, find $α^{2} + β^{2}$ , $\frac{1}{α} + \frac{1}{β}$ , $\frac{α}{β} + \frac{β}{α}$ or $α^{3} + β^{3}$ , without finding the zeros.	Write down $α + β$ and $α β$ from the coefficients. Rewrite the expression using the standard identities. Substitute and simplify.
Find an unknown when one zero is given	One zero of a quadratic with an unknown coefficient k is given. You must find k, and often the other zero too.	Substitute the given zero and set the result equal to 0 to find k. Then find the other zero by factorising, or quickly from the sum or the product of zeros.
Find an unknown from a relation between the zeros	You are given a condition, such as one zero being twice the other, the zeros being reciprocals, the zeros being negatives of each other, or the sum being equal to the product. You must find k.	Express the sum and the product in terms of k. Translate the condition into an equation: use $α$ and $m α$ , product $= 1$ , or sum $= 0$ . Solve, then check by factorising.
Build a new polynomial from old zeros	In a 3-mark or 5-mark question, $α$ and $β$ are the zeros of a given quadratic. You must find a quadratic whose zeros are, for example, $\frac{1}{α}$ and $\frac{1}{β}$ , $\frac{α}{β}$ and $\frac{β}{α}$ , or $2 α$ and $2 β$ .	Find the new sum and the new product using identities in $α + β$ and $α β$ . Write $k (x^{2} - S x + P)$ and choose $k$ to give integer coefficients.
Assertion-reason and case study on parabolic shapes	An assertion-reason item tests a claim about zeros, for example about their signs or their number. A case study describes an arch, a bridge or the path of a ball as a parabola, gives where it meets the ground, and asks for the zeros, the polynomial, or a value.	For assertion-reason, test each statement separately against the sum and product rules or the graph rule. Then decide whether R explains A. For a case study, the ground is the x-axis, so the points where the curve meets the ground give the zeros. Write $p (x) = a (x - α) (x - β)$ and use one more given point to find $a$ .

Tricks to remember Polynomials

Short memory aids. Say each one aloud once, then use it on a question the same day.

Sum flips, product stays
Sum of zeros: flip the sign of b and divide by a. Product of zeros: copy c and divide by a. Only the sum carries a minus sign.
Use it when: Every verification and every question that finds k uses this. Say it once before you write the relations.
x squared, minus Sum, plus Product
To form a quadratic, say 'x squared minus S x plus P'. The signs go minus then plus, always.
Use it when: Forming a polynomial from its zeros or from a sum and product. It stops the most common sign slip.
Only the floor counts
Think of the x-axis as the floor. Zeros are where the graph lands on the floor, by cutting through it or touching it. The y-axis is a wall, and a wall hit is never a zero.
Use it when: Graph-counting MCQs, where students often count y-axis crossings by mistake.
Smile or frown
If a is positive, the parabola smiles (opens upward). If a is negative, it frowns (opens downward).
Use it when: Matching a polynomial to a described graph, and finding the sign of a in case studies.
Two-second sign check
Look at the sign of c over a. If it is negative, the zeros are real and have opposite signs. If it is positive and the zeros exist, they have the same sign, and the sign of minus b over a tells you which.
Use it when: Spotting a wrong factorisation or a wrong MCQ option before you hand in the paper.
Plug one back in
After finding the zeros, substitute the easier one into p(x). If you do not get 0, there is a slip somewhere.
Use it when: A quick check before moving on, especially when there are surds or fractions.

Polynomials: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Finding unknown coefficients from given zeros · 1 mark · MCQBoard level

If $2$ and $- 3$ are the zeros of the polynomial $x^{2} + (a + 1) x + b$ , then
(a) $a = - 7$ , $b = - 1$
(b) $a = 5$ , $b = - 1$
(c) $a = 2$ , $b = - 6$
(d) $a = 0$ , $b = - 6$

Try it first, then show the solutionSolution

Sum of zeros $= 2 + (- 3) = - 1$ and product of zeros $= 2 \times (- 3) = - 6$ .
For $x^{2} + (a + 1) x + b$ , the coefficient of $x^{2}$ is 1. So the sum of zeros is $- (a + 1)$ and the product is $b$ .
$- (a + 1) = - 1$ gives $a = 0$ .
$b = - 6$ .
Check: $x^{2} + x - 6 = (x + 3) (x - 2)$ , which has zeros $2$ and $- 3$ .

Answer: (d) $a = 0$ , $b = - 6$

Example 2: Touching the x-axis: assertion-reason · 1 mark · Assertion-reasonBoard level

Assertion (A): The graph of $y = x^{2} - 4 x + 4$ meets the x-axis at exactly one point.
Reason (R): A quadratic polynomial can have at most two zeros.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$x^{2} - 4 x + 4 = (x - 2)^{2}$ , which is zero only when $x = 2$ .
So the graph meets the x-axis only at $(2, 0)$ . It touches the axis there and does not cross it. A is true.
A polynomial of degree 2 has at most 2 zeros. R is true.
R only says 'at most two'. It does not explain why there is exactly one. The real reason is that the polynomial is a perfect square. So R is not the correct explanation of A.

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Example 3: One zero given, find k and the other zero · 2 marks · Very short answerBoard level

If $- 3$ is a zero of the polynomial $p (x) = (k - 1) x^{2} + k x + 1$ , find the value of $k$ . Hence find the other zero.

Try it first, then show the solutionSolution

Since $- 3$ is a zero, $p (- 3) = 0$ : $(k - 1) (9) + k (- 3) + 1 = 0$ . [½ mark]
$9 k - 9 - 3 k + 1 = 0$ gives $6 k = 8$ , so $k = \frac{4}{3}$ . [½ mark]
Then $p (x) = \frac{1}{3} x^{2} + \frac{4}{3} x + 1 = \frac{1}{3} (x^{2} + 4 x + 3) = \frac{1}{3} (x + 1) (x + 3)$ . [½ mark]
So the other zero is $- 1$ . Check: the product of zeros should be $\frac{c}{a} = \frac{1}{1/3} = 3$ , and $(- 3) (- 1) = 3$ . [½ mark]

Answer: $k = \frac{4}{3}$ , and the other zero is $- 1$ .

Example 4: Zeros with surds, and verification · 3 marks · Short answerBoard level

Find the zeros of the quadratic polynomial $p (x) = 3 x^{2} + 10 x + 73$ and verify the relationship between the zeros and the coefficients.

Try it first, then show the solutionSolution

Here $a c = 3 \times 73 = 21$ . Split $10 x$ as $7 x + 3 x$ , since $7 \times 3 = 21$ and $7 + 3 = 10$ .
$3 x^{2} + 3 x + 7 x + 73 = 3 x (x + 3) + 7 (x + 3) = (x + 3) (3 x + 7)$ [1 mark]
The zeros are $- 3$ and $- \frac{7}{3} = - \frac{7 3}{3}$ . [½ mark]
Sum $= - 3 - \frac{7 3}{3} = - \frac{10 3}{3}$ , and $- \frac{b}{a} = - \frac{10}{3} = - \frac{10 3}{3}$ . These are equal. [1 mark]
Product $= (- 3) \times (- \frac{7}{3}) = 7$ , and $\frac{c}{a} = \frac{7 3}{3} = 7$ . These are equal. [½ mark]

Answer: The zeros are $- 3$ and $- \frac{7 3}{3}$ . Sum $= - \frac{10 3}{3} = - \frac{b}{a}$ and product $= 7 = \frac{c}{a}$ , so the relationship is verified.

Example 5: Reciprocal zeros: find k, then the zeros · 3 marks · Short answerHarder board level

If the zeros of the polynomial $p (x) = (k^{2} + 4) x^{2} + 20 x + 4 k$ are reciprocals of each other, find the value of $k$ . Hence find the zeros and check that they are reciprocals.

Try it first, then show the solutionSolution

Here $a = k^{2} + 4$ , which is never $0$ , and $c = 4 k$ . Product of zeros $= \frac{c}{a} = \frac{4 k}{k ^{2} + 4}$ . [½ mark]
If the zeros are $α$ and $\frac{1}{α}$ , their product is $1$ . So $\frac{4 k}{k ^{2} + 4} = 1$ , that is, $k^{2} + 4 = 4 k$ . [½ mark]
$k^{2} - 4 k + 4 = 0 \Rightarrow (k - 2)^{2} = 0 \Rightarrow k = 2$ . [1 mark]
With $k = 2$ : $p (x) = 8 x^{2} + 20 x + 8 = 4 (2 x^{2} + 5 x + 2) = 4 (2 x + 1) (x + 2)$ . [½ mark]
The zeros are $- \frac{1}{2}$ and $- 2$ . Check: $(- \frac{1}{2}) \times (- 2) = 1$ , so they are reciprocals of each other. [½ mark]

Answer: $k = 2$ . The zeros are $- \frac{1}{2}$ and $- 2$ , which are reciprocals of each other.

Example 6: Sum of zeros equals product: find k and the zeros · 3 marks · Short answerHarder board level

If the sum of the zeros of the polynomial $p (x) = 2 k x^{2} - 3 x + 9 k$ , where $k \neq = 0$ , is equal to their product, find the value of $k$ . Hence find the zeros of $p (x)$ and verify the condition.

Try it first, then show the solutionSolution

Here $a = 2 k$ , $b = - 3$ , $c = 9 k$ . Sum $= - \frac{- 3}{2 k} = \frac{3}{2 k}$ and product $= \frac{9 k}{2 k} = \frac{9}{2}$ . [1 mark]
Sum $=$ product: $\frac{3}{2 k} = \frac{9}{2} \Rightarrow 18 k = 6 \Rightarrow k = \frac{1}{3}$ . [½ mark]
With $k = \frac{1}{3}$ : $p (x) = \frac{2}{3} x^{2} - 3 x + 3 = \frac{1}{3} (2 x^{2} - 9 x + 9)$ . [½ mark]
$2 x^{2} - 9 x + 9 = 2 x^{2} - 6 x - 3 x + 9 = 2 x (x - 3) - 3 (x - 3) = (2 x - 3) (x - 3)$ , so the zeros are $\frac{3}{2}$ and $3$ . [½ mark]
Check: sum $= \frac{3}{2} + 3 = \frac{9}{2}$ and product $= \frac{3}{2} \times 3 = \frac{9}{2}$ . They are equal. [½ mark]

Answer: $k = \frac{1}{3}$ . The zeros are $\frac{3}{2}$ and $3$ , and both their sum and their product equal $\frac{9}{2}$ .

Example 7: Unknown from a condition, then a new polynomial · 5 marks · Long answerHarder board level

If $α$ and $β$ are the zeros of $p (x) = x^{2} - 8 x + k$ such that $α^{2} + β^{2} = 40$ , find the value of $k$ . Find the zeros of $p (x)$ and verify the relationship between the zeros and the coefficients. Also find a quadratic polynomial whose zeros are $α + 2$ and $β + 2$ .

Try it first, then show the solutionSolution

$α + β = - \frac{- 8}{1} = 8$ and $α β = \frac{k}{1} = k$ . [½ mark]
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 64 - 2 k$ [½ mark]
$64 - 2 k = 40$ gives $2 k = 24$ , so $k = 12$ . [1 mark]
$p (x) = x^{2} - 8 x + 12 = (x - 2) (x - 6)$ , so the zeros are $2$ and $6$ . [1 mark]
Verify: sum $= 8 = - \frac{- 8}{1}$ and product $= 12 = \frac{12}{1}$ . Also $2^{2} + 6^{2} = 40$ , which checks the condition. [1 mark]
The new zeros are $2 + 2 = 4$ and $6 + 2 = 8$ . Sum $= 12$ , product $= 32$ , so the polynomial is $x^{2} - 12 x + 32$ . [1 mark]

Answer: $k = 12$ . The zeros are $2$ and $6$ , and the relationship is verified. A polynomial with zeros $α + 2$ and $β + 2$ is $x^{2} - 12 x + 32$ (or any non-zero multiple of it).

Example 8: New polynomial without finding the zeros · 5 marks · Long answerHarder board level

If $α$ and $β$ are the zeros of $p (x) = x^{2} - 3 x - 2$ , find
(i) the value of $α^{3} + β^{3}$
(ii) a quadratic polynomial whose zeros are $\frac{1}{2 α + β}$ and $\frac{1}{α + 2 β}$ .

Try it first, then show the solutionSolution

$α + β = 3$ and $α β = - 2$ . The zeros themselves are irrational, so work only with the sum and product. [½ mark]
(i) $α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β) = 27 - 3 (- 2) (3) = 27 + 18 = 45$ [1½ marks]
(ii) $(2 α + β) + (α + 2 β) = 3 (α + β) = 9$ [½ mark]
$(2 α + β) (α + 2 β) = 2 α^{2} + 5 α β + 2 β^{2} = 2 (α^{2} + β^{2}) + 5 α β$ , where $α^{2} + β^{2} = 9 - 2 (- 2) = 13$ . [½ mark]
So $(2 α + β) (α + 2 β) = 2 (13) + 5 (- 2) = 16$ . [½ mark]
Sum of new zeros $= \frac{1}{2 α + β} + \frac{1}{α + 2 β} = \frac{9}{16}$ , and product of new zeros $= \frac{1}{16}$ . [1 mark]
Required polynomial $= k (x^{2} - \frac{9}{16} x + \frac{1}{16})$ . Take $k = 16$ to get $16 x^{2} - 9 x + 1$ . [½ mark]

Answer: (i) $α^{3} + β^{3} = 45$
(ii) $16 x^{2} - 9 x + 1$ (or any non-zero multiple of it)

Example 9: Find the error in a student's polynomial · 3 marks · Short answerHOTS / competency

Riya was asked to find a quadratic polynomial whose sum of zeros is $- \frac{3}{2}$ and product of zeros is $- 1$ . She wrote $2 x^{2} - 3 x - 2$ . Find her mistake, write the correct polynomial, and find its zeros to check your answer.

Try it first, then show the solutionSolution

For Riya's polynomial, sum of zeros $= - \frac{- 3}{2} = \frac{3}{2}$ . This is the wrong sign. Her product, $\frac{- 2}{2} = - 1$ , is correct. [1 mark]
Her mistake: she used $+ (sum) x$ instead of $- (sum) x$ . The correct form is $x^{2} - (- \frac{3}{2}) x + (- 1) = x^{2} + \frac{3}{2} x - 1$ . Multiply by 2 to get $2 x^{2} + 3 x - 2$ . [1 mark]
Check: $2 x^{2} + 3 x - 2 = (2 x - 1) (x + 2)$ , so the zeros are $\frac{1}{2}$ and $- 2$ . Sum $= - \frac{3}{2}$ and product $= - 1$ . [1 mark]

Answer: Riya used the wrong sign for the $x$ term. The correct polynomial is $2 x^{2} + 3 x - 2$ , with zeros $\frac{1}{2}$ and $- 2$ .

Example 10: Signs of coefficients from a graph · 3 marks · Short answerHOTS / competency

The graph of $y = a x^{2} + b x + c$ is a parabola that opens downward. It crosses the x-axis at two points: one to the left of the origin and one to the right. The point on the right is farther from the origin than the point on the left. The curve cuts the y-axis above the origin. Find the signs of $a$ , $b$ and $c$ . Give reasons.

Try it first, then show the solutionSolution

The parabola opens downward, so $a < 0$ . [½ mark]
The graph cuts the y-axis at $(0, p (0)) = (0, c)$ , which is above the origin, so $c > 0$ . [½ mark]
The zeros $α$ and $β$ have opposite signs, so $α β < 0$ . This agrees with $α β = \frac{c}{a}$ , which is a positive number divided by a negative number. [1 mark]
The positive zero is larger in size, so $α + β > 0$ . That means $- \frac{b}{a} > 0$ . Since $a < 0$ , this needs $b > 0$ . [1 mark]
Example check: $- x^{2} + x + 2 = - (x - 2) (x + 1)$ has zeros $2$ and $- 1$ , with $a = - 1$ , $b = 1$ , $c = 2$ .

Answer: $a < 0$ , $b > 0$ , $c > 0$

Case-study practice (Section E style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 1 + 2 marks.

Water jets at a musical fountain

A city park in Mysuru has a musical fountain. Engineers record the path of one water jet along a straight ground line. Let $x$ be the horizontal distance in metres from a lamp post, and $y$ the height of the water in metres. Their readings are: at $x = 1$ , $y = 0$ ; at $x = 2$ , $y = 3$ ; at $x = 3$ , $y = 4$ ; at $x = 4$ , $y = 3$ ; at $x = 5$ , $y = 0$ . The path is the graph of $y = p (x) = - x^{2} + 6 x - 5$ , a parabola that opens downward with its highest point at $(3, 4)$ .

(i) · 1 mark
What are the zeros of $p (x)$ , and what do they mean for the jet?
Answer
Answer: The zeros are $1$ and $5$ . The water leaves the ground 1 m from the lamp post and lands 5 m from it.
The zeros are the x-values where $y = 0$ , that is, where the graph meets the x-axis. The readings give $x = 1$ and $x = 5$ . Check: $p (1) = - 1 + 6 - 5 = 0$ and $p (5) = - 25 + 30 - 5 = 0$ .
(ii) · 1 mark
Verify the product of the zeros using the coefficients of $p (x)$ .
Answer
Answer: Product $= 5$ , which equals $\frac{c}{a} = 5$ .
Product of zeros $= 1 \times 5 = 5$ . Here $a = - 1$ and $c = - 5$ , so $\frac{c}{a} = \frac{- 5}{- 1} = 5$ . The two values are equal.
(iii) · 2 marks
A second jet also follows a quadratic path. It meets the ground 2 m and 8 m from the lamp post and is 9 m high at $x = 5$ . Find the polynomial for its path.
Answer
Answer: $- x^{2} + 10 x - 16$
The zeros are $2$ and $8$ , so the path is $q (x) = k (x - 2) (x - 8)$ . [½ mark] Put $x = 5$ : $k (3) (- 3) = 9$ , so $- 9 k = 9$ and $k = - 1$ . [1 mark] So $q (x) = - (x^{2} - 10 x + 16) = - x^{2} + 10 x - 16$ . Check: $q (5) = - 25 + 50 - 16 = 9$ . [½ mark]

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

If $α$ and $β$ are the zeros of $p (x) = x^{2} - 5 x + k$ such that $α - β = 1$ , find the value of $k$ . (2 marks)

Step	Marks
Sum of zeros: $α + β = - \frac{- 5}{1} = 5$	0.5
Solve $α + β = 5$ and $α - β = 1$ together: $2 α = 6$ , so $α = 3$ and $β = 2$	1
Product of zeros: $k = α β = 3 \times 2 = 6$	0.5

Find the zeros of $p (x) = 3 x^{2} - x - 4$ and verify the relationship between the zeros and the coefficients. (3 marks)

Step	Marks
Split the middle term ( $a c = - 12$ , numbers $- 4$ and $3$ ): $3 x^{2} - 4 x + 3 x - 4 = x (3 x - 4) + 1 (3 x - 4) = (x + 1) (3 x - 4)$	1
Zeros: $x + 1 = 0 \Rightarrow x = - 1$ and $3 x - 4 = 0 \Rightarrow x = \frac{4}{3}$	1
Sum $= - 1 + \frac{4}{3} = \frac{1}{3}$ and $- \frac{b}{a} = - \frac{- 1}{3} = \frac{1}{3}$ , so the sum is verified	0.5
Product $= - 1 \times \frac{4}{3} = - \frac{4}{3}$ and $\frac{c}{a} = \frac{- 4}{3}$ , so the product is verified	0.5

If $α$ and $β$ are the zeros of $p (x) = 2 x^{2} - 5 x + 1$ , find a quadratic polynomial whose zeros are $\frac{α}{β}$ and $\frac{β}{α}$ . (5 marks)

Step	Marks
$α + β = - \frac{- 5}{2} = \frac{5}{2}$ and $α β = \frac{1}{2}$	1
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{25}{4} - 1 = \frac{21}{4}$	1
New sum $= \frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β} = \frac{21}{4} \div \frac{1}{2} = \frac{21}{2}$	1
New product $= \frac{α}{β} \times \frac{β}{α} = 1$	0.5
The required polynomial is $k (x^{2} - \frac{21}{2} x + 1)$ , where $k \neq = 0$	1
Taking $k = 2$ , the polynomial is $2 x^{2} - 21 x + 2$	0.5

Where students lose marks in Polynomials

The slips examiners see most often, and what to write instead.

Mistake: Dropping the minus sign in the sum of zeros
Students remember 'b over a' and forget that the sum is the negative of b over a. This happens most when b is already negative and the double negative feels confusing.
Wrong working
For $2 x^{2} - 5 x + 3$ : $α + β = \frac{b}{a} = \frac{- 5}{2}$
Right working
For $2 x^{2} - 5 x + 3$ : $α + β = - \frac{b}{a} = - \frac{( - 5 )}{2} = \frac{5}{2}$
Do this: Write the formula $- \frac{b}{a}$ first. Then substitute b in brackets, including its sign.
Usually costs: The full 1 mark in an MCQ. In a verification question, the ½ to 1 mark for the sum step.
Mistake: Counting the y-axis crossing as a zero
Students count every point where the curve meets an axis, instead of only the x-axis.
Wrong working
The graph cuts the x-axis at 2 points and the y-axis at 1 point, so the number of zeros $= 3$ .
Right working
Zeros are where the graph meets the x-axis only, so the number of zeros $= 2$ .
Do this: Read the description again and count only the points of the form $(k, 0)$ .
Usually costs: 1 mark (the whole MCQ).
Mistake: Reading coefficients before writing standard form
The polynomial is given in a jumbled order, and students take a, b and c in the order the terms appear.
Wrong working
$p (x) = 5 x - 2 + 3 x^{2}$ , so $a = 5$ , $b = - 2$ , $c = 3$ , and the sum of zeros $= \frac{2}{5}$ .
Right working
Rearrange first: $p (x) = 3 x^{2} + 5 x - 2$ , so $a = 3$ , $b = 5$ , $c = - 2$ . Sum $= - \frac{5}{3}$ and product $= - \frac{2}{3}$ .
Do this: Always rewrite the polynomial as $a x^{2} + b x + c$ on a fresh line before using any formula. If a term is missing, its coefficient is 0. For example, $t^{2} - 15$ has $b = 0$ .
Usually costs: Usually every later step, so 1 to 2 marks in a 2-mark or 3-mark question.
Mistake: Getting the wrong sign of a zero from a factor
Students read the number inside the bracket directly, instead of setting the factor equal to 0.
Wrong working
$(2 x - 3) (x + 1) = 0$ , so the zeros are $- \frac{3}{2}$ and $1$ .
Right working
$2 x - 3 = 0 \Rightarrow x = \frac{3}{2}$ and $x + 1 = 0 \Rightarrow x = - 1$ , so the zeros are $\frac{3}{2}$ and $- 1$ .
Do this: Write 'factor $= 0$ ' for each factor and solve it. Then check that the sum matches $- \frac{b}{a}$ .
Usually costs: The ½ to 1 mark for the zeros, plus the marks for the verification that follows.
Mistake: Using plus for the sum when forming a polynomial
Students write $x^{2} + S x + P$ because it looks like the standard form.
Wrong working
Sum $= 4$ and product $= - 5$ , so the polynomial is $x^{2} + 4 x - 5$ .
Right working
The polynomial is $x^{2} - (4) x + (- 5) = x^{2} - 4 x - 5$ . Check: $(x - 5) (x + 1)$ has zeros $5$ and $- 1$ , with sum $4$ . Correct.
Do this: Write the template $x^{2} - (sum) x + (product)$ first, then substitute each value in brackets.
Usually costs: The full 1 to 2 marks for the final answer.
Mistake: Treating $α^{2} + β^{2}$ as $(α + β)^{2}$
Students forget the cross term $2 α β$ in the expansion of a square.
Wrong working
For $x^{2} - 6 x + 4$ : $α^{2} + β^{2} = (α + β)^{2} = 36$
Right working
For $x^{2} - 6 x + 4$ : $α^{2} + β^{2} = (α + β)^{2} - 2 α β = 36 - 8 = 28$
Do this: Learn the identity $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ as a single line, and write it out before substituting.
Usually costs: 1 to 2 marks, and everything that builds on it in a 5-mark question.
Mistake: Writing an equals sign between non-equal polynomials
When clearing fractions, students multiply through by a number but write '=' as if nothing changed.
Wrong working
$x^{2} - \frac{1}{4} x - 1 = 4 x^{2} - x - 4$
Right working
The required polynomial is $k (x^{2} - \frac{1}{4} x - 1)$ . Taking $k = 4$ gives $4 x^{2} - x - 4$ .
Do this: Bring in the constant $k$ in words, then state the value of $k$ you choose. Do not join different polynomials with '='.
Usually costs: Often ½ mark for incorrect working, even though the final polynomial is acceptable.

NCERT exercise guide: Polynomials

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Exercise 2.1

What it tests: Reading the number of zeros of a polynomial from its graph. Each question shows a curve $y = p (x)$ , and you count how many times it meets the x-axis.

How to approach it: A zero of $p (x)$ is the x-coordinate of a point where the graph meets the x-axis. Count every point where the curve crosses or just touches the x-axis. Ignore where it cuts the y-axis, because that point gives $p (0)$ , not a zero. A curve that never meets the x-axis has no zeros.

Like the graph-reading questions in Exercise 2.1

The graph of $y = p (x)$ is described as follows. Coming from the lower left, the curve crosses the x-axis upward at $(- 3, 0)$ and rises above it. It cuts the y-axis at $(0, 2)$ , then comes down and touches the x-axis at $(1, 0)$ without crossing it, staying above. It rises, turns, and comes down to cross the x-axis at $(4, 0)$ , then goes on downward. Find the number of zeros of $p (x)$ and name them. Can $p (x)$ be a quadratic polynomial? Give a reason.

Show the solutionSolution

The zeros of $p (x)$ are the x-coordinates of the points where the graph meets the x-axis.
The curve meets the x-axis at $(- 3, 0)$ , $(1, 0)$ and $(4, 0)$ . A touching point counts, just like a crossing point.
The point $(0, 2)$ is on the y-axis. It only tells you $p (0) = 2$ , so it is not a zero.
So $p (x)$ has 3 zeros: $- 3$ , $1$ and $4$ .
A polynomial of degree $n$ has at most $n$ zeros. A quadratic (degree 2) has at most 2 zeros, so this polynomial cannot be quadratic. Its degree is at least 3.

Answer: 3 zeros: $- 3$ , $1$ and $4$ . It cannot be quadratic, because a quadratic has at most 2 zeros.

Exercise 2.2

What it tests: Two kinds of question. (1) Find the zeros of a quadratic polynomial by factorisation and verify the relationship between the zeros and the coefficients. (2) Find a quadratic polynomial when the sum and product of its zeros are given.

How to approach it: For the first kind, split the middle term, write each factor equal to zero, then check that the sum of the zeros is $- \frac{b}{a}$ and the product is $\frac{c}{a}$ . Write both sides of each check, because the verification carries marks. For the second kind, write $k [x^{2} - (sum) x + product]$ and choose $k$ to clear fractions.

Like the find-the-zeros-and-verify questions in Exercise 2.2

Find the zeros of $p (x) = 4 x^{2} - 4 x - 3$ and verify the relationship between the zeros and the coefficients.

Show the solutionSolution

Here $a = 4$ , $b = - 4$ , $c = - 3$ , so $a c = - 12$ . You need two numbers with product $- 12$ and sum $- 4$ . These are $- 6$ and $2$ .
$4 x^{2} - 4 x - 3 = 4 x^{2} - 6 x + 2 x - 3 = 2 x (2 x - 3) + 1 (2 x - 3) = (2 x - 3) (2 x + 1)$
$p (x) = 0$ gives $2 x - 3 = 0$ or $2 x + 1 = 0$ , so the zeros are $α = \frac{3}{2}$ and $β = - \frac{1}{2}$ .
Sum of zeros $= \frac{3}{2} - \frac{1}{2} = 1$ , and $- \frac{b}{a} = - \frac{- 4}{4} = 1$ . These are equal.
Product of zeros $= \frac{3}{2} \times (- \frac{1}{2}) = - \frac{3}{4}$ , and $\frac{c}{a} = \frac{- 3}{4}$ . These are equal.
So the relationship is verified.

Answer: The zeros are $\frac{3}{2}$ and $- \frac{1}{2}$ . Sum $= 1 = - \frac{b}{a}$ and product $= - \frac{3}{4} = \frac{c}{a}$ , so the relationship is verified.

Polynomials: previous year questions

Standard: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 3.8 marks per paper; 52 questions in all, counting OR alternatives. In 4 of 4 sample papers.
Basic: Asked in 21 of the 21 board papers we hold (2023 to 2026), on average 4.2 marks per paper; 55 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Polynomials

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

2026 Basic, set 430/1/1, Q27 · 3 marks · Short answer

Find a quadratic polynomial whose sum and product of zeroes are $0$ and $- 9$ , respectively. Also, find the zeroes of the polynomial so obtained.

Try it first, then show our solutionSolution

A quadratic polynomial with sum of zeroes S and product P is $x^{2} - S x + P$ . So the polynomial is $x^{2} - 0 \cdot x + (- 9) = x^{2} - 9$ .
1 mark
For the zeroes: $x^{2} - 9 = (x + 3) (x - 3) = 0$ .
1 mark
Zeroes are $x = - 3$ and $x = 3$ . Check: sum $= 0$ and product $= - 9$ .
1 mark

Answer: x² − 9; zeroes −3 and 3

Where the marks are

Forms the polynomial $x^{2} - 9$ (1 mark)
Factorises it as $(x + 3) (x - 3)$ (1 mark)
States the zeroes $- 3, 3$ (1 mark)

See it in the full 2026 board paper, set 430/1/1

2026 Basic, set 430/4/1, Q6 · 1 mark · MCQ

The graph of polynomial $p (x) = k$ is shown here. Number of zeroes of polynomial $p (x)$ is :

(A) $0$
(B) $1$
(C) $2$
(D)infinitely many

Figure in the paper: Cartesian graph with axes XX′ (horizontal) and YY′ (vertical) meeting at origin 0; a horizontal line labelled p(x) = k drawn parallel to the x-axis, above it, not meeting the x-axis. See it in the official paper.

Try it first, then show our solutionSolution

Zeroes are the points where the graph meets the x-axis.
The line $y = k$ is parallel to the x-axis and lies above it, so it never meets the x-axis. There are 0 zeroes, option (A).
1 mark

Answer: 0, option (A)

Where the marks are

Notes that the graph does not cut the x-axis (1 mark)

See it in the full 2026 board paper, set 430/4/1

2026 Basic, set 430/4/1, Q22(a) · 2 marks · Very short answer

If $α, β$ are the zeroes of polynomial $p (x) = 6 x^{2} -5 x -3$ , then find the value of $\frac{1}{α} + \frac{1}{β}$ .

Try it first, then show our solutionSolution

For $6 x^{2} - 5 x - 3$ : $a = 6, b = - 5, c = - 3$ . So $α + β = - \frac{b}{a} = \frac{5}{6}$ and $α β = \frac{c}{a} = \frac{- 3}{6} = - \frac{1}{2}$ .
1 mark
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{5/6}{- 1/2} = - \frac{5}{3}$
1 mark

Answer: $\frac{1}{α} + \frac{1}{β} = - \frac{5}{3}$

Where the marks are

Writes sum $= \frac{5}{6}$ and product $= - \frac{3}{6}$ of zeroes (1 mark)
Uses $\frac{α + β}{α β}$ to get $- \frac{5}{3}$ (1 mark)

See it in the full 2026 board paper, set 430/4/1

Every Polynomials question, by year

Standard (62 questions, sample papers included)

For 2027

Sample paper 2026-27: Q2 (1 mark, solved), Q20 (1 mark, solved), Q27 (3 marks, solved)

2026

Set 30/1/1: Q4 (1 mark, solved), Q20 (1 mark, solved), Q21 (2 marks, solved)

Set 30/2/1: Q4 (1 mark, solved), Q5 (1 mark, solved), Q21 (2 marks, solved)

Set 30/3/1: Q14 (1 mark, solved), Q36 (4 marks, solved)

Set 30/4/1: Q5 (1 mark, solved), Q25 (2 marks, solved)

Set 30/5/1: Q3 (1 mark, solved), Q36 (4 marks, solved)

Sample paper 2025-26: Q12 (1 mark, solved), Q28 (3 marks, solved)

2025

Set 30/1/1: Q1 (1 mark, solved), Q10 (1 mark, solved), Q22 (2 marks, solved)

Set 30/2/1: Q16 (1 mark, solved), Q21 (2 marks, solved)

Set 30/3/1: Q4 (1 mark, solved), Q17 (1 mark, solved), Q24(b) (2 marks, solved)

Set 30/4/1: Q17 (1 mark, solved)

Set 30/5/1: Q27 (3 marks, solved)

Set 30/6/1: Q4 (1 mark, solved), Q27 (3 marks, solved)

Sample paper 2024-25: Q1 (1 mark, solved), Q8 (1 mark, solved), Q28 (3 marks, solved)

2024

Set 30/1/1: Q1 (1 mark, solved), Q10 (1 mark, solved), Q20 (1 mark, solved)

Set 30/2/1: Q9 (1 mark, solved), Q36 (4 marks, solved)

Set 30/3/1: Q6 (1 mark, solved), Q27 (3 marks, solved)

Set 30/4/1: Q3 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 30/5/1: Q15 (1 mark, solved), Q20 (1 mark, solved), Q23 (2 marks, solved)

Sample paper 2023-24: Q2 (1 mark, solved), Q27 (3 marks, solved)

2023

Set 30/1/1: Q1 (1 mark, solved), Q7 (1 mark, solved), Q14 (1 mark, solved)

Set 30/2/1: Q9 (1 mark, solved), Q14 (1 mark, solved), Q16 (1 mark, solved), Q20 (1 mark, solved)

Set 30/4/1: Q8 (1 mark, solved), Q17 (1 mark, solved), Q22 (2 marks, solved)

Set 30/5/1: Q1 (1 mark, solved), Q38 (4 marks, solved)

Set 30/6/1: Q3 (1 mark, solved), Q6 (1 mark, solved), Q17 (1 mark, solved)

Basic (62 questions, sample papers included)

For 2027

Sample paper 2026-27: Q2 (1 mark, solved), Q23 (2 marks, solved)

2026

Set 430/1/1: Q27 (3 marks, solved)

Set 430/2/1: Q27 (3 marks, solved)

Set 430/3/1: Q27 (3 marks, solved)

Set 430/4/1: Q6 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved)

Set 430/5/1: Q9 (1 mark, solved), Q12 (1 mark, solved), Q30(a) (3 marks, solved), Q30(b) (3 marks, solved)

Sample paper 2025-26: Q3 (1 mark, solved), Q23 (2 marks, solved)

2025

Set 430/1/1: Q27 (3 marks, solved)

Set 430/2/1: Q27 (3 marks, solved)

Set 430/3/1: Q27 (3 marks, solved)

Set 430/4/1: Q4 (1 mark, solved), Q9 (1 mark, solved), Q26 (3 marks, solved)

Set 430/5/1: Q4 (1 mark, solved), Q13 (1 mark, solved), Q26(a) (3 marks, solved), Q26(b) (3 marks, solved)

Set 430/6/1: Q1 (1 mark, solved), Q26(a) (3 marks, solved)

Sample paper 2024-25: Q18 (1 mark, solved), Q37 (4 marks, solved)

2024

Set 430/1/1: Q1 (1 mark, solved), Q15 (1 mark, solved), Q20 (1 mark, solved)

Set 430/2/1: Q9 (1 mark, solved), Q10 (1 mark, solved), Q29(a) (3 marks, solved), Q29(b) (3 marks, solved)

Set 430/3/1: Q4 (1 mark, solved), Q11 (1 mark, solved), Q29 (3 marks, solved)

Set 430/4/1: Q6 (1 mark, solved), Q7 (1 mark, solved), Q24(a) (2 marks, solved), Q24(b) (2 marks, solved)

Set 430/5/1: Q25(a) (2 marks, solved), Q25(b) (2 marks, solved), Q27 (3 marks, solved)

Sample paper 2023-24: Q27 (3 marks, solved)

2023

Set 430/1/1: Q7 (1 mark, solved), Q10 (1 mark, solved), Q29 (3 marks, solved)

Set 430/2/1: Q12 (1 mark, solved), Q19 (1 mark, solved), Q24 (2 marks, solved), Q30 (3 marks, solved)

Set 430/4/1: Q1 (1 mark, solved), Q5 (1 mark, solved), Q26 (3 marks, solved)

Set 430/5/1: Q7 (1 mark, solved), Q36 (4 marks, solved)

Set 430/6/1: Q7 (1 mark, solved), Q15 (1 mark, solved), Q22(a) (2 marks, solved), Q22(b) (2 marks, solved)

Practice questions on Polynomials

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The graph of $y = p (x)$ is described as follows. It crosses the x-axis at $(- 4, 0)$ going downward and cuts the y-axis at $(0, - 2)$ . It then turns and rises to touch the x-axis at $(1, 0)$ , and goes back down. Finally it turns again and crosses the x-axis going upward at $(3, 0)$ . The number of zeros of $p (x)$ is
(a) 1
(b) 2
(c) 3
(d) 4
Hint
Count every point where the curve meets the x-axis, including the touching point. Do not count the y-axis point.
Answer
(c) 3
The graph meets the x-axis at $x = - 4$ , $1$ and $3$ . The point $(0, - 2)$ is on the y-axis and is not a zero.
Q2 · 2 marks · Very short answerBoard level
Find a quadratic polynomial whose zeros are $2 + 3$ and $2 - 3$ .
Hint
Find the sum and the product first. The surds cancel in both.
Answer
Sum $= 4$ and product $= 4 - 3 = 1$ .
The polynomial is $x^{2} - 4 x + 1$ (or any non-zero multiple of it).
Q3 · 1 mark · Assertion-reasonBoard level
Assertion (A): The polynomial $p (x) = x^{2} + 2 x + 5$ has no zeros.
Reason (R): The graph of $y = x^{2} + 2 x + 5$ lies entirely above the x-axis.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Write $x^{2} + 2 x + 5$ as a perfect square plus a number.
Answer
(a)
$x^{2} + 2 x + 5 = (x + 1)^{2} + 4 \geq 4 > 0$ , so the graph never meets the x-axis. That is exactly why there are no zeros.
Q4 · 3 marks · Short answerBoard level
Find the zeros of $p (y) = 7 y^{2} - \frac{11}{3} y - \frac{2}{3}$ and verify the relationship between the zeros and the coefficients.
Hint
Write $p (y) = \frac{1}{3} (21 y^{2} - 11 y - 2)$ and factorise the bracket.
Answer
$21 y^{2} - 11 y - 2 = (7 y + 1) (3 y - 2)$ , so the zeros are $- \frac{1}{7}$ and $\frac{2}{3}$ .
Sum $= \frac{11}{21}$ , and $- \frac{b}{a} = \frac{11/3}{7} = \frac{11}{21}$ .
Product $= - \frac{2}{21}$ , and $\frac{c}{a} = \frac{- 2/3}{7} = - \frac{2}{21}$ . Verified.
Q5 · 3 marks · Short answerHarder board level
If $α$ and $β$ are the zeros of $p (x) = 3 x^{2} - 5 x - 2$ , find the value of $\frac{α ^{2}}{β} + \frac{β ^{2}}{α}$ without finding the zeros.
Hint
Combine the fractions to get $\frac{α ^{3} + β ^{3}}{α β}$ .
Answer
$α + β = \frac{5}{3}$ and $α β = - \frac{2}{3}$ .
$α^{3} + β^{3} = \frac{125}{27} - 3 (- \frac{2}{3}) (\frac{5}{3}) = \frac{215}{27}$
So the value is $\frac{215}{27} \div (- \frac{2}{3}) = - \frac{215}{18}$ . Check with the zeros $2$ and $- \frac{1}{3}$ : $- 12 + \frac{1}{18} = - \frac{215}{18}$ .
Q6 · 3 marks · Short answerHarder board level
One zero of the polynomial $3 x^{2} - 8 x + 2 k + 1$ is seven times the other. Find both zeros and the value of $k$ .
Hint
Call the zeros $α$ and $7 α$ . Use the sum first, then the product.
Answer
$α + 7 α = \frac{8}{3}$ gives $α = \frac{1}{3}$ , so the zeros are $\frac{1}{3}$ and $\frac{7}{3}$ .
Product: $\frac{7}{9} = \frac{2 k + 1}{3}$ gives $2 k + 1 = \frac{7}{3}$ , so $k = \frac{2}{3}$ .
Check: $9 x^{2} - 24 x + 7 = (3 x - 1) (3 x - 7)$ .
Q7 · 5 marks · Long answerHarder board level
If $α$ and $β$ are the zeros of $p (x) = x^{2} - 2 x - 8$ , find (i) $α^{2} + β^{2}$ , (ii) $\frac{1}{α} + \frac{1}{β}$ , and (iii) a quadratic polynomial whose zeros are $2 α + 1$ and $2 β + 1$ . Verify (iii) by finding $α$ and $β$ directly.
Hint
For (iii), the product is $(2 α + 1) (2 β + 1) = 4 α β + 2 (α + β) + 1$ .
Answer
$α + β = 2$ and $α β = - 8$ .
(i) $α^{2} + β^{2} = 4 + 16 = 20$
(ii) $\frac{1}{α} + \frac{1}{β} = \frac{2}{- 8} = - \frac{1}{4}$
(iii) New sum $= 2 (2) + 2 = 6$ and new product $= - 32 + 4 + 1 = - 27$ , giving $x^{2} - 6 x - 27$ . Check: the zeros of $p (x)$ are $4$ and $- 2$ , so the new zeros are $9$ and $- 3$ , and $(x - 9) (x + 3) = x^{2} - 6 x - 27$ .
Q8 · 3 marks · Short answerHarder board level
If $α$ and $β$ are the zeros of $x^{2} - p x + q$ , where $q \neq = 0$ , prove that $\frac{α ^{2}}{β ^{2}} + \frac{β ^{2}}{α ^{2}} = \frac{p ^{4}}{q ^{2}} - \frac{4 p ^{2}}{q} + 2$ .
Hint
Write the left side as $\frac{α ^{4} + β ^{4}}{( α β ) ^{2}}$ , and use $α^{4} + β^{4} = (α^{2} + β^{2})^{2} - 2 (α β)^{2}$ .
Answer
$α + β = p$ , $α β = q$ , so $α^{2} + β^{2} = p^{2} - 2 q$ .
$α^{4} + β^{4} = (p^{2} - 2 q)^{2} - 2 q^{2} = p^{4} - 4 p^{2} q + 2 q^{2}$
Divide by $q^{2}$ to get $\frac{p ^{4}}{q ^{2}} - \frac{4 p ^{2}}{q} + 2$ . Hence proved.
Q9 · 3 marks · Short answerHOTS / competency
A ball is thrown upward from the top of a building. Its height above the ground after $t$ seconds is $h (t) = - 5 t^{2} + 10 t + 15$ metres. (i) How tall is the building? (ii) Find the zeros of $h (t)$ . (iii) After how many seconds does the ball hit the ground? Explain why you reject one of the zeros.
Hint
Take out $- 5$ as a common factor, then factorise the quadratic that is left.
Answer
(i) $h (0) = 15$ , so the building is 15 m tall.
(ii) $h (t) = - 5 (t^{2} - 2 t - 3) = - 5 (t - 3) (t + 1)$ , so the zeros are $3$ and $- 1$ .
(iii) The ball hits the ground after 3 seconds. Time is measured from the moment of the throw, so $t = - 1$ has no meaning here and is rejected.
Q10 · 2 marks · Very short answerHOTS / competency
Aman says: 'If the constant term of a quadratic polynomial is negative, the polynomial must have two zeros of opposite signs.' Is he right? Justify your answer with a reason or a counterexample.
Hint
Try a polynomial whose $x^{2}$ coefficient is also negative.
Answer
He is not right. Counterexample: $- x^{2} - 1$ has $c = - 1 < 0$ , but $- x^{2} - 1 \leq - 1$ for every $x$ , so it has no zeros.
The correct statement: if $a$ and $c$ have opposite signs, then $\frac{c}{a} < 0$ , and the graph crosses the x-axis on both sides of the origin.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.When is a real number $k$ called a zero of $p (x)$ ?Show answer
When $p (k) = 0$ .
2.What is the zero of the linear polynomial $a x + b$ ?Show answer
$- \frac{b}{a}$ , where $a \neq = 0$ .
3.How many zeros can a polynomial of degree $n$ have at most?Show answer
At most $n$ zeros. So a quadratic has at most 2 and a cubic at most 3.
4.Where do you read the zeros of $p (x)$ on the graph of $y = p (x)$ ?Show answer
They are the x-coordinates of the points where the graph meets the x-axis.
5.For $a x^{2} + b x + c$ , write the sum and the product of the zeros.Show answer
$α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ .
6.Write the general quadratic polynomial with zeros $α$ and $β$ .Show answer
$k [x^{2} - (α + β) x + α β]$ , where $k \neq = 0$ .
7.What can you say about the zeros if the graph of a quadratic does not meet the x-axis?Show answer
It has no real zeros.
8.The zeros of a quadratic are reciprocals of each other. What does this tell you about $a$ and $c$ ?Show answer
The product of zeros is $1$ , so $\frac{c}{a} = 1$ , which means $c = a$ .

Best order to study it

Revise the Class 9 basics: degree, finding $p (k)$ , and factorising a quadratic by splitting the middle term. Practise until you can split confidently, including with surd coefficients.
Learn what a zero is and its geometrical meaning. Practise counting zeros from graphs described in words, for straight lines, parabolas and cubic curves.
Learn the shapes of quadratic graphs: upward or downward, and the three cases of two, one or no zeros.
Learn how the sum and product relations are derived from $a (x - α) (x - β)$ . Then practise finding zeros and verifying both relations.
Practise forming quadratics from given zeros and from a given sum and product, including surd zeros and an extra condition that fixes $k$ .
Finish with the harder problems: expressions such as $α^{2} + β^{2}$ , building new polynomials from old zeros, and finding $k$ from conditions on the zeros.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you state what a zero is and test whether $- 2$ is a zero of $x^{2} + x - 2$ ?
Can you count the zeros of a polynomial from a graph described in words, ignoring the y-axis crossing?
Can you say which way $y = - 3 x^{2} + x + 1$ opens, and why?
Can you write the sum and product of zeros of $a x^{2} + b x + c$ with the correct signs?
Can you factorise $6 x^{2} - 7 x - 3$ and verify both relations in under three minutes?
Can you form a quadratic whose zeros are $2 + 3$ and $2 - 3$ ? (Answer: $x^{2} - 4 x + 1$ .)
Can you find $α^{2} + β^{2}$ and $\frac{1}{α} + \frac{1}{β}$ for $x^{2} - 6 x + 4$ without finding the zeros?
Can you find $k$ if one zero of $2 x^{2} - 9 x + k$ is twice the other? (Answer: $k = 9$ .)
Can you write $p (x)$ from a parabola through $(- 1, 0)$ , $(3, 0)$ and $(1, 4)$ ?
Can you show that $x^{2} + 4 x + 5$ has no zeros by completing the square?

Polynomials: frequently asked questions

Is the division algorithm for polynomials in the Class 10 syllabus for 2026-27?

No. The division algorithm for polynomials and the relationship between the zeros and coefficients of a cubic polynomial are not in the 2026-27 syllabus. Only the quadratic relationship is assessed.

Do I have to draw graphs of polynomials in the board exam?

Usually you read a given or described graph rather than plotting one. You need to count the points where it meets the x-axis and read off the zeros. Sometimes you also form the polynomial from those points and one extra point.

Why is the answer to 'find a quadratic polynomial with these zeros' not unique?

Multiplying a polynomial by any non-zero number does not change its zeros, so there are infinitely many correct answers. Write the answer with a constant k, then pick a convenient value, usually one that clears the fractions. Any correct choice earns full marks.

What if I cannot split the middle term to find the zeros?

First check whether the polynomial is a difference of squares, like x squared minus 3, which works even with surds. You can also complete the square. The quadratic formula from the Quadratic Equations chapter also gives the zeros, because the zeros of p(x) are the roots of p(x) = 0.

Is it zeros or zeroes?

Both spellings are correct. The NCERT textbook uses zeroes and the syllabus uses zeros. Examiners accept either.

How can I quickly check my zeros in the exam?

Add your two zeros and compare the result with minus b over a, then multiply them and compare with c over a. If both match, your zeros are almost certainly right. For extra certainty, substitute one zero back into the polynomial and confirm that you get 0.

Based on: CBSE Mathematics syllabus 2026-27; NCERT Mathematics, Textbook for Class X (Reprint 2026-27)