MathsCBSE Class 10 · Polynomials

Zeros of a Polynomial Class 10: Meaning and Method

Q: Are zeros and roots the same thing?

Almost. Zeros belong to a polynomial p(x), and roots belong to the equation p(x) = 0. They are the same numbers, so you can use either method to find them.

Q: Can a zero be 0?

Yes. If the constant term is zero, as in 4x squared minus 12x, then x = 0 is a zero. A zero means the value of x that makes the polynomial zero, not the number 0 itself.

Q: Can a polynomial have no zeros?

Yes, it can have no real zeros. For example, x squared plus 1 is always at least 1, so it never equals zero. A non-zero constant polynomial also has no zero.

Q: Is the division algorithm needed to find zeros?

No. The division algorithm for polynomials is not in the 2026-27 syllabus. Use substitution and factorisation.

Updated 14 October 2026 · Fact-checked

A zero of a polynomial is a value of x that makes the polynomial equal to zero. You check a zero by substituting it, and you find zeros by solving p(x) = 0, usually by factorising. Board questions ask you to find zeros, test given values, and find an unknown coefficient when one zero is given.

Zeros are where the graph meets the x-axis, so p(2)=0 and p(3)=0, but p(1)=2 is above the axis and not a zero.

Illustration drawn for QuestionBuddy.

Understand Zeros of a Polynomial

A polynomial $p (x)$ gives a number for every value of $x$ you put in. For example, if $p (x) = x^{2} - 5 x + 6$ , then $p (1) = 1 - 5 + 6 = 2$ and $p (2) = 4 - 10 + 6 = 0$ . A real number $k$ is called a zero of $p (x)$ when $p (k) = 0$ . So $2$ is a zero of $x^{2} - 5 x + 6$ , but $1$ is not.

To check whether a number is a zero, substitute it and simplify. If the answer is exactly $0$ , it is a zero. If not, it is not. To find the zeros, solve the equation $p (x) = 0$ . So the zeros of $p (x)$ are the same as the roots of the equation $p (x) = 0$ . This is the bridge to the Quadratic Equations chapter.

Why does factorising work? If $p (x) = (x - 2) (x - 3)$ , the product is $0$ only when one factor is $0$ . So $x = 2$ or $x = 3$ . Each factor $(x - k)$ gives the zero $k$ . The reverse is also true: if $p (k) = 0$ , then $(x - k)$ is a factor. This is the factor theorem you met in Class 9, and it is why one known zero helps you find the others.

A linear polynomial $a x + b$ with $a \neq = 0$ has exactly one zero, $- \frac{b}{a}$ . A quadratic has at most two zeros, and a polynomial of degree $n$ has at most $n$ zeros. Some quadratics, like $x^{2} + 1$ , have no real zero, because $x^{2} + 1$ is always at least $1$ .

When the polynomial has an unknown coefficient such as $k$ and one zero is given, substitute that zero. The statement $p (zero) = 0$ becomes an equation in $k$ . Solve it, then factorise the complete polynomial to get the other zero. Watch the conditions: a quadratic needs $a \neq = 0$ , so reject any value of $k$ that makes the $x^{2}$ coefficient zero.

On the graph, a zero is the x-coordinate of a point where $y = p (x)$ meets the x-axis, because there $y = 0$ . So the number of zeros equals the number of points where the graph meets the x-axis. A straight line (not parallel to the x-axis) meets it once. A parabola can cut it twice, touch it once (a repeated zero, as in $(x - 1)^{2}$ ), or miss it altogether (no real zero, as in $x^{2} + 1$ ). The point where the graph cuts the y-axis is $(0, p (0))$ ; it is not a zero unless $p (0) = 0$ . For quadratics the zeros are also tied to the coefficients by the sum and product rules, which have their own page.

Zeros of a Polynomial: formulas to remember

Definition of a zero: $k is a zero of p (x) ⟺ p (k) = 0$; Substitute to test a value, or to set up an equation for an unknown coefficient.
Zero of a linear polynomial: $a x + b, a \neq = 0 \Rightarrow x = - \frac{b}{a}$; Exactly one zero. Watch the sign: the zero of 3x + 6 is -2.
Factor theorem link: $p (k) = 0 ⟺ (x - k) is a factor of p (x)$; A factor (x - k) gives zero k; a factor (ax - b) gives zero b/a.
Maximum number of zeros: $de g p = n \Rightarrow at most n real zeros$; A quadratic has 0, 1 or 2 distinct real zeros.
Zero product rule: $(x - α) (x - β) = 0 \Rightarrow x = α or x = β$; Why factorising gives the zeros.
Difference of squares: $x^{2} - a^{2} = (x - a) (x + a)$; Gives zeros a and -a, including surds such as x^2 - 5.

How to solve Zeros of a Polynomial questions

Every question on this topic is either 'test a value', 'find the zeros' or 'find an unknown using a zero'. This method covers all three.

1Rewrite the polynomial in standard form $a x^{2} + b x + c$ and note any restriction such as $a \neq = 0$ .
2If a value is to be tested, substitute it carefully with brackets, for example $p (- 2) = (- 2)^{2} + \dots$ , and see if the result is $0$ .
3If one zero contains an unknown $k$ , write $p (zero) = 0$ and solve for $k$ .
4Reject any value of $k$ that breaks a condition (for example makes the $x^{2}$ coefficient zero).
5Put $k$ back and factorise the polynomial: common factor, splitting the middle term (product $a c$ , sum $b$ ), or $a^{2} - b^{2}$ .
6Set each factor equal to $0$ and state the zeros as values of $x$ .
7Check: substitute each zero back into $p (x)$ to confirm you get $0$ .

Zeros of a Polynomial in real life

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

Imagine your cousin runs a small home tiffin service. Suppose her weekly profit in rupees, when she prices a tiffin at x rupees above a base price, is modelled by $P (x) = - 200 x^{2} + 3600 x - 16000$ . She breaks even (zero profit) at certain price levels.

Find the values of x at which the profit is zero, and check one of them by substitution. Between these values, is the profit positive or negative? Test x = 9.

Try it first, then show the solutionSolution

Set $P (x) = 0$ : $- 200 x^{2} + 3600 x - 16000 = 0$ .
Divide by $- 200$ : $x^{2} - 18 x + 80 = 0$ .
Two numbers with product $80$ and sum $- 18$ : $- 8$ and $- 10$ .
$(x - 8) (x - 10) = 0$ , so $x = 8$ or $x = 10$ .
Check $x = 8$ : $- 200 (64) + 3600 (8) - 16000 = - 12800 + 28800 - 16000 = 0$ .
The coefficient of $x^{2}$ is $- 200 < 0$ , so the graph opens downward and lies above the x-axis between its two zeros: the profit is positive for $8 < x < 10$ .
Test $x = 9$ : $- 200 (81) + 3600 (9) - 16000 = - 16200 + 32400 - 16000 = 200 > 0$ .

Answer: Profit is zero at x = 8 and x = 10 (rupees above base); between them the profit is positive, for example ₹200 at x = 9.

Memory hook: Break-even points are the zeros: profit hits exactly 0.

Tricks to remember Zeros of a Polynomial

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

Plug, then zero
Test a value: plug it in, and the answer must be zero. Not small, not close: exactly 0.
Use it when: Any 'is k a zero?' or 'find k if ... is a zero' question.
Factor flips the sign
Factor $(x + 3)$ means zero $- 3$ ; factor $(2 x - 5)$ means zero $\frac{5}{2}$ . Ask: what makes this bracket zero?
Use it when: Writing zeros from factors without sign errors.
Degree is the ceiling
Degree $n$ means at most $n$ zeros. More answers than the degree means a mistake.
Use it when: Checking your final list of zeros.

Zeros of a Polynomial: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Zeros with surd coefficients · 3 marks · Short answerBoard level

Find the zeros of $p (x) = 3 x^{2} + 10 x + 73$ and verify one of them by substitution.

Try it first, then show the solutionSolution

Product $a c = 3 \times 73 = 21$ and sum $b = 10$ , so split $10 x$ as $7 x + 3 x$ . (½)
$p (x) = 3 x^{2} + 7 x + 3 x + 73$ . (½)
Write $3 x = 3 \cdot 3 x$ : $p (x) = x (3 x + 7) + 3 (3 x + 7)$ . (½)
$p (x) = (3 x + 7) (x + 3)$ . (½)
$3 x + 7 = 0 \Rightarrow x = - \frac{7}{3}$ ; $x + 3 = 0 \Rightarrow x = - 3$ . (½)
Check: $p (- 3) = 3 (3) - 103 + 73 = 0$ . (½)

Answer: The zeros are $- 3$ and $- \frac{7}{3}$ (that is, $- \frac{7 3}{3}$ ).

Example 2: Unknown from a given zero · 3 marks · Short answerBoard level

If $x = - 2$ is a zero of $p (x) = 2 x^{2} + k x - 6$ , find $k$ and the other zero.

Try it first, then show the solutionSolution

$p (- 2) = 0$ : $2 (4) - 2 k - 6 = 0$ . (1)
$2 - 2 k = 0$ , so $k = 1$ . (½)
$p (x) = 2 x^{2} + x - 6 = 2 x^{2} + 4 x - 3 x - 6$ . (½)
$= 2 x (x + 2) - 3 (x + 2) = (x + 2) (2 x - 3)$ . (½)
Other zero: $x = \frac{3}{2}$ . (½)

Answer: $k = 1$ ; the other zero is $\frac{3}{2}$ .

Example 3: Two unknowns, then a new polynomial · 3 marks · Short answerBoard level

If $1$ and $- 3$ are zeros of $p (x) = x^{2} + a x + b$ , find $a$ and $b$ . Hence find the zeros of $q (x) = b x^{2} + a x + 1$ .

Try it first, then show the solutionSolution

$p (1) = 0$ : $1 + a + b = 0$ . (½)
$p (- 3) = 0$ : $9 - 3 a + b = 0$ . (½)
Subtract: $(9 - 3 a + b) - (1 + a + b) = 0$ , so $8 - 4 a = 0$ , $a = 2$ . (½)
Then $b = - 1 - 2 = - 3$ ; check $(x - 1) (x + 3) = x^{2} + 2 x - 3$ . (½)
$q (x) = - 3 x^{2} + 2 x + 1 = - (3 x^{2} - 3 x + x - 1) = - (3 x + 1) (x - 1)$ . (½)
Zeros of $q (x)$ : $1$ and $- \frac{1}{3}$ . (½)

Answer: $a = 2$ , $b = - 3$ ; the zeros of $q (x)$ are $1$ and $- \frac{1}{3}$ .

Example 4: Rejecting a value of k · 5 marks · Long answerHarder board level

If $x = 2$ is a zero of $p (x) = (k - 2) x^{2} + k x - k^{2}$ and $p (x)$ is a quadratic polynomial, find $k$ . Hence find both zeros of $p (x)$ and verify the second zero.

Try it first, then show the solutionSolution

$p (2) = 0$ : $4 (k - 2) + 2 k - k^{2} = 0$ . (1)
$- k^{2} + 6 k - 8 = 0$ , so $k^{2} - 6 k + 8 = 0$ . (½)
$(k - 2) (k - 4) = 0$ , so $k = 2$ or $k = 4$ . (1)
If $k = 2$ , the $x^{2}$ coefficient $k - 2 = 0$ and $p (x) = 2 x - 4$ is linear, so reject $k = 2$ ; hence $k = 4$ . (1)
$p (x) = 2 x^{2} + 4 x - 16 = 2 (x^{2} + 2 x - 8) = 2 (x + 4) (x - 2)$ . (½)
Zeros: $2$ and $- 4$ . (½)
Check: $p (- 4) = 2 (16) + 4 (- 4) - 16 = 32 - 16 - 16 = 0$ . (½)

Answer: $k = 4$ ( $k = 2$ is rejected); the zeros are $2$ and $- 4$ .

Example 5: Find the error · 3 marks · Short answerHOTS / competency

Aman says the zeros of $p (x) = 6 x^{2} - x - 2$ are $- \frac{2}{3}$ and $\frac{1}{2}$ , because $p (x) = (3 x + 2) (2 x - 1)$ . Check his claim by substitution, find his error and give the correct zeros.

Try it first, then show the solutionSolution

Expand his factors: $(3 x + 2) (2 x - 1) = 6 x^{2} + x - 2$ , not $6 x^{2} - x - 2$ . (1)
Test $\frac{1}{2}$ : $p (\frac{1}{2}) = \frac{6}{4} - \frac{1}{2} - 2 = - 1 \neq = 0$ , so it is not a zero. (½)
Correct split: product $- 12$ , sum $- 1$ : $- 4$ and $3$ . $6 x^{2} - 4 x + 3 x - 2 = 2 x (3 x - 2) + 1 (3 x - 2) = (3 x - 2) (2 x + 1)$ . (1)
Zeros $\frac{2}{3}$ and $- \frac{1}{2}$ ; check $p (\frac{2}{3}) = \frac{8}{3} - \frac{2}{3} - 2 = 0$ . (½)

Answer: Aman's signs are swapped; the correct zeros are $\frac{2}{3}$ and $- \frac{1}{2}$ .

Example 6: Assertion-reason: a quadratic with no real zero · 1 mark · Assertion-reasonBoard level

Assertion (A): The polynomial $x^{2} + 1$ has no real zero.
Reason (R): $x^{2} + 1 \geq 1$ for every real number $x$ .
(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

For every real $x$ , $x^{2} \geq 0$ , so $x^{2} + 1 \geq 1$ : R is true. (½)
A value that is always at least $1$ can never equal $0$ , so $x^{2} + 1$ has no real zero: A is true, and R is exactly why. (½)

Answer: (a) Both A and R are true and R is the correct explanation of A.

Example 7: Linear zeros, a zero at 0 and a repeated zero · 3 marks · Short answerBoard level

(i) The zero of the linear polynomial $(k - 1) x + 5$ is $- \frac{5}{2}$ . Find $k$ .
(ii) Find the zeros of $3 x^{2} - 12 x$ and of $4 x^{2} - 12 x + 9$ . State how many distinct zeros each has.

Try it first, then show the solutionSolution

(i) The zero of $a x + b$ is $- \frac{b}{a}$ , so $- \frac{5}{k - 1} = - \frac{5}{2}$ and $k - 1 = 2$ , giving $k = 3$ . Check: $2 (- \frac{5}{2}) + 5 = 0$ . (1)
(ii) $3 x^{2} - 12 x = 3 x (x - 4)$ , so the zeros are $0$ and $4$ : two distinct zeros. A zero can be $0$ whenever the constant term is $0$ . (1)
$4 x^{2} - 12 x + 9 = (2 x - 3)^{2}$ , so $2 x - 3 = 0$ and $x = \frac{3}{2}$ is the only zero. It is a repeated zero, so there is one distinct zero, and the graph only touches the x-axis there. (1)

Answer: (i) $k = 3$ . (ii) $3 x^{2} - 12 x$ : zeros $0$ and $4$ (two). $4 x^{2} - 12 x + 9$ : zero $\frac{3}{2}$ repeated (one distinct zero).

Example 8: Reading zeros from a graph · 3 marks · Short answerHarder board level

The graph of a quadratic polynomial $y = p (x)$ is a parabola opening upwards. It cuts the x-axis at $(- 3, 0)$ and $(1, 0)$ and the y-axis at $(0, - 6)$ .
(i) How many zeros does $p (x)$ have, and what are they?
(ii) Find $p (x)$ .
(iii) How many zeros does $p (x) + 8$ have? Explain using the graph.

Try it first, then show the solutionSolution

(i) The graph meets the x-axis at two points, so there are two zeros: $- 3$ and $1$ . The point $(0, - 6)$ is on the y-axis, so it is not a zero. (½)
(ii) Each zero gives a factor, so $p (x) = a (x + 3) (x - 1)$ . At $x = 0$ : $a (3) (- 1) = - 6$ , so $a = 2$ and $p (x) = 2 x^{2} + 4 x - 6$ . (1)
(iii) $p (x) + 8 = 2 x^{2} + 4 x + 2 = 2 (x + 1)^{2}$ . (1)
Its only zero is $- 1$ (repeated). Adding 8 lifts the parabola 8 units up, and its lowest point $(- 1, - 8)$ moves to $(- 1, 0)$ , so the graph just touches the x-axis: one zero. (½)

Answer: (i) Two zeros, $- 3$ and $1$ . (ii) $p (x) = 2 x^{2} + 4 x - 6$ . (iii) One zero, $- 1$ ; the shifted graph touches the x-axis at $(- 1, 0)$ .

Where students lose marks in Zeros of a Polynomial

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

Mistake: Wrong sign when reading a zero from a factor
Students copy the number in the bracket instead of solving bracket = 0.
Wrong working
$(x + 4) (x - 1) = 0$ , so zeros are $4$ and $- 1$ .
Right working
$x + 4 = 0 \Rightarrow x = - 4$ ; $x - 1 = 0 \Rightarrow x = 1$ . Zeros $- 4$ and $1$ .
Do this: Always write each factor $= 0$ and solve.
Usually costs: ½ to 1 mark per zero.
Mistake: Squaring a negative without brackets
Typing $- 2^{2}$ mentally as $- 4$ .
Wrong working
$p (- 2)$ for $x^{2} + 3 x$ : $- 4 - 6 = - 10$ .
Right working
$(- 2)^{2} + 3 (- 2) = 4 - 6 = - 2$ .
Do this: Put every substituted negative inside brackets.
Usually costs: Usually the whole value of k, so 1 to 2 marks.
Mistake: Stopping at factors
Thinking the factorised form is the answer.
Wrong working
Zeros of $x^{2} - 7 x + 10$ are $(x - 2) (x - 5)$ .
Right working
Zeros are $x = 2$ and $x = 5$ .
Do this: The final line must state values of x.
Usually costs: ½ mark.
Mistake: Not rejecting a k that kills the quadratic
Ignoring the condition $a \neq = 0$ .
Wrong working
Taking $k = 1$ for $(k - 1) x^{2} + \dots$ and still calling it quadratic.
Right working
State $k - 1 \neq = 0$ and reject $k = 1$ .
Do this: Check the leading coefficient after finding k.
Usually costs: ½ to 1 mark in long answers.

Exam tips

Show the substitution line, like $p (- 2) = 2 (- 2)^{2} + k (- 2) - 6 = 0$ ; step marks are given for it.
After finding zeros, a quick substitution check catches sign errors before the examiner does.
In case studies, ask whether each zero makes sense (negative time or length is rejected) and say why.
For 'find k and the other zero', write the full polynomial with k substituted before factorising.

Practice questions on Zeros of a Polynomial

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

Q1 · 3 marks · Short answerHarder board level
Find the zeros of $p (x) = 43 x^{2} + 5 x - 23$ and verify one of them by substitution. Hence, without factorising again, write the zeros of $q (x) = 43 x^{2} - 5 x - 23$ and justify your answer.
Hint
For (ii), compare $q (x)$ with $p (- x)$ .
Answer
$43 x^{2} + 8 x - 3 x - 23 = 4 x (3 x + 2) - 3 (3 x + 2) = (3 x + 2) (4 x - 3)$ . Zeros $- \frac{2}{3}$ and $\frac{3}{4}$ . Check: $p (\frac{3}{4}) = \frac{3 3}{4} + \frac{5 3}{4} - 23 = 0$ . Since $p (- x) = 43 x^{2} - 5 x - 23 = q (x)$ , $q (k) = 0$ exactly when $p (- k) = 0$ . So the zeros of $q (x)$ are the negatives: $\frac{2}{3}$ and $- \frac{3}{4}$ .
Q2 · 5 marks · Long answerHarder board level
If $x = 3$ is a zero of $p (x) = (k - 1) x^{2} - 2 k x + (5 + 2 k - k^{2})$ and $p (x)$ is a quadratic polynomial, find $k$ . Hence find both zeros of $p (x)$ , verify the second zero and find $p (- 1)$ .
Hint
$p (3) = 0$ gives a quadratic equation in $k$ ; one of its roots makes $k - 1 = 0$ .
Answer
$p (3) = 9 (k - 1) - 6 k + 5 + 2 k - k^{2} = 0$ gives $- k^{2} + 5 k - 4 = 0$ , so $k^{2} - 5 k + 4 = 0$ and $(k - 1) (k - 4) = 0$ . $k = 1$ makes the $x^{2}$ coefficient $0$ (then $p (x) = - 2 x + 6$ is linear), so reject it; $k = 4$ . Then $p (x) = 3 x^{2} - 8 x - 3 = 3 x^{2} - 9 x + x - 3 = (x - 3) (3 x + 1)$ ; zeros $3$ and $- \frac{1}{3}$ . Check: $p (- \frac{1}{3}) = \frac{1}{3} + \frac{8}{3} - 3 = 0$ . Also $p (- 1) = 3 + 8 - 3 = 8$ .
Q3 · 4 marks · Case-basedHOTS / competency
Case: suppose a ball is thrown up from a balcony. Its height above the ground in metres after t seconds is $h (t) = - 5 t^{2} + 10 t + 15$ . A student records: at $t = 0$ s the height is $15$ m, at $t = 1$ s it is $20$ m, and at $t = 3$ s it is $0$ m. (i) At what other time is the ball back at the starting height of $15$ m? (1) (ii) Find the zeros of $h (t)$ and say which one gives the landing time. (1) (iii) Riya says that, just like $15$ m, the ball must pass $20$ m at two different times. Check her claim by solving $h (t) = 20$ and explain what $t = 1$ s represents. (2)
Hint
For (i) and (iii), set $h (t)$ equal to the height and simplify; for (ii), take out $- 5$ before factorising.
Answer
(i) $- 5 t^{2} + 10 t + 15 = 15$ gives $- 5 t (t - 2) = 0$ , so $t = 0$ or $t = 2$ ; the ball is back at $15$ m at $t = 2$ s. (ii) $h (t) = - 5 (t^{2} - 2 t - 3) = - 5 (t - 3) (t + 1)$ ; zeros $3$ and $- 1$ . The ball lands at $t = 3$ s; $t = - 1$ is rejected because time after the throw cannot be negative. (iii) $- 5 t^{2} + 10 t + 15 = 20$ gives $t^{2} - 2 t + 1 = 0$ , so $(t - 1)^{2} = 0$ and $t = 1$ only. Riya is wrong: $20$ m is reached just once, at $t = 1$ s, which is the highest point, since $h (t) = 20 - 5 (t - 1)^{2} \leq 20$ .
Q4 · 3 marks · Short answerHarder board level
Find the zeros of $p (x) = 2 x^{2} - (1 + 22) x + 2$ . If $α$ and $β$ are its zeros, find $α^{2} + β^{2}$ .
Hint
Product $a c = 22$ and sum $- (1 + 22)$ : use $- 1$ and $- 22$ .
Answer
$2 x^{2} - x - 22 x + 2 = x (2 x - 1) - 2 (2 x - 1) = (2 x - 1) (x - 2)$ . Zeros $\frac{1}{2}$ and $2$ . Then $α^{2} + β^{2} = \frac{1}{4} + 2 = \frac{9}{4}$ .
Q5 · 1 mark · MCQBoard level
If $1$ is a zero of both $a x^{2} + a x + 2$ and $x^{2} + x + b$ , then $ab$ equals
(a) $- 2$
(b) $2$
(c) $1$
(d) $- 1$
Hint
Substitute $x = 1$ in each polynomial.
Answer
(b) $2$ . $a + a + 2 = 0$ gives $a = - 1$ ; $1 + 1 + b = 0$ gives $b = - 2$ ; so $ab = 2$ .
Q6 · 1 mark · MCQBoard level
If $x = - 2$ is a zero of $x^{2} + k x - 8$ , then its other zero is
(a) $4$
(b) $- 4$
(c) $2$
(d) $8$
Hint
Find $k$ first, then factorise.
Answer
(a) $4$ . $4 - 2 k - 8 = 0$ gives $k = - 2$ , and $x^{2} - 2 x - 8 = (x - 4) (x + 2)$ .
Q7 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): If $2$ is a zero of $x^{2} + p x + q$ , where $p$ and $q$ are integers, then $q$ is even.
Reason (R): Every zero of a quadratic polynomial is a factor of its constant term.
(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
Substitute $x = 2$ for A. For R, look for a quadratic with a fractional zero.
Answer
(c). A: $4 + 2 p + q = 0$ gives $q = - 4 - 2 p = - 2 (2 + p)$ , which is even. So A is true. R is false: $2 x^{2} + x - 6 = (2 x - 3) (x + 2)$ has the zero $\frac{3}{2}$ , which is not a factor of $- 6$ . A is proved by substitution, not by R.
Q8 · 5 marks · Long answerHarder board level
$3$ is a zero of $p (x) = x^{2} - k x + k^{2} - 13$ , and both zeros of $p (x)$ are positive. Find $k$ , both zeros, and $p (- 1)$ . Explain why the other value of $k$ is rejected.
Hint
Setting $p (3) = 0$ gives a quadratic in $k$ . Test each value against the condition.
Answer
$p (3) = 9 - 3 k + k^{2} - 13 = 0$ , so $k^{2} - 3 k - 4 = 0$ and $(k - 4) (k + 1) = 0$ : $k = 4$ or $k = - 1$ . For $k = - 1$ : $p (x) = x^{2} + x - 12 = (x + 4) (x - 3)$ , zeros $3$ and $- 4$ . One zero is negative, so reject $k = - 1$ . For $k = 4$ : $p (x) = x^{2} - 4 x + 3 = (x - 1) (x - 3)$ , zeros $1$ and $3$ , both positive. So $k = 4$ . Check: $p (1) = 1 - 4 + 3 = 0$ . Finally $p (- 1) = 1 + 4 + 3 = 8$ .
Q9 · 4 marks · Case-basedHOTS / competency
Case study: Suppose the arch under a footbridge has the shape $h (x) = - \frac{1}{4} x^{2} + 2 x$ , where $h$ is the height of the arch in metres at a horizontal distance $x$ m from one foot of the arch.
(i) Find the zeros of $h (x)$ and say what they tell you about the arch. (1 mark)
(ii) Find the height of the arch 2 m from the foot. (1 mark)
(iii) A boat with a flat roof 3 m above the water must pass under the arch (take the water level as $h = 0$ ). Between which values of $x$ can the roof pass, and how wide is that gap? (2 marks)
Hint
For (iii), solve $h (x) = 3$ and use the fact that the arch is higher than 3 m between the two answers.
Answer
(i) $h (x) = - \frac{1}{4} x (x - 8)$ , so the zeros are $0$ and $8$ : the arch meets the water at its two feet, 8 m apart. (1)
(ii) $h (2) = - 1 + 4 = 3$ m. (1)
(iii) $- \frac{1}{4} x^{2} + 2 x = 3$ gives $x^{2} - 8 x + 12 = 0$ , so $(x - 2) (x - 6) = 0$ and $x = 2$ or $6$ . The arch opens downward, so it is higher than 3 m for $2 < x < 6$ . The roof can pass in that gap, which is 4 m wide. (2)
Q10 · 1 mark · MCQBoard level
The graph of $y = p (x)$ touches the x-axis at $(2, 0)$ , crosses it at $(- 1, 0)$ and $(4, 0)$ , and cuts the y-axis at $(0, 3)$ . The number of distinct zeros of $p (x)$ is
(a) $3$
(b) $4$
(c) $2$
(d) $1$
Hint
Count only the points on the x-axis.
Answer
(a) $3$ : the zeros are $- 1$ , $2$ and $4$ . The point $(0, 3)$ is on the y-axis, not the x-axis, so it is not a zero.

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 k a zero of p(x)?Show answer
When p(k) = 0.
2.Zero of $5 x - 10$ ?Show answer
$x = 2$ .
3.Zeros of $(x - 3) (2 x + 1)$ ?Show answer
$3$ and $- \frac{1}{2}$ .
4.Maximum zeros of a quadratic?Show answer
2.
5.Is $1$ a zero of $x^{2} + x - 2$ ?Show answer
Yes, $1 + 1 - 2 = 0$ .

Zeros of a Polynomial: frequently asked questions

Are zeros and roots the same thing?

Almost. Zeros belong to a polynomial p(x), and roots belong to the equation p(x) = 0. They are the same numbers, so you can use either method to find them.

Can a zero be 0?

Yes. If the constant term is zero, as in 4x squared minus 12x, then x = 0 is a zero. A zero means the value of x that makes the polynomial zero, not the number 0 itself.

Can a polynomial have no zeros?

Yes, it can have no real zeros. For example, x squared plus 1 is always at least 1, so it never equals zero. A non-zero constant polynomial also has no zero.

Is the division algorithm needed to find zeros?

No. The division algorithm for polynomials is not in the 2026-27 syllabus. Use substitution and factorisation.