MathsCBSE Class 10 · Polynomials

Forming a Quadratic Polynomial from Its Zeros

Q: Is there only one correct quadratic polynomial for given zeros?

No. Any non-zero multiple of x squared minus the sum times x plus the product has the same zeros. The board accepts any correct one unless the question fixes the polynomial with an extra condition.

Q: Should I find alpha and beta first when forming a new polynomial?

Usually not. Find the new sum and new product using the sum and product of the original zeros. This is faster and avoids surd arithmetic.

Q: Why do I multiply by a number at the end?

Multiplying by the LCM of the denominators gives integer coefficients without changing the zeros. It makes the answer neat and easier to check.

Q: Do I need the relation between zeros and coefficients of a cubic?

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

Updated 13 October 2026 · Fact-checked

To form a quadratic polynomial from its zeros, find the sum S and the product P of the zeros and write k times (x squared minus S x plus P), with k not zero. Board questions give the zeros or their sum and product, or ask for a new polynomial whose zeros are built from a given one's.

Only the sum and product of the zeros matter: write x² − (sum)x + product, then multiply by any non-zero k.

Illustration drawn for QuestionBuddy.

Understand Forming a Quadratic Polynomial from Its Zeros

You already know how to go forwards. You start with a quadratic $a x^{2} + b x + c$ and read off its zeros, with sum $α + β = - \frac{b}{a}$ and product $α β = \frac{c}{a}$ . This topic runs the same idea backwards. You are given the zeros, or just their sum and product, and you must build the polynomial.

Why does it work? If $α$ and $β$ are zeros, then $(x - α)$ and $(x - β)$ are factors. So the polynomial is a constant times $(x - α) (x - β)$ . Expanding gives $x^{2} - (α + β) x + α β$ . In words: ** $x^{2}$ minus (sum) $x$ plus (product)**. So you only ever need two numbers, the sum $S$ and the product $P$ .

Why the constant $k$ ? Multiplying by a non-zero number does not change where the polynomial equals $0$ . So $x^{2} - 5 x + 6$ and $3 x^{2} - 15 x + 18$ have the same zeros, $2$ and $3$ . The general answer is $k (x^{2} - S x + P)$ with $k \neq = 0$ . The board accepts any correct one. Usually you choose $k$ to clear fractions. If the question gives an extra fact, such as the value of $p (0)$ or $p (1)$ , that fact fixes $k$ .

The harder board questions give a quadratic with zeros $α$ and $β$ and ask for a polynomial whose zeros are new expressions, such as $α^{2}$ and $β^{2}$ , or $\frac{α}{β}$ and $\frac{β}{α}$ . You do not find $α$ and $β$ . You find the new sum and new product using identities in $α + β$ and $α β$ , then use the same rule.

Conditions to remember: the method is for quadratic polynomials only, the coefficient $k$ must be non-zero, and for real zeros the numbers must satisfy $(α - β)^{2} = S^{2} - 4 P \geq 0$ . If $S^{2} - 4 P < 0$ , the polynomial still exists, but it has no real zeros.

Forming a Quadratic Polynomial from Its Zeros: formulas to remember

Quadratic from its zeros: $k [x^{2} - (α + β) x + α β], k \neq = 0$; x squared minus (sum) x plus (product). Choose k to clear fractions.
Quadratic from sum S and product P: $k (x^{2} - S x + P), k \neq = 0$; Use directly when the question gives the sum and product.
Factored form: $a (x - α) (x - β)$; Same polynomial, useful when a graph or case gives the zeros and one more point.
Sum and product from coefficients: $α + β = - \frac{b}{a}, α β = \frac{c}{a}$; Your starting point when the zeros come from a given polynomial.
Sum of squares: $α^{2} + β^{2} = (α + β)^{2} - 2 α β$; Needed for zeros like alpha squared and beta squared, or alpha over beta.
Square of difference: $(α - β)^{2} = (α + β)^{2} - 4 α β$; Use when the difference of zeros is given, to find the product.
Ratio pair: $\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β}, \frac{α}{β} \cdot \frac{β}{α} = 1$; The product of this pair is always 1.
Shifted and scaled zeros: $(m α + n) + (m β + n) = m (α + β) + 2 n, (m α + n) (m β + n) = m^{2} α β + mn (α + β) + n^{2}$; For new zeros such as 2 alpha + 1 and 2 beta + 1.

How to solve Forming a Quadratic Polynomial from Its Zeros questions

Every question on this topic comes down to finding two numbers, the sum and the product of the required zeros. Use this method:

1Identify what is given: the zeros themselves, their sum and product, or a polynomial whose zeros are alpha and beta.
2If a polynomial is given, write $α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$ first.
3Write the sum $S$ of the required zeros, rewriting it with identities in $α + β$ and $α β$ .
4Write the product $P$ of the required zeros the same way, expanding carefully.
5Write $k (x^{2} - S x + P)$ and choose $k$ (usually the LCM of denominators) for integer coefficients.
6If a condition like $p (0)$ or $p (1)$ is given, substitute it to find the exact $k$ .
7Check: for a simple case, factorise or recompute $- \frac{b}{a}$ and $\frac{c}{a}$ to confirm they equal $S$ and $P$ .

Forming a Quadratic Polynomial from Its Zeros 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.

Forming a quadratic polynomial from its zeros

Imagine a school cricket match where two opening batters together score 90 runs. Your scorer friend remembers only that the product of their two scores is 1800. You want to recover each score.

Form a quadratic polynomial whose zeros are the two scores, and find the scores. If each batter had scored 10 more runs, form the quadratic polynomial whose zeros are the new scores without using the individual scores.

Try it first, then show the solutionSolution

Sum of zeros $S = 90$ , product $P = 1800$ .
Polynomial: $x^{2} - 90 x + 1800$ (taking $k = 1$ ).
Split: two numbers with product $1800$ and sum $- 90$ are $- 30$ and $- 60$ .
$x^{2} - 90 x + 1800 = (x - 30) (x - 60)$ , so the scores are 30 and 60 runs.
New sum: $(α + 10) + (β + 10) = 90 + 20 = 110$ .
New product: $α β + 10 (α + β) + 100 = 1800 + 900 + 100 = 2800$ .
New polynomial: $x^{2} - 110 x + 2800$ .
Check: $40 + 70 = 110$ and $40 \times 70 = 2800$ .

Answer: Polynomial $x^{2} - 90 x + 1800$ ; scores 30 runs and 60 runs; new polynomial $x^{2} - 110 x + 2800$ .

Memory hook: Total and product of two scores: x squared minus total x plus product.

Tricks to remember Forming a Quadratic Polynomial from Its Zeros

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

Minus sum, plus product
Say it as a chant: "x squared, minus the sum, plus the product." The sign in front of the sum is always minus.
Use it when: Writing $x^{2} - S x + P$ without a sign slip.
Two numbers only
Whatever the new zeros look like, your only job is to find S and P. Never try to find alpha and beta first.
Use it when: Questions on zeros like $α^{2}$ , $\frac{1}{α}$ , $α + 2$ .
k clears the clutter
Fractions in S or P? Multiply everything by the LCM of the denominators; that is your k.
Use it when: Getting a neat integer polynomial the examiner expects.
Swap test
If swapping alpha and beta leaves an expression unchanged, it can be written in terms of the sum and product.
Use it when: Knowing when the alpha-beta identities will work.

Forming a Quadratic Polynomial from Its Zeros: solved examples

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

Example 1: Polynomial from a sum and product with surds · 3 marks · Short answerBoard level

Find a quadratic polynomial whose sum of zeros is $2$ and product of zeros is $- 12$ . Hence find its zeros.

Try it first, then show the solutionSolution

Required polynomial is $k (x^{2} - S x + P)$ with $S = 2$ , $P = - 12$ . (½ mark)
Taking $k = 1$ : $p (x) = x^{2} - 2 x - 12$ . (½ mark)
Split the middle term: need two numbers with product $- 12$ and sum $- 2$ : these are $- 32$ and $22$ . (½ mark)
$x^{2} - 32 x + 22 x - 12 = x (x - 32) + 22 (x - 32)$ . (½ mark)
$= (x - 32) (x + 22)$ , so zeros are $32$ and $- 22$ . (½ mark)
Check: sum $= 2$ , product $= 32 \times (- 22) = - 12$ . (½ mark)

Answer: $x^{2} - 2 x - 12$ (or any non-zero multiple); zeros $32$ and $- 22$ .

Example 2: Zeros alpha/beta and beta/alpha · 3 marks · Short answerHarder board level

If $α$ and $β$ are the zeros of $p (x) = 3 x^{2} - 5 x + 1$ , find a quadratic polynomial whose zeros are $\frac{α}{β}$ and $\frac{β}{α}$ .

Try it first, then show the solutionSolution

$α + β = \frac{5}{3}$ , $α β = \frac{1}{3}$ . (½ mark)
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{25}{9} - \frac{2}{3} = \frac{19}{9}$ . (½ mark)
New sum $S = \frac{α ^{2} + β ^{2}}{α β} = \frac{19}{9} \div \frac{1}{3} = \frac{19}{3}$ . (½ mark)
New product $P = \frac{α}{β} \cdot \frac{β}{α} = 1$ . (½ mark)
Polynomial: $k (x^{2} - \frac{19}{3} x + 1)$ . (½ mark)
Take $k = 3$ : $3 x^{2} - 19 x + 3$ . (½ mark)

Answer: $3 x^{2} - 19 x + 3$

Example 3: Zeros alpha + 1/beta and beta + 1/alpha · 3 marks · Short answerHarder board level

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

Try it first, then show the solutionSolution

$α + β = 4$ , $α β = 2$ . (½ mark)
New sum $S = (α + β) + \frac{α + β}{α β} = 4 + \frac{4}{2} = 6$ . (1 mark)
New product $P = α β + 1 + 1 + \frac{1}{α β} = 2 + 2 + \frac{1}{2} = \frac{9}{2}$ . (1 mark)
Polynomial $k (x^{2} - 6 x + \frac{9}{2})$ ; with $k = 2$ : $2 x^{2} - 12 x + 9$ . (½ mark)

Answer: $2 x^{2} - 12 x + 9$

Example 4: Sum and difference given, polynomial fixed by a value · 5 marks · Long answerHarder board level

The sum of the zeros of a quadratic polynomial $p (x)$ is $6$ and the difference of its zeros is $23$ . Also $p (1) = 2$ .
(i) Find the product of the zeros.
(ii) Find $p (x)$ .
(iii) Find the zeros of $p (x)$ and verify the relationship between the zeros and the coefficients.
(iv) Find $α^{2} + β^{2}$ .

Try it first, then show the solutionSolution

Let the zeros be $α, β$ : $α + β = 6$ , $α - β = 23$ .
$(α - β)^{2} = (α + β)^{2} - 4 α β$ , so $12 = 36 - 4 α β$ . (½ mark)
$α β = 6$ . (½ mark)
$p (x) = k (x^{2} - 6 x + 6)$ . (½ mark)
$p (1) = k (1 - 6 + 6) = k = 2$ . (½ mark)
$p (x) = 2 x^{2} - 12 x + 12$ . (½ mark)
Solving $α + β = 6$ , $α - β = 23$ : $α = 3 + 3$ , $β = 3 - 3$ . (1 mark)
Sum $= 6 = - \frac{- 12}{2}$ ; product $= 9 - 3 = 6 = \frac{12}{2}$ . Verified. (½ mark)
$α^{2} + β^{2} = 36 - 12 = 24$ . (1 mark)

Answer: (i) $6$ (ii) $2 x^{2} - 12 x + 12$ (iii) $3 + 3$ , $3 - 3$ , relations verified (iv) $24$

Example 5: Case study: a fountain jet · 4 marks · Case-basedHOTS / competency

Suppose a garden fountain shoots a jet of water that follows a parabola. Take the ground as the x-axis and the nozzle wall as the y-axis, with distances in metres. The jet's curve, extended, meets the ground at $x = - 2$ and $x = 4$ , and it passes through the point $(0, 4)$ on the y-axis. Its height is $y = p (x)$ .
(i) Write the sum and the product of the zeros of $p (x)$ . (1 mark)
(ii) Find $p (x)$ . (1 mark)
(iii) The highest point of the jet is directly above the midpoint of the zeros. Find this greatest height. (2 marks)

Try it first, then show the solutionSolution

(i) Zeros $- 2$ and $4$ : sum $= 2$ , product $= - 8$ . (1 mark)
(ii) $p (x) = k (x^{2} - 2 x - 8)$ ; $p (0) = - 8 k = 4$ , so $k = - \frac{1}{2}$ ; $p (x) = - \frac{1}{2} x^{2} + x + 4$ . (1 mark)
(iii) Midpoint of zeros $= \frac{- 2 + 4}{2} = 1$ . (½ mark)
$p (1) = - \frac{1}{2} + 1 + 4$ . (1 mark)
$= 4.5$ , so the greatest height is $4.5$ m. (½ mark)

Answer: (i) Sum $2$ , product $- 8$ (ii) $p (x) = - \frac{1}{2} x^{2} + x + 4$ (iii) $4.5$ m at $x = 1$ m

Where students lose marks in Forming a Quadratic Polynomial from Its Zeros

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

Mistake: Wrong sign on the sum term
Students copy the sum straight in as $+ S x$ instead of $- S x$ .
Wrong working
Sum $- \frac{3}{2}$ , product $- 1$ : $x^{2} - \frac{3}{2} x - 1$ , so $2 x^{2} - 3 x - 2$ .
Right working
$x^{2} - (- \frac{3}{2}) x + (- 1) = x^{2} + \frac{3}{2} x - 1$ , so $2 x^{2} + 3 x - 2$ .
Do this: Write $x^{2} - (S) x + (P)$ with S and P in brackets, then simplify signs.
Usually costs: Usually the whole answer mark, as every later step is wrong.
Mistake: Treating sum of squares as square of sum
Forgetting the cross term when finding a new sum.
Wrong working
$α^{2} + β^{2} = (α + β)^{2} = 25$
Right working
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 25 - 6 = 19$
Do this: Always subtract $2 α β$ ; check with simple zeros like 1 and 2.
Usually costs: 1 to 2 marks in a 3-mark question.
Mistake: Incomplete expansion of the new product
Rushing products like $(α + \frac{1}{β}) (β + \frac{1}{α})$ .
Wrong working
$(α + \frac{1}{β}) (β + \frac{1}{α}) = α β + \frac{1}{α β}$
Right working
$= α β + 1 + 1 + \frac{1}{α β}$
Do this: Multiply all four pairs of terms, writing each one.
Usually costs: 1 mark plus the final answer.
Mistake: Ignoring the given condition on k
Students write $k = 1$ out of habit even when $p (0)$ or $p (1)$ is given.
Wrong working
Sum 6, product 6, $p (1) = 2$ : answer $x^{2} - 6 x + 6$ .
Right working
$p (1) = k (1 - 6 + 6) = k = 2$ , so $p (x) = 2 x^{2} - 12 x + 12$ .
Do this: Keep k as a letter until every condition in the question has been used.
Usually costs: 1 mark in a long answer.

Exam tips

Show the sum and the product on separate labelled lines before writing the polynomial; these lines carry step marks.
Write the general form $k (x^{2} - S x + P)$ once, then state your choice of k; it shows you know infinitely many answers exist.
For new zeros built from alpha and beta, never solve for alpha and beta, especially when they are surds; use identities.
When the numbers allow, factorise your final polynomial or recompute $- \frac{b}{a}$ and $\frac{c}{a}$ to check; it takes a few seconds.
In case studies, the ground is the x-axis, so landing points are the zeros, and one more point fixes k.

Practice questions on Forming a Quadratic Polynomial from Its Zeros

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

Q1 · 2 marks · Very short answerBoard level
Find a quadratic polynomial whose sum and product of zeros are $- \frac{2}{3}$ and $- \frac{1}{3}$ respectively. Find its zeros and verify.
Hint
Write $x^{2} - S x + P$ carefully with the signs, then multiply by 3.
Answer
$x^{2} + \frac{2}{3} x - \frac{1}{3}$ , times 3: $3 x^{2} + 2 x - 1 = (3 x - 1) (x + 1)$ . Zeros $\frac{1}{3}$ and $- 1$ : sum $- \frac{2}{3}$ , product $- \frac{1}{3}$ .
Q2 · 3 marks · Short answerHarder board level
If $α$ and $β$ are the zeros of $x^{2} - 5 x + 3$ , find a quadratic polynomial whose zeros are $α^{2}$ and $β^{2}$ .
Hint
New sum is $α^{2} + β^{2}$ ; new product is $(α β)^{2}$ .
Answer
$α + β = 5$ , $α β = 3$ . Sum $= 25 - 6 = 19$ , product $= 9$ . Polynomial $x^{2} - 19 x + 9$ .
Q3 · 3 marks · Short answerHarder board level
If $α$ and $β$ are the zeros of $3 x^{2} - 6 x + 1$ , find a quadratic polynomial whose zeros are $2 α + 1$ and $2 β + 1$ .
Hint
Expand $(2 α + 1) (2 β + 1) = 4 α β + 2 (α + β) + 1$ .
Answer
$α + β = 2$ , $α β = \frac{1}{3}$ . Sum $= 4 + 2 = 6$ ; product $= \frac{4}{3} + 4 + 1 = \frac{19}{3}$ . Polynomial $3 x^{2} - 18 x + 19$ .
Q4 · 5 marks · Long answerHarder board level
The sum of the zeros of a quadratic polynomial is $8$ and the sum of their squares is $34$ . (i) Find the polynomial with leading coefficient 1. (ii) Find its zeros and verify the relationship with the coefficients. (iii) Find a quadratic polynomial whose zeros are the reciprocals of these zeros.
Hint
Use $α^{2} + β^{2} = (α + β)^{2} - 2 α β$ to get the product first.
Answer
$34 = 64 - 2 α β$ , so $α β = 15$ . (i) $x^{2} - 8 x + 15$ . (ii) $(x - 3) (x - 5)$ : zeros 3, 5; sum $8 = - \frac{- 8}{1}$ , product $15 = \frac{15}{1}$ . (iii) Reciprocal sum $\frac{8}{15}$ , product $\frac{1}{15}$ : $15 x^{2} - 8 x + 1$ .
Q5 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): There is no real number pair whose sum is 4 and product is 5, so the quadratic polynomial $x^{2} - 4 x + 5$ has no real zeros.
Reason (R): $x^{2} - 4 x + 5 = (x - 2)^{2} + 1$ , which is at least 1 for every real $x$ .
(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.
Hint
Expand $(x - 2)^{2} + 1$ and ask whether it can ever equal 0.
Answer
(a). $(x - 2)^{2} + 1 = x^{2} - 4 x + 5$ is always at least 1, so it never equals 0: no real zeros, hence no real pair with sum 4 and product 5. R explains A.

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.Polynomial with zeros alpha and beta?Show answer
$k [x^{2} - (α + β) x + α β]$ , $k \neq = 0$
2.Polynomial with sum 5 and product 6?Show answer
$x^{2} - 5 x + 6$
3.Product of the zeros alpha/beta and beta/alpha?Show answer
Always $1$ .
4.Sum of squares of zeros in terms of sum and product?Show answer
$(α + β)^{2} - 2 α β$
5.How many quadratics have zeros 2 and 3?Show answer
Infinitely many: $k (x^{2} - 5 x + 6)$ for any non-zero k.

Forming a Quadratic Polynomial from Its Zeros: frequently asked questions

Is there only one correct quadratic polynomial for given zeros?

No. Any non-zero multiple of x squared minus the sum times x plus the product has the same zeros. The board accepts any correct one unless the question fixes the polynomial with an extra condition.

Should I find alpha and beta first when forming a new polynomial?

Usually not. Find the new sum and new product using the sum and product of the original zeros. This is faster and avoids surd arithmetic.

Why do I multiply by a number at the end?

Multiplying by the LCM of the denominators gives integer coefficients without changing the zeros. It makes the answer neat and easier to check.

Do I need the relation between zeros and coefficients of a cubic?

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