CBSE Maths › Class 10 PYQs › Polynomials

Polynomials: 3 marks Questions (CBSE Class 10)

49 different 3 marks questions on Polynomials from CBSE Class 10 Maths board exams 2022–2026, newest first.

1 mark (104)2 marks (32)3 marks (49)4 marks (5)

Q303 marksShort AnswerPolynomials CBSE 2026 · Basic 430/5/1

If $α, β$ are zeroes of the polynomial $p (x) = 5 x^{2} - 7 x - 3$ , then form a quadratic polynomial whose zeroes are $\frac{2}{α}$ and $\frac{2}{β}$ .

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Answer:

3 x^{2} + 14 x - 20

(or any non-zero multiple of it)

$α + β = \frac{7}{5}$ , $α β = - \frac{3}{5}$
Sum $= \frac{2}{α} + \frac{2}{β} = \frac{2 ( α + β )}{α β} = \frac{14/5}{- 3/5} = - \frac{14}{3}$
Product $= \frac{4}{α β} = \frac{4}{- 3/5} = - \frac{20}{3}$
Polynomial: $k (x^{2} + \frac{14}{3} x - \frac{20}{3})$ ; taking $k = 3$ , $3 x^{2} + 14 x - 20$

Also asked in: 2026 Basic 430/5/3

Q30 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2026 · Basic 430/5/1

Find the zeroes of the polynomial $p (x) = 3 x^{2} + 7 x - 20$ and verify the relationship between its zeroes and the coefficients.

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Answer: Zeroes:

\frac{5}{3}

and

- 4

; relationship verified.

$3 x^{2} + 7 x - 20 = 3 x^{2} + 12 x - 5 x - 20 = (3 x - 5) (x + 4)$
Zeroes: $x = \frac{5}{3}$ and $x = - 4$
Sum $= \frac{5}{3} - 4 = - \frac{7}{3} = - \frac{b}{a}$
Product $= \frac{5}{3} \times (- 4) = - \frac{20}{3} = \frac{c}{a}$ . Verified.

Also asked in: 2026 Basic 430/5/3

Q263 marksShort AnswerPolynomials CBSE 2026 · Basic 430/5/2

Find the zeroes of the polynomial $p (x) = 4 x^{2} - 8 x + 3$ and verify the relationship between its zeroes and co-efficients.

Show answer & solution

Answer: Zeroes:

\frac{1}{2}

and

\frac{3}{2}

; relationship verified.

$4 x^{2} - 8 x + 3 = 4 x^{2} - 2 x - 6 x + 3 = (2 x - 1) (2 x - 3)$
Zeroes: $\frac{1}{2}$ and $\frac{3}{2}$
Sum $= \frac{1}{2} + \frac{3}{2} = 2 = \frac{8}{4} = - \frac{b}{a}$
Product $= \frac{3}{4} = \frac{c}{a}$ . Verified.

Q26 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2026 · Basic 430/5/2

Form a polynomial whose zeroes are $α^{2}$ and $β^{2}$ , where $α$ and $β$ are zeroes of the polynomial $p (x) = x^{2} - 32 x + 4$ .

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Answer:

x^{2} - 10 x + 16

(or any non-zero multiple of it)

$α + β = 32$ , $α β = 4$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 18 - 8 = 10$
$α^{2} β^{2} = (α β)^{2} = 16$
Required polynomial: $x^{2} - 10 x + 16$

Q273 marksShort AnswerPolynomials CBSE 2025 · Basic 430/1/1

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

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Answer:

x^{2} - 9

; zeroes 3 and

- 3

A quadratic polynomial is $x^{2} - (sum) x + product$ .
$p (x) = x^{2} - 0 \cdot x + (- 9) = x^{2} - 9$ .
$x^{2} - 9 = (x - 3) (x + 3) = 0$ gives zeroes 3 and $- 3$ .
Check: sum $= 0$ , product $= - 9$ .

Q313 marksShort AnswerPolynomials CBSE 2025 · Basic 430/1/2

Find a quadratic polynomial, sum and product of whose zeroes are 5 and $- 6$ , respectively. Also, find the zeroes of the polynomial so obtained.

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Answer:

x^{2} - 5 x - 6

; zeroes 6 and

- 1

A quadratic polynomial is $x^{2} - (sum) x + product$ .
$p (x) = x^{2} - 5 x - 6$ .
$x^{2} - 5 x - 6 = (x - 6) (x + 1)$ , so the zeroes are 6 and $- 1$ .
Check: $6 + (- 1) = 5$ , $6 \times (- 1) = - 6$ .

Q293 marksShort AnswerPolynomials CBSE 2025 · Basic 430/1/3

Determine a quadratic polynomial, sum and product of whose zeroes are $- 10$ and 24, respectively. Also, determine the zeroes of the polynomial so obtained.

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Answer:

x^{2} + 10 x + 24

; zeroes

- 4

and

- 6

A quadratic polynomial is $x^{2} - (sum) x + product$ .
$p (x) = x^{2} - (- 10) x + 24 = x^{2} + 10 x + 24$ .
$x^{2} + 10 x + 24 = (x + 4) (x + 6)$ , so the zeroes are $- 4$ and $- 6$ .
Check: $- 4 + (- 6) = - 10$ , $(- 4) (- 6) = 24$ .

Q273 marksShort AnswerPolynomials CBSE 2025 · Basic 430/2/1

Find the zeroes of the polynomial $p (x) = 3 x^{2} - 2 x - 1$ and verify the relationship between the zeroes of p(x) and the coefficients of p(x).

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Answer: Zeroes:

1

and

- \frac{1}{3}

$3 x^{2} - 2 x - 1 = 3 x^{2} - 3 x + x - 1 = 3 x (x - 1) + 1 (x - 1) = (3 x + 1) (x - 1)$ .
Zeroes: $x = 1$ and $x = - \frac{1}{3}$ .
Sum $= 1 - \frac{1}{3} = \frac{2}{3}$ and $- \frac{b}{a} = - \frac{- 2}{3} = \frac{2}{3}$ . Verified.
Product $= 1 \times (- \frac{1}{3}) = - \frac{1}{3}$ and $\frac{c}{a} = - \frac{1}{3}$ . Verified.

Q313 marksShort AnswerPolynomials CBSE 2025 · Basic 430/2/2

Find the zeroes of the polynomial $p (x) = 2 x^{2} + 5 x + 2$ and verify the relationship between zeroes of p(x) and its coefficients.

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Answer: Zeroes:

- \frac{1}{2}

and

- 2

$2 x^{2} + 5 x + 2 = 2 x^{2} + 4 x + x + 2 = 2 x (x + 2) + 1 (x + 2) = (2 x + 1) (x + 2)$ .
Zeroes: $x = - \frac{1}{2}$ and $x = - 2$ .
Sum $= - \frac{1}{2} - 2 = - \frac{5}{2}$ and $- \frac{b}{a} = - \frac{5}{2}$ . Verified.
Product $= (- \frac{1}{2}) (- 2) = 1$ and $\frac{c}{a} = \frac{2}{2} = 1$ . Verified.

Q293 marksShort AnswerPolynomials CBSE 2025 · Basic 430/2/3

Find the zeroes of the polynomial $q (x) = 6 x^{2} - 5 x - 1$ and verify the relationship between the zeroes of q(x) and its coefficients.

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Answer: Zeroes:

1

and

- \frac{1}{6}

$6 x^{2} - 5 x - 1 = 6 x^{2} - 6 x + x - 1 = 6 x (x - 1) + 1 (x - 1) = (6 x + 1) (x - 1)$ .
Zeroes: $x = 1$ and $x = - \frac{1}{6}$ .
Sum $= 1 - \frac{1}{6} = \frac{5}{6}$ and $- \frac{b}{a} = - \frac{- 5}{6} = \frac{5}{6}$ . Verified.
Product $= - \frac{1}{6}$ and $\frac{c}{a} = - \frac{1}{6}$ . Verified.

Q273 marksShort AnswerPolynomials CBSE 2025 · Basic 430/3/1

Find the zeroes of the polynomial $9 s^{2} - 6 s + 1$ and verify the relationship between the zeroes and the coefficients of the given polynomial.

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Answer: Zeroes:

\frac{1}{3}

\frac{1}{3}

$9 s^{2} - 6 s + 1 = (3 s - 1)^{2}$ , so the zeroes are $\frac{1}{3}$ and $\frac{1}{3}$ .
Sum of zeroes $= \frac{2}{3}$ and $- \frac{b}{a} = - \frac{- 6}{9} = \frac{2}{3}$ . Verified.
Product of zeroes $= \frac{1}{9}$ and $\frac{c}{a} = \frac{1}{9}$ . Verified.

Q313 marksShort AnswerPolynomials CBSE 2025 · Basic 430/3/2

Find the zeroes of the polynomial $4 x^{2} + 4 x + 1$ and verify the relationship between the zeroes and the coefficients of the given polynomial.

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Answer: Zeroes:

- \frac{1}{2}

- \frac{1}{2}

$4 x^{2} + 4 x + 1 = (2 x + 1)^{2}$ , so the zeroes are $- \frac{1}{2}$ and $- \frac{1}{2}$ .
Sum of zeroes $= - 1$ and $- \frac{b}{a} = - \frac{4}{4} = - 1$ . Verified.
Product of zeroes $= \frac{1}{4}$ and $\frac{c}{a} = \frac{1}{4}$ . Verified.

Q293 marksShort AnswerPolynomials CBSE 2025 · Basic 430/3/3

Find the zeroes of the polynomial $25 a^{2} - 10 a + 1$ and verify the relationship between the zeroes and coefficients of the given polynomial.

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Answer: Zeroes:

\frac{1}{5}

\frac{1}{5}

$25 a^{2} - 10 a + 1 = (5 a - 1)^{2}$ , so the zeroes are $\frac{1}{5}$ and $\frac{1}{5}$ .
Sum of zeroes $= \frac{2}{5}$ and $- \frac{b}{a} = - \frac{- 10}{25} = \frac{2}{5}$ . Verified.
Product of zeroes $= \frac{1}{25}$ and $\frac{c}{a} = \frac{1}{25}$ . Verified.

Q263 marksShort AnswerPolynomials CBSE 2025 · Basic 430/4/1

Find the zeroes of the polynomial $p (x) = 6 x^{2} + 13 x - 5$ and verify the relationship between its zeroes and the coefficients.

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Answer: Zeroes

\frac{1}{3}

and

- \frac{5}{2}

; relationship verified.

$6 x^{2} + 13 x - 5 = 6 x^{2} + 15 x - 2 x - 5 = 3 x (2 x + 5) - (2 x + 5) = (3 x - 1) (2 x + 5)$ .
Zeroes: $x = \frac{1}{3}$ and $x = - \frac{5}{2}$ .
Sum $= \frac{1}{3} - \frac{5}{2} = - \frac{13}{6}$ and $- \frac{b}{a} = - \frac{13}{6}$ . Verified.
Product $= \frac{1}{3} \times (- \frac{5}{2}) = - \frac{5}{6}$ and $\frac{c}{a} = - \frac{5}{6}$ . Verified.

Q303 marksShort AnswerPolynomials CBSE 2025 · Basic 430/4/2

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

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Answer: Zeroes

\frac{3}{2}

and

- \frac{1}{2}

; relationship verified.

$4 x^{2} - 4 x - 3 = 4 x^{2} - 6 x + 2 x - 3 = 2 x (2 x - 3) + (2 x - 3) = (2 x + 1) (2 x - 3)$ .
Zeroes: $x = \frac{3}{2}$ and $x = - \frac{1}{2}$ .
Sum $= \frac{3}{2} - \frac{1}{2} = 1$ and $- \frac{b}{a} = \frac{4}{4} = 1$ . Verified.
Product $= \frac{3}{2} \times (- \frac{1}{2}) = - \frac{3}{4}$ and $\frac{c}{a} = - \frac{3}{4}$ . Verified.

Q283 marksShort AnswerPolynomials CBSE 2025 · Basic 430/4/3

Find the zeroes of the polynomial $p (x) = 9 x^{2} - 6 x - 35$ and verify the relationship between zeroes and its coefficients.

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Answer: Zeroes

\frac{7}{3}

and

- \frac{5}{3}

; relationship verified.

$9 x^{2} - 6 x - 35 = 9 x^{2} - 21 x + 15 x - 35 = 3 x (3 x - 7) + 5 (3 x - 7) = (3 x + 5) (3 x - 7)$ .
Zeroes: $x = \frac{7}{3}$ and $x = - \frac{5}{3}$ .
Sum $= \frac{7}{3} - \frac{5}{3} = \frac{2}{3}$ and $- \frac{b}{a} = \frac{6}{9} = \frac{2}{3}$ . Verified.
Product $= \frac{7}{3} \times (- \frac{5}{3}) = - \frac{35}{9}$ and $\frac{c}{a} = - \frac{35}{9}$ . Verified.

Q263 marksShort AnswerPolynomials CBSE 2025 · Basic 430/5/1

If $α$ , $β$ are zeroes of the polynomial $3 x^{2} - 8 x + 4$ , then form a quadratic polynomial in $x$ whose zeroes are $\frac{1}{α}$ and $\frac{1}{β}$ .

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Answer:

x^{2} - 2 x + \frac{3}{4}

, or

k (4 x^{2} - 8 x + 3)

$α + β = \frac{8}{3}$ , $α β = \frac{4}{3}$ .
Sum of new zeroes $= \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{8/3}{4/3} = 2$ .
Product of new zeroes $= \frac{1}{α β} = \frac{3}{4}$ .
Required polynomial $= x^{2} - 2 x + \frac{3}{4}$ , or $4 x^{2} - 8 x + 3$ (any non-zero multiple).

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q26 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2025 · Basic 430/5/1

Find zeroes of the polynomial $6 x^{2} - 7 x - 3$ and verify the relationship between zeroes and its coefficients.

Show answer & solution

Answer: Zeroes

\frac{3}{2}

and

- \frac{1}{3}

; relationship verified.

$6 x^{2} - 7 x - 3 = 6 x^{2} - 9 x + 2 x - 3 = 3 x (2 x - 3) + 1 (2 x - 3) = (2 x - 3) (3 x + 1)$ .
Zeroes: $x = \frac{3}{2}$ and $x = - \frac{1}{3}$ .
Sum $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6}$ and $- \frac{b}{a} = - \frac{- 7}{6} = \frac{7}{6}$ .
Product $= \frac{3}{2} \times (- \frac{1}{3}) = - \frac{1}{2}$ and $\frac{c}{a} = \frac{- 3}{6} = - \frac{1}{2}$ .
Both relations hold.

Also asked in: 2025 Basic 430/5/2, 2025 Basic 430/5/3

Q263 marksShort AnswerPolynomials CBSE 2025 · Basic 430/6/1

If $α$ , $β$ are zeroes of the polynomial $8 x^{2} - 5 x - 1$ , then form a quadratic polynomial in $x$ whose zeroes are $\frac{2}{α}$ and $\frac{2}{β}$ .

Show answer & solution

Answer:

x^{2} + 10 x - 32

(or any non-zero multiple

k (x^{2} + 10 x - 32)

)

$α + β = \frac{5}{8}$ , $α β = - \frac{1}{8}$ .
Sum of new zeroes $= \frac{2}{α} + \frac{2}{β} = \frac{2 ( α + β )}{α β} = \frac{2 \times \frac{5}{8}}{- \frac{1}{8}} = - 10$ .
Product of new zeroes $= \frac{4}{α β} = \frac{4}{- \frac{1}{8}} = - 32$ .
Required polynomial $= x^{2} - (sum) x + product = x^{2} + 10 x - 32$ .

Also asked in: 2025 Basic 430/6/2

Q26 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2025 · Basic 430/6/1

Find the zeroes of the polynomial $p (x) = 3 x^{2} + x - 10$ and verify the relationship between zeroes and its coefficients.

Show answer & solution

Answer: Zeroes:

\frac{5}{3}

and

- 2

; relationship verified.

$3 x^{2} + x - 10 = 3 x^{2} + 6 x - 5 x - 10 = 3 x (x + 2) - 5 (x + 2) = (3 x - 5) (x + 2)$ .
Zeroes: $x = \frac{5}{3}$ , $x = - 2$ .
Sum $= \frac{5}{3} - 2 = - \frac{1}{3}$ and $- \frac{b}{a} = - \frac{1}{3}$ .
Product $= \frac{5}{3} \times (- 2) = - \frac{10}{3}$ and $\frac{c}{a} = - \frac{10}{3}$ .
Hence the relationship is verified.

Also asked in: 2025 Basic 430/6/2

Q313 marksShort AnswerPolynomials CBSE 2025 · Basic 430/6/3

$α$ , $β$ are zeroes of the polynomial $3 x^{2} - 8 x + k$ . Find the value of k, if $α^{2} + β^{2} = \frac{40}{9}$ .

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Answer: k = 4

$α + β = \frac{8}{3}$ , $α β = \frac{k}{3}$ .
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{64}{9} - \frac{2 k}{3}$ .
$\frac{64}{9} - \frac{2 k}{3} = \frac{40}{9}$ , so $\frac{2 k}{3} = \frac{24}{9} = \frac{8}{3}$ and $k = 4$ .

Q31 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2025 · Basic 430/6/3

Find the zeroes of the polynomial $2 x^{2} + 7 x + 5$ and verify the relationship between its zeroes and co-efficients.

Show answer & solution

Answer: Zeroes:

- 1

and

- \frac{5}{2}

; relationship verified.

$2 x^{2} + 7 x + 5 = 2 x^{2} + 2 x + 5 x + 5 = 2 x (x + 1) + 5 (x + 1) = (2 x + 5) (x + 1)$ .
Zeroes: $x = - 1$ , $x = - \frac{5}{2}$ .
Sum $= - 1 - \frac{5}{2} = - \frac{7}{2}$ and $- \frac{b}{a} = - \frac{7}{2}$ .
Product $= (- 1) (- \frac{5}{2}) = \frac{5}{2}$ and $\frac{c}{a} = \frac{5}{2}$ .
Hence verified.

Q303 marksShort AnswerPolynomials CBSE 2025 · Standard 30/4/1

$α$ and $β$ are zeroes of a quadratic polynomial $p x^{2} + q x + 1$ . Form a quadratic polynomial whose zeroes are $\frac{2}{α}$ and $\frac{2}{β}$ .

Show answer & solution

Answer:

x^{2} + 2 q x + 4 p

(or any non-zero multiple

k (x^{2} + 2 q x + 4 p)

)

$α + β = - \frac{q}{p}$ , $α β = \frac{1}{p}$
Sum of new zeroes $= \frac{2}{α} + \frac{2}{β} = \frac{2 ( α + β )}{α β} = \frac{2 ( - q / p )}{1/ p} = - 2 q$
Product $= \frac{4}{α β} = 4 p$
Required polynomial $= x^{2} - (- 2 q) x + 4 p = x^{2} + 2 q x + 4 p$

Q263 marksShort AnswerPolynomials CBSE 2025 · Standard 30/4/2

$α$ and $β$ are zeroes of a quadratic polynomial $x^{2} - a x - b$ . Obtain a quadratic polynomial whose zeroes are $3 α + 1$ and $3 β + 1$ .

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Answer:

x^{2} - (3 a + 2) x + (3 a - 9 b + 1)

(or any non-zero multiple)

$α + β = a$ , $α β = - b$
Sum of new zeroes $= 3 (α + β) + 2 = 3 a + 2$
Product $= 9 α β + 3 (α + β) + 1 = - 9 b + 3 a + 1$
Required polynomial $= x^{2} - (3 a + 2) x + (3 a - 9 b + 1)$

Q283 marksShort AnswerPolynomials CBSE 2025 · Standard 30/4/3

If $α$ and $β$ are the zeroes of the polynomial $a x^{2} - x + c$ . Obtain a polynomial whose zeroes are $α - 3$ and $β - 3$ .

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Answer:

a x^{2} + (6 a - 1) x + (9 a + c - 3)

(or any non-zero multiple)

$α + β = \frac{1}{a}$ , $α β = \frac{c}{a}$
Sum of new zeroes $= (α + β) - 6 = \frac{1 - 6 a}{a}$
Product $= α β - 3 (α + β) + 9 = \frac{c - 3 + 9 a}{a}$
Polynomial $= x^{2} - \frac{1 - 6 a}{a} x + \frac{9 a + c - 3}{a}$
Multiplying by $a$ : $a x^{2} + (6 a - 1) x + (9 a + c - 3)$

Q273 marksShort AnswerPolynomials CBSE 2025 · Standard 30/5/1

Obtain the zeroes of the polynomial $7 x^{2} + 18 x - 9$ . Hence, write a polynomial each of whose zeroes is twice the zeroes of given polynomial.

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Answer: Zeroes:

\frac{3}{7}

and

- 3

; required polynomial:

k (7 x^{2} + 36 x - 36)

, e.g.

7 x^{2} + 36 x - 36

$7 x^{2} + 18 x - 9 = 7 x^{2} + 21 x - 3 x - 9 = (7 x - 3) (x + 3)$
Zeroes: $\frac{3}{7}$ and $- 3$
New zeroes: $\frac{6}{7}$ and $- 6$
Sum $= \frac{6}{7} - 6 = - \frac{36}{7}$ , product $= - \frac{36}{7}$
Polynomial: $x^{2} + \frac{36}{7} x - \frac{36}{7}$ , i.e. $k (7 x^{2} + 36 x - 36)$

Q293 marksShort AnswerPolynomials CBSE 2025 · Standard 30/5/2

Obtain the zeroes of the polynomial $p (x) = 2 x^{2} - 5 x - 3$ . Hence, obtain a polynomial each of whose zeroes is one less than each of the zero of $p (x)$ .

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Answer: Zeroes:

3

and

- \frac{1}{2}

; required polynomial:

k (2 x^{2} - x - 6)

, e.g.

2 x^{2} - x - 6

$2 x^{2} - 5 x - 3 = 2 x^{2} - 6 x + x - 3 = (2 x + 1) (x - 3)$
Zeroes: $3$ and $- \frac{1}{2}$
New zeroes: $2$ and $- \frac{3}{2}$
Sum $= \frac{1}{2}$ , product $= - 3$
Polynomial: $x^{2} - \frac{1}{2} x - 3$ , i.e. $k (2 x^{2} - x - 6)$

Q263 marksShort AnswerPolynomials CBSE 2025 · Standard 30/5/3

Find the zeroes of the polynomial $p (x) = 6 x^{2} - 5 x - 1$ . Hence, obtain a polynomial each of whose zeroes is three times the zeroes of $p (x)$ .

Show answer & solution

Answer: Zeroes:

1

and

- \frac{1}{6}

; required polynomial:

k (2 x^{2} - 5 x - 3)

, e.g.

2 x^{2} - 5 x - 3

$6 x^{2} - 5 x - 1 = 6 x^{2} - 6 x + x - 1 = (6 x + 1) (x - 1)$
Zeroes: $1$ and $- \frac{1}{6}$
New zeroes: $3$ and $- \frac{1}{2}$
Sum $= \frac{5}{2}$ , product $= - \frac{3}{2}$
Polynomial: $x^{2} - \frac{5}{2} x - \frac{3}{2}$ , i.e. $k (2 x^{2} - 5 x - 3)$

Q273 marksShort AnswerPolynomials CBSE 2025 · Standard 30/6/1

Find the zeroes of the polynomial $p (x) = 3 x^{2} - 4 x - 4$ . Hence, write a polynomial whose each of the zeroes is 2 more than zeroes of $p (x)$ .

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Answer: Zeroes:

2

and

- \frac{2}{3}

; required polynomial:

3 x^{2} - 16 x + 16

(or any non-zero multiple)

$3 x^{2} - 4 x - 4 = 3 x^{2} - 6 x + 2 x - 4 = (3 x + 2) (x - 2)$
Zeroes: $x = 2$ and $x = - \frac{2}{3}$
New zeroes: $4$ and $\frac{4}{3}$
Sum $= \frac{16}{3}$ , product $= \frac{16}{3}$
Polynomial: $k (x^{2} - \frac{16}{3} x + \frac{16}{3})$ ; with $k = 3$ : $3 x^{2} - 16 x + 16$

Q293 marksShort AnswerPolynomials CBSE 2025 · Standard 30/6/2

Find the zeroes of the polynomial $q (x) = 8 x^{2} - 2 x - 3$ . Hence, find a polynomial whose zeroes are 2 less than the zeroes of $q (x)$ .

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Answer: Zeroes:

\frac{3}{4}

and

- \frac{1}{2}

; required polynomial:

8 x^{2} + 30 x + 25

(or any non-zero multiple)

$8 x^{2} - 2 x - 3 = 8 x^{2} - 6 x + 4 x - 3 = (4 x - 3) (2 x + 1)$
Zeroes: $x = \frac{3}{4}$ and $x = - \frac{1}{2}$
New zeroes: $\frac{3}{4} - 2 = - \frac{5}{4}$ and $- \frac{1}{2} - 2 = - \frac{5}{2}$
Sum $= - \frac{15}{4}$ , product $= \frac{25}{8}$
Polynomial: $k (x^{2} + \frac{15}{4} x + \frac{25}{8})$ ; with $k = 8$ : $8 x^{2} + 30 x + 25$

Q313 marksShort AnswerPolynomials CBSE 2025 · Standard 30/6/3

Find the zeroes of the polynomial $r (x) = 4 x^{2} + 3 x - 1$ . Hence, write a polynomial whose zeroes are reciprocal of the zeroes of polynomial $r (x)$ .

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Answer: Zeroes:

\frac{1}{4}

and

- 1

; required polynomial:

x^{2} - 3 x - 4

(or any non-zero multiple)

$4 x^{2} + 3 x - 1 = 4 x^{2} + 4 x - x - 1 = (4 x - 1) (x + 1)$
Zeroes: $x = \frac{1}{4}$ and $x = - 1$
Reciprocals: $4$ and $- 1$
Sum $= 3$ , product $= - 4$
Polynomial: $x^{2} - 3 x - 4$

Q293 marksShort AnswerPolynomials CBSE 2024 · Basic 430/2/1

Zeroes of the quadratic polynomial $x^{2} - 3 x + 2$ are $α$ and $β$ . Construct a quadratic polynomial whose zeroes are $2 α + 1$ and $2 β + 1$ .

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Answer:

x^{2} - 8 x + 15

(or any non-zero multiple)

$α + β = 3$ and $α β = 2$
Sum of new zeroes $= 2 (α + β) + 2 = 6 + 2 = 8$
Product $= 4 α β + 2 (α + β) + 1 = 8 + 6 + 1 = 15$
Required polynomial $= x^{2} - 8 x + 15$ (or $k (x^{2} - 8 x + 15)$ , $k \neq = 0$ )

Also asked in: 2024 Basic 430/2/3

Q29 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2024 · Basic 430/2/1

Find the zeroes of the polynomial $4 x^{2} - 4 x + 1$ and verify the relationship between the zeroes and the coefficients.

Show answer & solution

Answer: Zeroes:

\frac{1}{2}, \frac{1}{2}

; relationship verified.

$4 x^{2} - 4 x + 1 = (2 x - 1)^{2}$ , so the zeroes are $\frac{1}{2}$ and $\frac{1}{2}$ .
Sum of zeroes $= 1$ and $- \frac{b}{a} = - \frac{- 4}{4} = 1$ .
Product of zeroes $= \frac{1}{4}$ and $\frac{c}{a} = \frac{1}{4}$ .
Both relations hold.

Also asked in: 2024 Basic 430/2/3

Q273 marksShort AnswerPolynomials CBSE 2024 · Basic 430/2/2

Zeroes of the quadratic polynomial $x^{2} + x - 6$ are ‘ $α$ ’ and ‘ $β$ ’. Construct a quadratic polynomial whose zeroes are $\frac{1}{α}$ and $\frac{1}{β}$ .

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Answer:

6 x^{2} - x - 1

(or any non-zero multiple)

$α + β = - 1$ and $α β = - 6$
Sum of new zeroes $= \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- 1}{- 6} = \frac{1}{6}$
Product $= \frac{1}{α β} = - \frac{1}{6}$
Polynomial $= x^{2} - \frac{1}{6} x - \frac{1}{6}$ , i.e. $6 x^{2} - x - 1$ (or $k (6 x^{2} - x - 1)$ , $k \neq = 0$ )

Q27 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2024 · Basic 430/2/2

Find the zeroes of the polynomial $2 x^{2} + 3 x - 2$ and verify the relationship between the zeroes and the coefficients.

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Answer: Zeroes:

\frac{1}{2}, - 2

; relationship verified.

$2 x^{2} + 3 x - 2 = 2 x^{2} + 4 x - x - 2 = (2 x - 1) (x + 2)$ , so zeroes are $\frac{1}{2}$ and $- 2$ .
Sum $= \frac{1}{2} - 2 = - \frac{3}{2}$ and $- \frac{b}{a} = - \frac{3}{2}$ .
Product $= \frac{1}{2} \times (- 2) = - 1$ and $\frac{c}{a} = \frac{- 2}{2} = - 1$ .
Both relations hold.

Q293 marksShort AnswerPolynomials CBSE 2024 · Basic 430/3/1

Find the zeroes of the quadratic polynomial $5 x^{2} + 3 x - 2$ and verify the relationship between the zeroes and the co-efficients.

Show answer & solution

Answer: Zeroes are

\frac{2}{5}

and

- 1

; relationship verified.

$5 x^{2} + 3 x - 2 = 5 x^{2} + 5 x - 2 x - 2 = (5 x - 2) (x + 1)$ .
Zeroes: $x = \frac{2}{5}$ and $x = - 1$ .
Sum $= \frac{2}{5} - 1 = - \frac{3}{5} = - \frac{b}{a}$ .
Product $= \frac{2}{5} \times (- 1) = - \frac{2}{5} = \frac{c}{a}$ .
Hence the relationship is verified.

Q263 marksShort AnswerPolynomials CBSE 2024 · Basic 430/3/2

Find the zeroes of the polynomial $2 x^{2} + 3 x - 9$ and verify the relationship between the zeroes and the co-efficients of polynomial.

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Answer: Zeroes are

\frac{3}{2}

and

- 3

; relationship verified.

$2 x^{2} + 3 x - 9 = 2 x^{2} + 6 x - 3 x - 9 = (2 x - 3) (x + 3)$ .
Zeroes: $x = \frac{3}{2}$ and $x = - 3$ .
Sum $= \frac{3}{2} - 3 = - \frac{3}{2} = - \frac{b}{a}$ .
Product $= \frac{3}{2} \times (- 3) = - \frac{9}{2} = \frac{c}{a}$ .
Hence the relationship is verified.

Q313 marksShort AnswerPolynomials CBSE 2024 · Basic 430/3/3

If one zero of the polynomial $x^{2} - 8 x + k$ exceeds the other by 2, then find the zeroes and the value of k.

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Answer: Zeroes are 3 and 5;

k = 15

Let the zeroes be $α$ and $α + 2$ .
Sum of zeroes $= 2 α + 2 = 8$ , so $α = 3$ ; the zeroes are 3 and 5.
Product of zeroes $= k = 3 \times 5 = 15$ .

Q273 marksShort AnswerPolynomials CBSE 2024 · Basic 430/5/1

Find the zeroes of the quadratic polynomial $6 x^{2} - 7 x - 3$ and verify the relationship between the zeroes and the coefficients of the polynomial.

Show answer & solution

Answer: Zeroes are

\frac{3}{2}

and

- \frac{1}{3}

; relationship verified.

$6 x^{2} - 7 x - 3 = 6 x^{2} - 9 x + 2 x - 3 = 3 x (2 x - 3) + 1 (2 x - 3) = (2 x - 3) (3 x + 1)$ .
Zeroes: $x = \frac{3}{2}$ and $x = - \frac{1}{3}$ .
Sum $= \frac{3}{2} - \frac{1}{3} = \frac{7}{6}$ and $- \frac{b}{a} = - \frac{- 7}{6} = \frac{7}{6}$ .
Product $= \frac{3}{2} \times (- \frac{1}{3}) = - \frac{1}{2}$ and $\frac{c}{a} = \frac{- 3}{6} = - \frac{1}{2}$ .
Hence the relationship is verified.

Also asked in: 2024 Basic 430/5/2

Q263 marksShort AnswerPolynomials CBSE 2024 · Basic 430/5/3

Find the zeroes of the polynomial $3 x^{2} - 5 x - 2$ and verify the relationship between the zeroes and the coefficients of the polynomial.

Show answer & solution

Answer: Zeroes are 2 and

- \frac{1}{3}

; relationship verified.

$3 x^{2} - 5 x - 2 = 3 x^{2} - 6 x + x - 2 = 3 x (x - 2) + 1 (x - 2) = (x - 2) (3 x + 1)$ .
Zeroes: $x = 2$ and $x = - \frac{1}{3}$ .
Sum $= 2 - \frac{1}{3} = \frac{5}{3}$ and $- \frac{b}{a} = - \frac{- 5}{3} = \frac{5}{3}$ .
Product $= 2 \times (- \frac{1}{3}) = - \frac{2}{3}$ and $\frac{c}{a} = \frac{- 2}{3}$ .
Hence the relationship is verified.

Q273 marksShort AnswerPolynomials CBSE 2024 · Standard 30/3/1

Find the zeroes of the quadratic polynomial $x^{2} - 15$ and verify the relationship between the zeroes and the coefficients of the polynomial.

Show answer & solution

Answer: Zeroes:

15

and

- 15

; relationship verified.

$x^{2} - 15 = (x - 15) (x + 15)$ , so the zeroes are $15$ and $- 15$ .
Here $a = 1$ , $b = 0$ , $c = - 15$ .
Sum of zeroes = $15 + (- 15) = 0 = \frac{- b}{a} = \frac{- 0}{1}$ .
Product of zeroes = $15 \times (- 15) = - 15 = \frac{c}{a} = \frac{- 15}{1}$ .
Hence the relationship is verified.

Also asked in: 2024 Standard 30/3/2, 2024 Standard 30/3/3

Q293 marksShort AnswerPolynomials CBSE 2024 · Standard 30/4/1

Find the zeroes of the polynomial $4 x^{2} + 4 x - 3$ and verify the relationship between zeroes and coefficients of the polynomial.

Show answer & solution

Answer: Zeroes are

\frac{1}{2}

and

- \frac{3}{2}

; relationship verified.

$4 x^{2} + 4 x - 3 = 4 x^{2} + 6 x - 2 x - 3 = 2 x (2 x + 3) - 1 (2 x + 3) = (2 x - 1) (2 x + 3)$ .
Zeroes: $x = \frac{1}{2}$ and $x = - \frac{3}{2}$ .
Sum = $\frac{1}{2} - \frac{3}{2} = - 1$ and $- \frac{b}{a} = - \frac{4}{4} = - 1$ .
Product = $\frac{1}{2} \times (- \frac{3}{2}) = - \frac{3}{4}$ and $\frac{c}{a} = - \frac{3}{4}$ .
Hence the relationship is verified.

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q29 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2024 · Standard 30/4/1

If $α$ and $β$ are the zeroes of the polynomial $x^{2} + x - 2$ , then find the value of $\frac{α}{β} + \frac{β}{α}$ .

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Answer:

- \frac{5}{2}

$α + β = - 1$ and $α β = - 2$ .
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 1 + 4 = 5$ .
$\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β} = \frac{5}{- 2} = - \frac{5}{2}$ .

Also asked in: 2024 Standard 30/4/2, 2024 Standard 30/4/3

Q293 marksShort AnswerPolynomials CBSE 2023 · Basic 430/1/1

If $α$ , $β$ are zeroes of the quadratic polynomial $x^{2} + 3 x + 2$ , find a quadratic polynomial whose zeroes are $α + 1$ , $β + 1$ .

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Answer:

x^{2} + x

$α + β = - 3$ , $α β = 2$
Sum of new zeroes $= (α + 1) + (β + 1) = - 3 + 2 = - 1$
Product $= (α + 1) (β + 1) = α β + (α + β) + 1 = 2 - 3 + 1 = 0$
Required polynomial $= x^{2} - (sum) x + product = x^{2} + x$

Also asked in: 2023 Basic 430/1/2, 2023 Basic 430/1/3

Q303 marksShort AnswerPolynomials CBSE 2023 · Basic 430/2/1

Find the zeroes of the polynomial $p (x) = 2 x^{2} - 7 x - 15$ and verify the relationship between its coefficients and zeroes.

Show answer & solution

Answer: Zeroes: 5 and

- \frac{3}{2}

$2 x^{2} - 7 x - 15 = 2 x^{2} - 10 x + 3 x - 15 = (x - 5) (2 x + 3)$
Zeroes: x = 5 and x = $- \frac{3}{2}$
Sum $= 5 - \frac{3}{2} = \frac{7}{2}$ and $- \frac{b}{a} = - \frac{- 7}{2} = \frac{7}{2}$ ✓
Product $= 5 \times (- \frac{3}{2}) = - \frac{15}{2}$ and $\frac{c}{a} = - \frac{15}{2}$ ✓

Also asked in: 2023 Basic 430/2/3

Q273 marksShort AnswerPolynomials CBSE 2023 · Basic 430/2/2

Find the zeroes of the polynomial $p (x) = 3 x^{2} + 5 x - 28$ and verify the relationship between its coefficients and zeroes.

Show answer & solution

Answer: Zeroes:

\frac{7}{3}

and

- 4

$3 x^{2} + 5 x - 28 = 3 x^{2} + 12 x - 7 x - 28 = (x + 4) (3 x - 7)$
Zeroes: x = $- 4$ and x = $\frac{7}{3}$
Sum $= \frac{7}{3} - 4 = - \frac{5}{3}$ and $- \frac{b}{a} = - \frac{5}{3}$ ✓
Product $= \frac{7}{3} \times (- 4) = - \frac{28}{3}$ and $\frac{c}{a} = - \frac{28}{3}$ ✓

Q263 marksShort AnswerPolynomials CBSE 2023 · Basic 430/4/1

Find the zeroes of the quadratic polynomial $x^{2} + 6 x + 8$ and verify the relationship between the zeroes and the coefficients.

Show answer & solution

Answer: Zeroes:

- 2

and

- 4

(sum

= - 6 = - \frac{b}{a}

, product

= 8 = \frac{c}{a}

$x^{2} + 6 x + 8 = x^{2} + 2 x + 4 x + 8 = (x + 2) (x + 4)$ .
Zeroes are $- 2$ and $- 4$ .
Sum of zeroes $= - 6$ and $- \frac{b}{a} = - \frac{6}{1} = - 6$ . Verified.
Product of zeroes $= 8$ and $\frac{c}{a} = \frac{8}{1} = 8$ . Verified.

Also asked in: 2023 Basic 430/4/3

Q273 marksShort AnswerPolynomials CBSE 2023 · Basic 430/4/2

If $α$ , $β$ are zeroes of the quadratic polynomial $x^{2} - 5 x + 6$ , form another quadratic polynomial whose zeroes are $\frac{1}{α}, \frac{1}{β}$ .

Show answer & solution

Answer:

x^{2} - \frac{5}{6} x + \frac{1}{6}

, or

k (6 x^{2} - 5 x + 1)

for any non-zero

k

$α + β = 5$ , $α β = 6$ .
Sum of new zeroes $= \frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{5}{6}$ .
Product of new zeroes $= \frac{1}{α β} = \frac{1}{6}$ .
Required polynomial $= x^{2} - \frac{5}{6} x + \frac{1}{6}$ , i.e. $\frac{1}{6} (6 x^{2} - 5 x + 1)$ ; so $6 x^{2} - 5 x + 1$ (or any non-zero multiple).

Q29 (OR) (OR)3 marksShort AnswerPolynomials CBSE 2023 · Standard 30/6/1

If $α$ and $β$ are roots of the quadratic equation $x^{2} - 7 x + 10 = 0$ , find the quadratic equation whose roots are $α^{2}$ and $β^{2}$ .

Show answer & solution

Answer:

x^{2} - 29 x + 100 = 0

$α + β = 7$ , $α β = 10$ .
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 49 - 20 = 29$
$α^{2} β^{2} = 100$
Required equation: $x^{2} - 29 x + 100 = 0$

Also asked in: 2023 Standard 30/6/2

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