CBSE Maths › Class 10 PYQs › Polynomials

Polynomials: 2 marks Questions (CBSE Class 10)

32 different 2 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)

Q222 marksVery Short AnswerPolynomials CBSE 2026 · Basic 430/4/1

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

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Answer:

- \frac{5}{3}

$α + β = \frac{5}{6}$ , $α β = - \frac{3}{6} = - \frac{1}{2}$ .
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{5}{6} \div (- \frac{1}{2}) = - \frac{5}{3}$ .

Q22 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2026 · Basic 430/4/1

One zero of the polynomial $4 x^{2} - 12 x + (2 k + 1)$ is five times the other. Find the value of k.

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Answer: k = 2

Let the zeroes be $α$ and $5 α$ .
Sum: $6 α = \frac{12}{4} = 3$ , so $α = \frac{1}{2}$ .
Product: $5 α^{2} = \frac{2 k + 1}{4} \Rightarrow \frac{5}{4} = \frac{2 k + 1}{4}$ .
$2 k + 1 = 5$ , so $k = 2$ .

Q222 marksVery Short AnswerPolynomials CBSE 2026 · Basic 430/4/2

If $α$ , $β$ are the zeroes of polynomial p( $x$ ) = $- 9 x^{2} - 6 x + 1$ . Find the value of $α^{2} + β^{2}$ .

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Answer:

\frac{2}{3}

$α + β = - \frac{- 6}{- 9} = - \frac{2}{3}$ , $α β = \frac{1}{- 9} = - \frac{1}{9}$ .
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{4}{9} + \frac{2}{9} = \frac{2}{3}$ .

Q22 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2026 · Basic 430/4/2

Form a quadratic polynomial whose zeroes are twice the zeroes of polynomial p( $x$ ) = $x^{2} - 3 x - 5$ .

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Answer:

x^{2} - 6 x - 20

For $x^{2} - 3 x - 5$ : $α + β = 3$ , $α β = - 5$ .
New zeroes $2 α$ , $2 β$ : sum = 6, product = $4 α β = - 20$ .
Required polynomial: $x^{2} - 6 x - 20$ (or any non-zero multiple of it).

Q242 marksVery Short AnswerPolynomials CBSE 2026 · Basic 430/4/3

Form a quadratic polynomial whose sum and product of the zeroes are $\frac{1}{2}$ and $- \frac{1}{9}$ respectively. Hence, find the zeroes of the polynomial.

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Answer:

18 x^{2} - 9 x - 2

; zeroes

\frac{2}{3}

and

- \frac{1}{6}

Polynomial: $x^{2} - (sum) x + product = x^{2} - \frac{1}{2} x - \frac{1}{9}$ , or $18 x^{2} - 9 x - 2$ .
$18 x^{2} - 9 x - 2 = 18 x^{2} - 12 x + 3 x - 2 = 6 x (3 x - 2) + 1 (3 x - 2) = (6 x + 1) (3 x - 2)$ .
Zeroes: $x = \frac{2}{3}$ and $x = - \frac{1}{6}$ .

Q24 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2026 · Basic 430/4/3

If $α$ , $β$ are zeroes of the polynomial $x^{2} - 6 x + 7$ , then find the value of $4 (\frac{1}{α ^{2}} + \frac{1}{β ^{2}})$ .

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Answer:

\frac{88}{49}

$α + β = 6$ , $α β = 7$ .
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 36 - 14 = 22$ .
$\frac{1}{α ^{2}} + \frac{1}{β ^{2}} = \frac{α ^{2} + β ^{2}}{( α β ) ^{2}} = \frac{22}{49}$ .
Required value = $4 \times \frac{22}{49} = \frac{88}{49}$ .

Q212 marksVery Short AnswerPolynomials CBSE 2026 · Standard 30/1/1

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

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Answer:

- 3

$α + β = - \frac{- 3}{1} = 3$ and $α β = \frac{- 1}{1} = - 1$ .
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{3}{- 1} = - 3$ .

Q252 marksVery Short AnswerPolynomials CBSE 2026 · Standard 30/1/2

Find the zeroes of the quadratic polynomial $x^{2} + 7 x + 10$ , and verify the relationship between the zeroes and its coefficients.

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Answer: Zeroes

- 2

and

- 5

; relationship verified.

$x^{2} + 7 x + 10 = (x + 2) (x + 5)$ , so the zeroes are $- 2$ and $- 5$ .
Sum of zeroes $= - 7 = - \frac{7}{1} = - \frac{b}{a}$ .
Product of zeroes $= 10 = \frac{10}{1} = \frac{c}{a}$ .
Hence the relationship is verified.

Q242 marksVery Short AnswerPolynomials CBSE 2026 · Standard 30/1/3

Find a quadratic polynomial whose zeroes are $(5 - 23)$ and $(5 + 23)$ .

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Answer:

x^{2} - 10 x + 13

Sum of zeroes $= (5 - 23) + (5 + 23) = 10$ .
Product of zeroes $= 25 - 12 = 13$ .
Required polynomial $= x^{2} - (sum) x + product = x^{2} - 10 x + 13$ (or any non-zero multiple of it).

Q212 marksVery Short AnswerPolynomials CBSE 2026 · Standard 30/2/1

If $α$ , $β$ are the zeroes of the quadratic polynomial $p x^{2} + q x + r$ , then find the value of $α^{3} β + β^{3} α$ .

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Answer:

\frac{r ( q ^{2} - 2 p r )}{p ^{3}}

$α + β = - \frac{q}{p}$ , $α β = \frac{r}{p}$
$α^{3} β + β^{3} α = α β (α^{2} + β^{2}) = α β [(α + β)^{2} - 2 α β]$
$= \frac{r}{p} (\frac{q ^{2}}{p ^{2}} - \frac{2 r}{p}) = \frac{r}{p} \cdot \frac{q ^{2} - 2 p r}{p ^{2}}$
$= \frac{r ( q ^{2} - 2 p r )}{p ^{3}}$

Also asked in: 2026 Standard 30/2/3

Q242 marksVery Short AnswerPolynomials CBSE 2026 · Standard 30/2/2

Find the value of p, for which one zero of the quadratic polynomial $p x^{2} - 14 x + 8$ is 6 times the other.

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Answer: p = 3

Let the zeroes be $α$ and $6 α$ .
Sum: $7 α = \frac{14}{p}$ , so $α = \frac{2}{p}$ .
Product: $6 α^{2} = \frac{8}{p}$ , so $6 \times \frac{4}{p ^{2}} = \frac{8}{p}$ .
$\frac{24}{p ^{2}} = \frac{8}{p}$ gives p = 3.

Q252 marksVery Short AnswerPolynomials CBSE 2026 · Standard 30/4/1

$α$ and $β$ are the zeroes of the polynomial $5 x^{2} - 16 x - 10$ . Find the value of $\frac{α}{β} + \frac{β}{α}$ .

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Answer:

- \frac{178}{25}

$α + β = \frac{16}{5}$ , $α β = \frac{- 10}{5} = - 2$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = \frac{256}{25} + 4 = \frac{356}{25}$
$\frac{α}{β} + \frac{β}{α} = \frac{α ^{2} + β ^{2}}{α β} = \frac{356}{25} \div (- 2) = - \frac{178}{25}$

Also asked in: 2026 Standard 30/4/3

Q242 marksVery Short AnswerPolynomials CBSE 2026 · Standard 30/4/2

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

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Answer:

\frac{66}{25}

$α + β = \frac{6}{3} = 2$ , $α β = - \frac{5}{3}$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 4 + \frac{10}{3} = \frac{22}{3}$
$\frac{1}{α ^{2}} + \frac{1}{β ^{2}} = \frac{α ^{2} + β ^{2}}{( α β ) ^{2}} = \frac{22}{3} \div \frac{25}{9} = \frac{66}{25}$

Q222 marksVery Short AnswerPolynomials CBSE 2025 · Standard 30/1/1

Find the zeroes of the polynomial $p (x) = x^{2} + \frac{4}{3} x - \frac{4}{3}$ .

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Answer:

\frac{2}{3}

and

- 2

$p (x) = \frac{1}{3} (3 x^{2} + 4 x - 4)$ .
$3 x^{2} + 4 x - 4 = 3 x^{2} + 6 x - 2 x - 4 = 3 x (x + 2) - 2 (x + 2) = (3 x - 2) (x + 2)$ .
$p (x) = 0$ gives $x = \frac{2}{3}$ or $x = - 2$ .
Check: sum $= \frac{2}{3} - 2 = - \frac{4}{3}$ and product $= - \frac{4}{3}$ , as expected.

Also asked in: 2025 Standard 30/1/3

Q212 marksVery Short AnswerPolynomials CBSE 2025 · Standard 30/1/2

If the zeroes of the polynomial $x^{2} + a x + b$ are in the ratio $3 : 4$ , then prove that $12 a^{2} = 49 b$ .

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Answer: Proved.

Let the zeroes be $3 k$ and $4 k$ .
Sum of zeroes: $3 k + 4 k = 7 k = - a$ , so $k = - \frac{a}{7}$ .
Product of zeroes: $3 k \times 4 k = 12 k^{2} = b$ .
So $b = 12 \times \frac{a ^{2}}{49}$ , i.e. $49 b = 12 a^{2}$ .
Hence $12 a^{2} = 49 b$ .

Q212 marksVery Short AnswerPolynomials CBSE 2025 · Standard 30/2/1

If p and q are zeroes of the polynomial $p (y) = 21 y^{2} - y - 2$ , then find the value of $(1 - p) \cdot (1 - q)$ .

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Answer:

\frac{6}{7}

$p + q = \frac{1}{21}$ , $pq = \frac{- 2}{21}$
$(1 - p) (1 - q) = 1 - (p + q) + pq$
$= 1 - \frac{1}{21} - \frac{2}{21} = \frac{18}{21} = \frac{6}{7}$

Also asked in: 2025 Standard 30/2/3

Q212 marksVery Short AnswerPolynomials CBSE 2025 · Standard 30/2/2

Find a quadratic polynomial whose zeroes are 2 and $- \frac{7}{5}$ .

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Answer:

5 x^{2} - 3 x - 14

(or any non-zero multiple

k (5 x^{2} - 3 x - 14)

)

Sum of zeroes $= 2 - \frac{7}{5} = \frac{3}{5}$
Product of zeroes $= 2 \times (- \frac{7}{5}) = - \frac{14}{5}$
Polynomial: $x^{2} - \frac{3}{5} x - \frac{14}{5}$ , or multiplying by 5, $5 x^{2} - 3 x - 14$

Q24 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2025 · Standard 30/3/1

If ' $α$ ' and ' $β$ ' are the zeroes of the polynomial $p (y) = y^{2} - 5 y + 3$ , then find the value of $α^{4} β^{3} + α^{3} β^{4}$ .

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Answer: 135

$α + β = 5$ , $α β = 3$ .
$α^{4} β^{3} + α^{3} β^{4} = α^{3} β^{3} (α + β) = (α β)^{3} (α + β)$ .
$= 3^{3} \times 5 = 135$ .

Also asked in: 2025 Standard 30/3/3

Q212 marksVery Short AnswerPolynomials CBSE 2025 · Standard 30/3/2

If the sum of the zeroes of the polynomial $p (x) = (p + 1) x^{2} + (2 p + 3) x + (3 p + 4)$ is $- 1$ , then find the value of 'p'.

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Answer:

p = - 2

Sum of zeroes $= - \frac{2 p + 3}{p + 1} = - 1$ .
$2 p + 3 = p + 1$ , so $p = - 2$ .
Check: $p + 1 = - 1 \neq = 0$ , so the polynomial is quadratic.

Q21 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2025 · Standard 30/3/2

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

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Answer:

- 4

$α + β = 2$ , $α β = - 1$ .
$\frac{1}{2 α} + \frac{1}{2 β} = \frac{α + β}{2 α β} = \frac{2}{- 2} = - 1$ .
$3 α β = - 3$ .
Required value $= - 1 - 3 = - 4$ .

Q242 marksVery Short AnswerPolynomials CBSE 2024 · Basic 430/4/1

If $α$ , $β$ are zeroes of the polynomial $8 x^{2} + 14 x + 3$ , then find the value of $(\frac{1}{α} + \frac{1}{β})$ .

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Answer:

- \frac{14}{3}

$α + β = - \frac{14}{8}$ and $α β = \frac{3}{8}$ .
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{- 14/8}{3/8} = - \frac{14}{3}$ .

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q24 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2024 · Basic 430/4/1

Find a quadratic polynomial whose zeroes are $- 9$ and 6.

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Answer:

x^{2} + 3 x - 54

(or any non-zero multiple of it)

Sum of zeroes $= - 9 + 6 = - 3$ ; product $= (- 9) (6) = - 54$ .
Required polynomial $= x^{2} - (sum) x + product = x^{2} + 3 x - 54$ .

Also asked in: 2024 Basic 430/4/2, 2024 Basic 430/4/3

Q252 marksVery Short AnswerPolynomials CBSE 2024 · Basic 430/5/1

If $α$ , $β$ are zeroes of the quadratic polynomial $2 x^{2} + 7 x + 5$ , then find the value of $α^{2} + β^{2} + α β$ .

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Answer:

\frac{39}{4}

$α + β = - \frac{7}{2}$ and $α β = \frac{5}{2}$ .
$α^{2} + β^{2} + α β = (α + β)^{2} - α β$ .
$= \frac{49}{4} - \frac{5}{2} = \frac{49 - 10}{4} = \frac{39}{4}$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q25 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2024 · Basic 430/5/1

If one zero of the quadratic polynomial $6 x^{2} + 37 x - (p - 2)$ is reciprocal of the other, then find the value of p.

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Answer:

p = - 4

Let the zeroes be $α$ and $\frac{1}{α}$ ; their product is 1.
Product of zeroes $= \frac{c}{a} = \frac{- ( p - 2 )}{6}$ .
$\frac{- ( p - 2 )}{6} = 1$ , so $p - 2 = - 6$ .
$p = - 4$ .

Also asked in: 2024 Basic 430/5/2, 2024 Basic 430/5/3

Q232 marksVery Short AnswerPolynomials CBSE 2024 · Standard 30/5/1

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

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Answer:

\frac{7}{5}

$α + β = - \frac{- 6}{5} = \frac{6}{5}$ and $α β = \frac{1}{5}$ .
$α + β + α β = \frac{6}{5} + \frac{1}{5} = \frac{7}{5}$ .

Also asked in: 2024 Standard 30/5/3

Q212 marksVery Short AnswerPolynomials CBSE 2024 · Standard 30/5/2

If $α$ and $β$ are zeroes of the quadratic polynomial $p (x) = x^{2} - 5 x + 4$ , then find the value of $\frac{1}{α} + \frac{1}{β} - 2 α β$ .

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Answer:

- \frac{27}{4}

$α + β = 5$ and $α β = 4$ .
$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{5}{4}$ .
$\frac{1}{α} + \frac{1}{β} - 2 α β = \frac{5}{4} - 8 = - \frac{27}{4}$ .

Q242 marksVery Short AnswerPolynomials CBSE 2023 · Basic 430/2/1

Find the value of ‘k’ such that the polynomial $p (x) = 3 x^{2} + 2 k x + x - k - 5$ has the sum of zeroes equal to half of their product.

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Answer: k = 1

$p (x) = 3 x^{2} + (2 k + 1) x - (k + 5)$
Sum of zeroes $= - \frac{2 k + 1}{3}$ , product $= - \frac{k + 5}{3}$
Sum = $\frac{1}{2}$ product: $- \frac{2 k + 1}{3} = - \frac{k + 5}{6}$
$2 (2 k + 1) = k + 5$ , so $4 k + 2 = k + 5$
3k = 3, k = 1

Q222 marksVery Short AnswerPolynomials CBSE 2023 · Basic 430/2/2

$α, β$ are the zeroes of the quadratic polynomial $p (x) = x^{2} - 8 x + k$ , such that $α^{2} + β^{2} = 40$ . Find the value of k.

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Answer: k = 12

$α + β = 8$ and $α β = k$
$α^{2} + β^{2} = (α + β)^{2} - 2 α β = 64 - 2 k$
$64 - 2 k = 40$ , so 2k = 24 and k = 12

Q212 marksVery Short AnswerPolynomials CBSE 2023 · Basic 430/2/3

If $(- 3)$ is one of the zeroes of the polynomial $(k - 1) x^{2} + k x + 1$ , find the value of k.

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Answer:

k = \frac{4}{3}

Since $- 3$ is a zero, $p (- 3) = 0$ .
$(k - 1) (9) + k (- 3) + 1 = 0$
$9 k - 9 - 3 k + 1 = 0$
$6 k = 8$ , so $k = \frac{4}{3}$

Q222 marksVery Short AnswerPolynomials CBSE 2023 · Basic 430/6/1

Find a quadratic polynomial whose zeroes are 6 and $- 3$ .

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Answer:

x^{2} - 3 x - 18

Sum of zeroes $= 6 + (- 3) = 3$ ; product $= 6 \times (- 3) = - 18$ .
Polynomial $= x^{2} - (sum) x + product = x^{2} - 3 x - 18$ (or any non-zero multiple of it).

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q22 (OR) (OR)2 marksVery Short AnswerPolynomials CBSE 2023 · Basic 430/6/1

Find the zeroes of the polynomial $x^{2} + 4 x - 12$ .

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Answer: 2 and

- 6

$x^{2} + 4 x - 12 = x^{2} + 6 x - 2 x - 12 = (x + 6) (x - 2)$ .
Zeroes: $x = 2$ and $x = - 6$ .

Also asked in: 2023 Basic 430/6/2, 2023 Basic 430/6/3

Q222 marksVery Short AnswerPolynomials CBSE 2023 · Standard 30/4/1

If one zero of the polynomial $p (x) = 6 x^{2} + 37 x - (k - 2)$ is reciprocal of the other, then find the value of k.

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Answer:

k = - 4

Let the zeroes be $α$ and $\frac{1}{α}$ ; their product is 1.
Product of zeroes $= \frac{- ( k - 2 )}{6}$
$\frac{- ( k - 2 )}{6} = 1 \Rightarrow k - 2 = - 6 \Rightarrow k = - 4$

Also asked in: 2023 Standard 30/4/2, 2023 Standard 30/4/3

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