Board Previous Year Questions (Board PYQs)

How to Approach Board PYQs

This section collects the types of questions on Polynomials that repeatedly appear in CBSE and State Board examinations, with full step-by-step solutions. Working through these 26 problems is the best way to predict what your exam will ask.

What examiners love in this chapter:

1. Find the zeroes and verify the relation with coefficients (2–3 marks).
2. Form a quadratic from given sum and product, or from given zeroes (2 marks).
3. Find a parameter $k$ using the sum/product of zeroes (2–3 marks).
4. Find all zeroes of a cubic/quartic given some, using division (3–4 marks).

Tag note: Questions are tagged by exam (e.g. [CBSE Board]). Where the exact year of a specific question could not be confirmed, only the exam name is given, to keep the content trustworthy.

Exam Tip: Always show the factorisation or the substitution into $-b/a$ and $c/a$. Method marks are awarded generously in this chapter.

Solved Previous Year Questions

PYQ 1: Find zeroes and verify (3 marks)

Find the zeroes of $p(x) = x^2 - 2x - 8$ and verify the relationship between the zeroes and the coefficients. [CBSE Board]

Solution:

1. Factorise: $x^2 - 2x - 8 = (x-4)(x+2)$, so zeroes are $4$ and $-2$.
2. Sum $= 4 + (-2) = 2 = -\dfrac{-2}{1} = -\dfrac{b}{a}$. ✓
3. Product $= (4)(-2) = -8 = \dfrac{-8}{1} = \dfrac{c}{a}$. ✓

Final Answer: Zeroes $4, -2$; relations verified.

Takeaway: Always present both computed and formula values.

PYQ 2: Form a quadratic (2 marks)

Find a quadratic polynomial whose zeroes are $-3$ and $4$. [CBSE Board]

Solution:

1. Sum $= 1$, product $= -12$.
2. $p(x) = x^2 - x - 12$.

Final Answer: $x^2 - x - 12$.

Takeaway: Template $x^2 - (\text{sum})x + (\text{product})$.

PYQ 3: Sum and product of zeroes (1 mark)

If $\alpha, \beta$ are zeroes of $x^2 - 6x + 8$, write the values of $\alpha + \beta$ and $\alpha\beta$. [CBSE Board]

Solution:

1. $\alpha + \beta = -\dfrac{-6}{1} = 6$.
2. $\alpha\beta = \dfrac{8}{1} = 8$.

Final Answer: Sum 6, product 8.

Takeaway: Read off $-b/a$ and $c/a$ directly.

PYQ 4: Quadratic from sum and product (2 marks)

Find a quadratic polynomial the sum and product of whose zeroes are $\dfrac{1}{4}$ and $-1$ respectively. [CBSE Board]

Solution:

1. $p(x) = x^2 - \dfrac{1}{4}x - 1$.
2. Multiply by 4: $4x^2 - x - 4$.

Final Answer: $4x^2 - x - 4$.

Takeaway: Scale to clear fractions for a neat answer.

PYQ 5: Find $k$ from a zero (2 marks)

If one zero of $p(x) = 3x^2 + 8x + k$ is the reciprocal of the other, find $k$. [CBSE Board]

Solution:

1. If zeroes are $\alpha$ and $\dfrac{1}{\alpha}$, product $= \alpha \cdot \dfrac{1}{\alpha} = 1$.
2. Product $= \dfrac{c}{a} = \dfrac{k}{3}$, so $\dfrac{k}{3} = 1 \Rightarrow k = 3$.

Final Answer: $k = 3$.

Takeaway: Reciprocal zeroes ⇒ product = 1.

PYQ 6: Find $\alpha^2 + \beta^2$ (2 marks)

If $\alpha, \beta$ are zeroes of $x^2 - 7x + 10$, find $\alpha^2 + \beta^2$. [CBSE Board]

Solution:

1. Sum $= 7$, product $= 10$.
2. $\alpha^2+\beta^2 = 7^2 - 2(10) = 49 - 20 = 29$.

Final Answer: 29.

Takeaway: $(\alpha+\beta)^2 - 2\alpha\beta$.

PYQ 7: Find all zeroes of a quartic (4 marks)

Find all the zeroes of $p(x) = x^4 - 5x^2 + 4$. [CBSE Board]

Solution:

1. Let $y = x^2$: $y^2 - 5y + 4 = (y-1)(y-4)$.
2. So $x^2 = 1$ or $x^2 = 4$.
3. Zeroes: $x = 1, -1, 2, -2$.

Final Answer: $1, -1, 2, -2$.

Takeaway: A biquadratic in $x^2$ can be solved by substitution.

PYQ 8: Verify cubic relations (3 marks)

Verify that 3, $-1$, $-\dfrac{1}{3}$ are the zeroes of the cubic $p(x) = 3x^3 - 5x^2 - 11x - 3$, and verify the relationship between the zeroes and coefficients. [CBSE Board]

Solution:

1. Here $a=3, b=-5, c=-11, d=-3$.
2. Sum $= 3 - 1 - \dfrac{1}{3} = \dfrac{5}{3} = -\dfrac{-5}{3} = -\dfrac{b}{a}$. ✓
3. Pair-sum $= (3)(-1) + (-1)(-\tfrac{1}{3}) + (-\tfrac{1}{3})(3) = -3 + \tfrac{1}{3} - 1 = -\dfrac{11}{3} = \dfrac{c}{a}$. ✓
4. Product $= (3)(-1)(-\tfrac{1}{3}) = 1 = -\dfrac{-3}{3} = -\dfrac{d}{a}$. ✓

Final Answer: All three relations verified.

Takeaway: Match each symmetric function to $-b/a$, $c/a$, $-d/a$.

PYQ 9: Find all zeroes given two (4 marks)

Find all zeroes of $p(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2$, if two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$. [CBSE Board]

Solution:

1. $(x-\sqrt2)(x+\sqrt2) = x^2 - 2$ is a factor.
2. Divide: $2x^4 - 3x^3 - 3x^2 + 6x - 2 \div (x^2 - 2)$ gives quotient $2x^2 - 3x + 1$.
3. $2x^2 - 3x + 1 = (2x-1)(x-1)$, so zeroes $\dfrac{1}{2}, 1$.

Final Answer: $\sqrt2, -\sqrt2, \dfrac{1}{2}, 1$.

Takeaway: Divide by the known quadratic factor, then solve the quotient.

PYQ 10: Find $k$ using sum of zeroes (2 marks)

If the sum of the zeroes of $p(x) = kx^2 - 3x + 5$ is 1, find $k$. [CBSE Board]

Solution:

1. Sum $= -\dfrac{b}{a} = -\dfrac{-3}{k} = \dfrac{3}{k}$.
2. Set $= 1$: $\dfrac{3}{k} = 1 \Rightarrow k = 3$.

Final Answer: $k = 3$.

Takeaway: Set the sum formula equal to the given value.

PYQ 11: Zeroes of $\sqrt3 x^2 + 10x + 7\sqrt3$ (3 marks)

Find the zeroes of $p(x) = \sqrt{3}x^2 + 10x + 7\sqrt{3}$. [CBSE Board]

Solution:

1. Product $a \cdot c = \sqrt3 \cdot 7\sqrt3 = 21$; split 10 as $3 + 7$.
2. $\sqrt3 x^2 + 3x + 7x + 7\sqrt3 = \sqrt3 x(x + \sqrt3) + 7(x + \sqrt3) = (x+\sqrt3)(\sqrt3 x + 7)$.
3. Zeroes: $x = -\sqrt3$ and $x = -\dfrac{7}{\sqrt3} = -\dfrac{7\sqrt3}{3}$.

Final Answer: $-\sqrt3$ and $-\dfrac{7\sqrt3}{3}$.

Takeaway: Split the middle term even when coefficients are surds.

PYQ 12: Form quadratic with given conditions (2 marks)

Find a quadratic polynomial whose zeroes are 5 and $-3$. [State Board]

Solution:

1. Sum $= 2$, product $= -15$.
2. $p(x) = x^2 - 2x - 15$.

Final Answer: $x^2 - 2x - 15$.

Takeaway: Apply the standard template.

PYQ 13: Reciprocal-sum question (2 marks)

If $\alpha, \beta$ are zeroes of $x^2 - 5x + 4$, find $\dfrac{1}{\alpha} + \dfrac{1}{\beta} - 2\alpha\beta$. [CBSE Board]

Solution:

1. Sum $= 5$, product $= 4$.
2. $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{5}{4}$.
3. $\dfrac{5}{4} - 2(4) = \dfrac{5}{4} - 8 = -\dfrac{27}{4}$.

Final Answer: $-\dfrac{27}{4}$.

Takeaway: Build each piece from sum and product, then combine.

PYQ 14: Find $k$ for a given zero of a cubic (3 marks)

If $x = 2$ is a zero of $p(x) = x^3 - 3x^2 + kx - 2$, find $k$ and the other zeroes. [CBSE Board]

Solution:

1. $p(2) = 8 - 12 + 2k - 2 = 0 \Rightarrow 2k - 6 = 0 \Rightarrow k = 3$.
2. So $p(x) = x^3 - 3x^2 + 3x - 2$. Divide by $(x-2)$: quotient $x^2 - x + 1$.
3. $x^2 - x + 1 = 0$ has discriminant $1 - 4 = -3 < 0$, so no other real zeroes.

Final Answer: $k = 3$; the only real zero is 2.

Takeaway: Substitute the zero to find $k$, then divide to study the rest.

PYQ 15: Zeroes of $x^2 - (\sqrt3 + 1)x + \sqrt3$ (3 marks)

Find the zeroes of $p(x) = x^2 - (\sqrt3 + 1)x + \sqrt3$. [CBSE Board]

Solution:

1. Split: $x^2 - \sqrt3 x - x + \sqrt3 = x(x - \sqrt3) - 1(x - \sqrt3) = (x-\sqrt3)(x-1)$.
2. Zeroes: $x = \sqrt3$ and $x = 1$.

Final Answer: $\sqrt3$ and $1$.

Takeaway: Group the surd terms together when splitting the middle term.

PYQ 16: Polynomial division (3 marks)

Divide $p(x) = 3x^3 + x^2 + 2x + 5$ by $g(x) = x^2 + 2x + 1$ and write the quotient and remainder. [State Board]

Solution:

1. $\dfrac{3x^3}{x^2} = 3x$; $3x(x^2+2x+1) = 3x^3 + 6x^2 + 3x$; subtract: $-5x^2 - x + 5$.
2. $\dfrac{-5x^2}{x^2} = -5$; $-5(x^2+2x+1) = -5x^2 - 10x - 5$; subtract: $9x + 10$.
3. Degree 1 < 2, stop.

Final Answer: Quotient $3x - 5$, remainder $9x + 10$.

Takeaway: Stop when the remainder degree drops below the divisor's.

PYQ 17: Both relations for a quadratic (2 marks)

If $\alpha, \beta$ are zeroes of $2x^2 - 4x + 5$, find $\alpha^2 + \beta^2$. [CBSE Board]

Solution:

1. Sum $= \dfrac{4}{2} = 2$, product $= \dfrac{5}{2}$.
2. $\alpha^2+\beta^2 = 2^2 - 2\cdot\dfrac{5}{2} = 4 - 5 = -1$.

Final Answer: $-1$.

Takeaway: A negative value here signals the zeroes are not real — but the identity still computes the value of $\alpha^2 + \beta^2$.

PYQ 18: Form a polynomial from shifted zeroes (3 marks)

If $\alpha, \beta$ are zeroes of $x^2 - 2x + 3$, find a quadratic whose zeroes are $\alpha + 2$ and $\beta + 2$. [CBSE Board]

Solution:

1. Sum $= 2$, product $= 3$.
2. New sum $= (\alpha + \beta) + 4 = 2 + 4 = 6$.
3. New product $= \alpha\beta + 2(\alpha+\beta) + 4 = 3 + 4 + 4 = 11$.
4. Polynomial: $x^2 - 6x + 11$.

Final Answer: $x^2 - 6x + 11$.

Takeaway: Compute the new sum and product after the shift.

PYQ 19: Find $a$ and $b$ (3 marks)

If 1 and $-2$ are zeroes of $p(x) = x^3 + ax^2 - x + b$, find $a$ and $b$.

Solution:

1. $p(1) = 1 + a - 1 + b = 0 \Rightarrow a + b = 0$.
2. $p(-2) = -8 + 4a + 2 + b = 0 \Rightarrow 4a + b = 6$.
3. Subtract: $3a = 6 \Rightarrow a = 2$, then $b = -2$.

Final Answer: $a = 2$, $b = -2$.

Takeaway: Each known zero gives one equation; solve the system.

PYQ 20: Equal zeroes condition (2 marks)

Find the value of $k$ for which $x^2 - 4x + k$ has two equal real zeroes. [CBSE Board]

Solution:

1. Equal zeroes ⇒ both zeroes equal $\alpha$, so sum $= 2\alpha = 4 \Rightarrow \alpha = 2$.
2. Product $= \alpha^2 = 4 = \dfrac{k}{1}$, so $k = 4$.

Final Answer: $k = 4$.

Takeaway: Equal zeroes means product equals the square of half the sum.

PYQ 21: Sum of products and a parameter (3 marks)

If $\alpha, \beta$ are zeroes of $x^2 - 6x + a$ and $3\alpha + 2\beta = 20$, find $a$.

Solution:

1. $\alpha + \beta = 6$ and $\alpha\beta = a$.
2. From $3\alpha + 2\beta = 20$ and $\alpha + \beta = 6$ (so $2\alpha + 2\beta = 12$): subtract to get $\alpha = 8$, then $\beta = -2$.
3. $a = \alpha\beta = (8)(-2) = -16$.

Final Answer: $a = -16$.

Takeaway: Use the sum relation plus the extra condition to find each zero, then the product.

PYQ 22: Zeroes of $6x^2 - 7x - 3$ (3 marks)

Find the zeroes of $p(x) = 6x^2 - 7x - 3$ and verify the relations. [CBSE Board]

Solution:

1. Split: $6x^2 - 9x + 2x - 3 = 3x(2x-3) + 1(2x-3) = (3x+1)(2x-3)$.
2. Zeroes: $-\dfrac{1}{3}$ and $\dfrac{3}{2}$.
3. Sum $= \dfrac{7}{6} = -\dfrac{-7}{6}$ ✓; product $= -\dfrac{1}{2} = \dfrac{-3}{6}$ ✓.

Final Answer: Zeroes $-\dfrac{1}{3}, \dfrac{3}{2}$; verified.

Takeaway: Split the middle term, then read off the zeroes and verify them against the sum and product.

PYQ 23: Quadratic from a single given relation (2 marks)

Find a quadratic polynomial whose zeroes are $2$ and $\dfrac{1}{2}$. [State Board]

Solution:

1. Sum $= 2 + \dfrac{1}{2} = \dfrac{5}{2}$; product $= 2 \times \dfrac{1}{2} = 1$.
2. $p(x) = x^2 - \dfrac{5}{2}x + 1$; multiply by 2: $2x^2 - 5x + 2$.

Final Answer: $2x^2 - 5x + 2$.

Takeaway: Scale the polynomial to remove fractions.

PYQ 24: Find the polynomial from zero conditions (3 marks)

The sum and product of the zeroes of a quadratic polynomial are $-3$ and $2$ respectively. Find the polynomial and its zeroes. [CBSE Board]

Solution:

1. $p(x) = x^2 - (-3)x + 2 = x^2 + 3x + 2$.
2. Factorise: $(x+1)(x+2)$, so zeroes $-1$ and $-2$.

Final Answer: $x^2 + 3x + 2$; zeroes $-1, -2$.

Takeaway: Form the polynomial, then factor to get the zeroes.

PYQ 25: Geometrical / conceptual (1 mark)

The graph of a polynomial $p(x)$ cuts the x-axis at 3 points and touches it at 2 points. How many zeroes does $p(x)$ have? [CBSE Board]

Solution:

1. The number of zeroes equals the number of points where the graph meets the x-axis.
2. The graph meets the axis at $3 + 2 = 5$ points, so $p(x)$ has $5$ zeroes.

Final Answer: 5 zeroes.

Takeaway: Count every point where the graph meets the x-axis — crossings and touches alike.

PYQ 26: Find all zeroes of a cubic given one (4 marks)

Find all zeroes of $p(x) = x^3 - 3x^2 - 10x + 24$, given that $x = 2$ is a zero. [State Board]

Solution:

1. $(x - 2)$ is a factor. Divide: quotient $x^2 - x - 12$.
2. $x^2 - x - 12 = (x-4)(x+3)$, so zeroes $4$ and $-3$.

Final Answer: $2, 4, -3$.

Takeaway: Use the given zero to reduce the cubic to a quadratic, then factor.