CBSE Class 10 · Maths
CBSE Class 10 Maths Polynomials Previous Year Questions
Help your child master CBSE Class 10 Maths Polynomials Previous Year Questions with this curated collection sourced from real board papers spanning 2015–2025. Every question comes with a detailed step-by-step solution, helping your child confidently tackle zeroes of polynomials, Vieta’s formulas, and forming polynomials — topics that consistently carry marks in the board exam.
CBSE Class 10 Maths Polynomials — Questions with Solutions
The graph of a polynomial p(x) cuts the X-axis at 3 points and touches it at 2 other points. The number of zeroes of p(x) is
Solution
Answer: Option (D) is correct.
Explanation: According to the property of polynomials, the number of zeroes of a polynomial is equal to the number of points where the graph intersects the X-axis.
The graph cuts the X-axis at 3 points — each corresponding to a distinct zero. The graph touches the X-axis at 2 other points — these are repeated zeroes.
Thus, the total number of zeroes = 3 + 2 = 5.
Explanation: According to the property of polynomials, the number of zeroes of a polynomial is equal to the number of points where the graph intersects the X-axis.
The graph cuts the X-axis at 3 points — each corresponding to a distinct zero. The graph touches the X-axis at 2 other points — these are repeated zeroes.
Thus, the total number of zeroes = 3 + 2 = 5.
A quadratic polynomial, the product and sum of whose zeroes are 5 and 8 respectively, is
Solution
Answer: Option (A) is correct.
Explanation: For any quadratic polynomial $ax^2 + bx + c$:
• Sum of zeroes $= -\dfrac{b}{a}$
• Product of zeroes $= \dfrac{c}{a}$
Given sum $= 8$, so $b = -8a$. Given product $= 5$, so $c = 5a$.
Assuming $a = k$, the polynomial becomes $k(x^2 – 8x + 5)$, which matches Option (A).
Explanation: For any quadratic polynomial $ax^2 + bx + c$:
• Sum of zeroes $= -\dfrac{b}{a}$
• Product of zeroes $= \dfrac{c}{a}$
Given sum $= 8$, so $b = -8a$. Given product $= 5$, so $c = 5a$.
Assuming $a = k$, the polynomial becomes $k(x^2 – 8x + 5)$, which matches Option (A).
If $x – 1$ is a factor of the polynomial $p(x) = x^3 + ax^2 + 2b$ and $a + b = 4$, then
Solution
Answer: Option (B) is correct.
Explanation: Since $x – 1$ is a factor, $p(1) = 0$.
Substituting $x = 1$: $1 + a + 2b = 0 \Rightarrow a + 2b = -1$ …(2)
Given: $a + b = 4$ …(1)
Subtracting (1) from (2): $b = -5$
Substituting back: $a = 9$
Therefore $a = 9,\ b = -5$, which corresponds to Option (B).
Explanation: Since $x – 1$ is a factor, $p(1) = 0$.
Substituting $x = 1$: $1 + a + 2b = 0 \Rightarrow a + 2b = -1$ …(2)
Given: $a + b = 4$ …(1)
Subtracting (1) from (2): $b = -5$
Substituting back: $a = 9$
Therefore $a = 9,\ b = -5$, which corresponds to Option (B).
If the zeroes of the quadratic polynomial $x^2 + (a + 1)x + b$ are 2 and $-3$, then
Solution
Answer: Option (D) is correct.
Explanation: Let $\alpha = 2$ and $\beta = -3$.
Sum of zeroes: $\alpha + \beta = 2 + (-3) = -1$
Product of zeroes: $\alpha\beta = 2 \times (-3) = -6$
The polynomial is $x^2 – (\alpha + \beta)x + \alpha\beta = x^2 + x – 6$
Comparing with $x^2 + (a+1)x + b$:
$a + 1 = 1 \Rightarrow a = 0$ and $b = -6$
Therefore, $a = 0,\ b = -6$, which corresponds to Option (D).
Explanation: Let $\alpha = 2$ and $\beta = -3$.
Sum of zeroes: $\alpha + \beta = 2 + (-3) = -1$
Product of zeroes: $\alpha\beta = 2 \times (-3) = -6$
The polynomial is $x^2 – (\alpha + \beta)x + \alpha\beta = x^2 + x – 6$
Comparing with $x^2 + (a+1)x + b$:
$a + 1 = 1 \Rightarrow a = 0$ and $b = -6$
Therefore, $a = 0,\ b = -6$, which corresponds to Option (D).
Which of these is the polynomial whose zeroes are $\dfrac{1}{3}$ and $-\dfrac{3}{4}$?
Solution
Answer: Option (A) is correct.
Explanation:
Sum of zeroes $= \dfrac{1}{3} + \left(-\dfrac{3}{4}\right) = \dfrac{4}{12} – \dfrac{9}{12} = -\dfrac{5}{12}$
Product of zeroes $= \dfrac{1}{3} \times \left(-\dfrac{3}{4}\right) = -\dfrac{1}{4}$
The polynomial is $x^2 + \dfrac{5}{12}x – \dfrac{1}{4}$
Multiplying by 12: $12x^2 + 5x – 3$, which corresponds to Option (A).
Explanation:
Sum of zeroes $= \dfrac{1}{3} + \left(-\dfrac{3}{4}\right) = \dfrac{4}{12} – \dfrac{9}{12} = -\dfrac{5}{12}$
Product of zeroes $= \dfrac{1}{3} \times \left(-\dfrac{3}{4}\right) = -\dfrac{1}{4}$
The polynomial is $x^2 + \dfrac{5}{12}x – \dfrac{1}{4}$
Multiplying by 12: $12x^2 + 5x – 3$, which corresponds to Option (A).
The number of quadratic polynomials having zeroes $-5$ and $-3$ is
Solution
Answer: Option (D) is correct.
Explanation: The general form of a quadratic polynomial with zeroes $-5$ and $-3$ is:
$$k[x^2 – (-5 + (-3))x + (-5)(-3)] = k(x^2 + 8x + 15)$$
where $k$ can be any non-zero real number. Since $k$ can take infinitely many values, the number of such polynomials is more than 3, corresponding to Option (D).
Explanation: The general form of a quadratic polynomial with zeroes $-5$ and $-3$ is:
$$k[x^2 – (-5 + (-3))x + (-5)(-3)] = k(x^2 + 8x + 15)$$
where $k$ can be any non-zero real number. Since $k$ can take infinitely many values, the number of such polynomials is more than 3, corresponding to Option (D).
$p$ and $q$ are the zeroes of the polynomial $4y^2 – 4y + 1$. What is the value of $\dfrac{1}{p} + \dfrac{1}{q} + pq$?
Solution
Answer: Option (D) is correct.
Explanation: For $4y^2 – 4y + 1$:
• $p + q = \dfrac{4}{4} = 1$
• $pq = \dfrac{1}{4}$
Now, $\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{p + q}{pq} = \dfrac{1}{\frac{1}{4}} = 4$
Therefore: $\dfrac{1}{p} + \dfrac{1}{q} + pq = 4 + \dfrac{1}{4} = \dfrac{17}{4}$, corresponding to Option (D).
Explanation: For $4y^2 – 4y + 1$:
• $p + q = \dfrac{4}{4} = 1$
• $pq = \dfrac{1}{4}$
Now, $\dfrac{1}{p} + \dfrac{1}{q} = \dfrac{p + q}{pq} = \dfrac{1}{\frac{1}{4}} = 4$
Therefore: $\dfrac{1}{p} + \dfrac{1}{q} + pq = 4 + \dfrac{1}{4} = \dfrac{17}{4}$, corresponding to Option (D).
If one zero of the quadratic polynomial $x^2 + 3x + k$ is 2, then the value of $k$ is
Solution
Answer: Option (B) is correct.
Explanation: Since 2 is a zero, substitute $x = 2$ into $p(x) = x^2 + 3x + k$:
$(2)^2 + 3(2) + k = 0 \Rightarrow 4 + 6 + k = 0 \Rightarrow k = -10$
Thus, $k = -10$, which corresponds to Option (B).
Explanation: Since 2 is a zero, substitute $x = 2$ into $p(x) = x^2 + 3x + k$:
$(2)^2 + 3(2) + k = 0 \Rightarrow 4 + 6 + k = 0 \Rightarrow k = -10$
Thus, $k = -10$, which corresponds to Option (B).
The zeroes of the polynomial $x^2 – 3x – m(m + 3)$ are
Solution
Answer: Option (B) is correct.
Explanation:
Step 1: Put $x = -m$:
$(-m)^2 – 3(-m) – m(m+3) = m^2 + 3m – m^2 – 3m = 0$ ✓
Step 2: Put $x = m + 3$:
$(m+3)^2 – 3(m+3) – m(m+3) = (m+3)[m+3-3-m] = 0$ ✓
Therefore, the zeroes are $-m$ and $m+3$, corresponding to Option (B).
Explanation:
Step 1: Put $x = -m$:
$(-m)^2 – 3(-m) – m(m+3) = m^2 + 3m – m^2 – 3m = 0$ ✓
Step 2: Put $x = m + 3$:
$(m+3)^2 – 3(m+3) – m(m+3) = (m+3)[m+3-3-m] = 0$ ✓
Therefore, the zeroes are $-m$ and $m+3$, corresponding to Option (B).
The degree of the polynomial having zeroes $-3$ and $4$ only is
Solution
Answer: Option (A) is correct.
Explanation: A polynomial of degree $n$ has at most $n$ zeroes. Since this polynomial has exactly two distinct zeroes ($-3$ and $4$), its degree is 2, corresponding to Option (A).
Explanation: A polynomial of degree $n$ has at most $n$ zeroes. Since this polynomial has exactly two distinct zeroes ($-3$ and $4$), its degree is 2, corresponding to Option (A).
If $\alpha$ and $\beta$ are the zeroes of a polynomial $f(x) = px^2 – 2x + 3p$ and $\alpha + \beta = \alpha\beta$, then $p$ is
Solution
Answer: Option (B) is correct.
Explanation: For $f(x) = px^2 – 2x + 3p$:
• $\alpha + \beta = \dfrac{2}{p}$
• $\alpha\beta = \dfrac{3p}{p} = 3$
Given $\alpha + \beta = \alpha\beta$:
$$\frac{2}{p} = 3 \Rightarrow p = \frac{2}{3}$$ This corresponds to Option (B).
Explanation: For $f(x) = px^2 – 2x + 3p$:
• $\alpha + \beta = \dfrac{2}{p}$
• $\alpha\beta = \dfrac{3p}{p} = 3$
Given $\alpha + \beta = \alpha\beta$:
$$\frac{2}{p} = 3 \Rightarrow p = \frac{2}{3}$$ This corresponds to Option (B).
Assertion (A): The values of $k$ for which the quadratic polynomial $kx^2 + x + k$ has equal zeroes are $\pm \dfrac{1}{2}$.
Reason (R): If all the three zeroes of a cubic polynomial $x^3 + ax^2 – bx + c$ are positive, then at least one of $a$, $b$, and $c$ is non-negative.
(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
Reason (R): If all the three zeroes of a cubic polynomial $x^3 + ax^2 – bx + c$ are positive, then at least one of $a$, $b$, and $c$ is non-negative.
(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
Answer
Answer: Option (C) is correct.
For Assertion (A):
For $f(x) = kx^2 + x + k$, equal zeroes require discriminant $= 0$:
$$(1)^2 – 4(k)(k) = 0 \Rightarrow 1 – 4k^2 = 0 \Rightarrow k = \pm\frac{1}{2}$$ Hence Assertion (A) is true.
For Reason (R):
The statement is false. For example, $x^3 – 3x^2 + 3x – 1$ has all positive zeroes, yet $a = -3$ (negative). So the reason does not hold universally.
Therefore, A is true but R is false — Option (C).
For Assertion (A):
For $f(x) = kx^2 + x + k$, equal zeroes require discriminant $= 0$:
$$(1)^2 – 4(k)(k) = 0 \Rightarrow 1 – 4k^2 = 0 \Rightarrow k = \pm\frac{1}{2}$$ Hence Assertion (A) is true.
For Reason (R):
The statement is false. For example, $x^3 – 3x^2 + 3x – 1$ has all positive zeroes, yet $a = -3$ (negative). So the reason does not hold universally.
Therefore, A is true but R is false — Option (C).
Form a quadratic polynomial, the sum and product of whose zeroes are $-3$ and $2$, respectively.
Answer
We use the general form: $f(x) = x^2 – (\alpha + \beta)x + \alpha\beta$
Given: $\alpha + \beta = -3$ and $\alpha\beta = 2$
$$f(x) = x^2 – (-3)x + 2 = x^2 + 3x + 2$$
The required quadratic polynomial is $f(x) = x^2 + 3x + 2$.
Given: $\alpha + \beta = -3$ and $\alpha\beta = 2$
$$f(x) = x^2 – (-3)x + 2 = x^2 + 3x + 2$$
The required quadratic polynomial is $f(x) = x^2 + 3x + 2$.
Find a quadratic polynomial where the zeroes are $5\dfrac{3}{2}$ and $5\dfrac{3}{2} + 2$.
Answer
Let $\alpha = 5\dfrac{3}{2}$ and $\beta = 5\dfrac{3}{2} + 2$.
Sum of zeroes: $\alpha + \beta = 10\dfrac{1}{2} = \dfrac{21}{2}$
Product of zeroes: $\alpha\beta = 5\dfrac{3}{2} \times \left(5\dfrac{3}{2} + 2\right) = 25 – 18 = 7$
The required quadratic polynomial is: $$p(x) = x^2 – \frac{21}{2}x + 7$$
Sum of zeroes: $\alpha + \beta = 10\dfrac{1}{2} = \dfrac{21}{2}$
Product of zeroes: $\alpha\beta = 5\dfrac{3}{2} \times \left(5\dfrac{3}{2} + 2\right) = 25 – 18 = 7$
The required quadratic polynomial is: $$p(x) = x^2 – \frac{21}{2}x + 7$$
Find the value of $p$ for which one root of the quadratic polynomial $px^2 – 14x + 8 = 0$ is 6 times the other.
Answer
Let the roots be $\alpha$ and $6\alpha$.
By Vieta’s formulas:
• Sum: $7\alpha = \dfrac{14}{p} \Rightarrow \alpha = \dfrac{2}{p}$
• Product: $6\alpha^2 = \dfrac{8}{p}$
Substituting $\alpha = \dfrac{2}{p}$: $$6 \times \frac{4}{p^2} = \frac{8}{p} \Rightarrow \frac{24}{p^2} = \frac{8}{p} \Rightarrow p = 3$$ The value of $p$ is $\boxed{3}$.
By Vieta’s formulas:
• Sum: $7\alpha = \dfrac{14}{p} \Rightarrow \alpha = \dfrac{2}{p}$
• Product: $6\alpha^2 = \dfrac{8}{p}$
Substituting $\alpha = \dfrac{2}{p}$: $$6 \times \frac{4}{p^2} = \frac{8}{p} \Rightarrow \frac{24}{p^2} = \frac{8}{p} \Rightarrow p = 3$$ The value of $p$ is $\boxed{3}$.
If $a$ and $b$ are the zeroes of the polynomial $x^2 – \dfrac{4}{3}x + 3$, then find the value of $a + b – ab$.
Answer
From the polynomial: $a + b = \dfrac{4}{3}$ and $ab = 3$
$$a + b – ab = \frac{4}{3} – 3 = \frac{4}{3} – \frac{9}{3} = \boxed{-\frac{5}{3}}$$
$$a + b – ab = \frac{4}{3} – 3 = \frac{4}{3} – \frac{9}{3} = \boxed{-\frac{5}{3}}$$
If $a$ and $b$ are the zeroes of $x^2 – x – 2$, form a quadratic polynomial whose zeroes are $2a + 1$ and $2b + 1$.
Answer
From $x^2 – x – 2$: $a + b = 1$ and $ab = -2$
Sum of new zeroes:
$(2a + 1) + (2b + 1) = 2(a + b) + 2 = 2(1) + 2 = 4$
Product of new zeroes:
$(2a + 1)(2b + 1) = 4ab + 2(a + b) + 1 = 4(-2) + 2(1) + 1 = -5$
Required polynomial: $k(x^2 – 4x – 5)$
Sum of new zeroes:
$(2a + 1) + (2b + 1) = 2(a + b) + 2 = 2(1) + 2 = 4$
Product of new zeroes:
$(2a + 1)(2b + 1) = 4ab + 2(a + b) + 1 = 4(-2) + 2(1) + 1 = -5$
Required polynomial: $k(x^2 – 4x – 5)$
If one root of the quadratic equation $3x^2 + px + 4 = 0$ is $\dfrac{2}{3}$, then find the value of $p$ and the other root of the equation.
Answer
Substitute $x = \dfrac{2}{3}$:
$$3\left(\frac{4}{9}\right) + p\left(\frac{2}{3}\right) + 4 = 0 \Rightarrow \frac{4}{3} + \frac{2p}{3} + 4 = 0$$
Multiplying by 3: $4 + 2p + 12 = 0 \Rightarrow p = -8$
Substituting $p = -8$: $3x^2 – 8x + 4 = 0 \Rightarrow (x – 2)(3x – 2) = 0$
The value of $p$ is $-8$, and the other root is $x = 2$.
Substituting $p = -8$: $3x^2 – 8x + 4 = 0 \Rightarrow (x – 2)(3x – 2) = 0$
The value of $p$ is $-8$, and the other root is $x = 2$.
The roots $a$ and $b$ of the quadratic equation $x^2 – 5x + 3(k – 1) = 0$ are such that $a – b = 1$. Find the value of $k$.
Answer
From Vieta’s formulas: $a + b = 5$ and $ab = 3(k – 1)$
Using $a – b = \sqrt{(a+b)^2 – 4ab}$: $$1 = \sqrt{25 – 12(k-1)} = \sqrt{37 – 12k}$$ Squaring: $1 = 37 – 12k \Rightarrow 12k = 36 \Rightarrow k = 3$
The value of $k$ is $\boxed{3}$.
Using $a – b = \sqrt{(a+b)^2 – 4ab}$: $$1 = \sqrt{25 – 12(k-1)} = \sqrt{37 – 12k}$$ Squaring: $1 = 37 – 12k \Rightarrow 12k = 36 \Rightarrow k = 3$
The value of $k$ is $\boxed{3}$.
If $a$ and $b$ are the zeroes of the quadratic polynomial $5x^2 + 5x + 1$, find the value of: (1) $a^2 + b^2$ (2) $a^{-1} + b^{-1}$
Answer
From $5x^2 + 5x + 1$: $a + b = -1$ and $ab = \dfrac{1}{5}$
1. Finding $a^2 + b^2$: $$a^2 + b^2 = (a+b)^2 – 2ab = 1 – \frac{2}{5} = \frac{3}{5}$$ 2. Finding $a^{-1} + b^{-1}$: $$a^{-1} + b^{-1} = \frac{a+b}{ab} = \frac{-1}{\frac{1}{5}} = -5$$ Final Answers: $a^2 + b^2 = \dfrac{3}{5}$, $\ a^{-1} + b^{-1} = -5$
1. Finding $a^2 + b^2$: $$a^2 + b^2 = (a+b)^2 – 2ab = 1 – \frac{2}{5} = \frac{3}{5}$$ 2. Finding $a^{-1} + b^{-1}$: $$a^{-1} + b^{-1} = \frac{a+b}{ab} = \frac{-1}{\frac{1}{5}} = -5$$ Final Answers: $a^2 + b^2 = \dfrac{3}{5}$, $\ a^{-1} + b^{-1} = -5$
Find the value of $k$ such that the polynomial $x^2 – (k + 6)x + 2(2k – 1)$ has the sum of its zeroes equal to half of their product.
Answer
Sum of zeroes $= k + 6$ and Product of zeroes $= 2(2k – 1)$
Given: Sum $= \dfrac{1}{2} \times$ Product: $$k + 6 = \frac{1}{2} \times 2(2k-1) = 2k – 1$$ $$-k = -7 \Rightarrow k = 7$$ The value of $k$ is $\boxed{7}$.
Given: Sum $= \dfrac{1}{2} \times$ Product: $$k + 6 = \frac{1}{2} \times 2(2k-1) = 2k – 1$$ $$-k = -7 \Rightarrow k = 7$$ The value of $k$ is $\boxed{7}$.
Find the zeroes of the quadratic polynomial $7y^2 – 11y – \dfrac{3}{2}$ and verify the relationship between the zeroes and the coefficients.
Answer
Working with $\dfrac{1}{3}(21y^2 – 11y – 2)$:
$$= \frac{1}{3}(3y – 2)(7y + 1)$$ Zeroes: $y = \dfrac{2}{3}$ and $y = -\dfrac{1}{7}$
Verification (using $a=21,\ b=-11,\ c=-2$):
Sum: $\dfrac{2}{3} + \left(-\dfrac{1}{7}\right) = \dfrac{11}{21} = -\dfrac{b}{a}$ ✓
Product: $\dfrac{2}{3} \times \left(-\dfrac{1}{7}\right) = -\dfrac{2}{21} = \dfrac{c}{a}$ ✓
The zeroes are $y = \dfrac{2}{3}$ and $y = -\dfrac{1}{7}$, and the relationship is verified.
$$= \frac{1}{3}(3y – 2)(7y + 1)$$ Zeroes: $y = \dfrac{2}{3}$ and $y = -\dfrac{1}{7}$
Verification (using $a=21,\ b=-11,\ c=-2$):
Sum: $\dfrac{2}{3} + \left(-\dfrac{1}{7}\right) = \dfrac{11}{21} = -\dfrac{b}{a}$ ✓
Product: $\dfrac{2}{3} \times \left(-\dfrac{1}{7}\right) = -\dfrac{2}{21} = \dfrac{c}{a}$ ✓
The zeroes are $y = \dfrac{2}{3}$ and $y = -\dfrac{1}{7}$, and the relationship is verified.
If the zeroes of the polynomial $x^2 + px + q$ are double in value to the zeroes of the polynomial $2x^2 – 5x – 3$, then find the values of $p$ and $q$.
Answer
Let zeroes of $2x^2 – 5x – 3$ be $\alpha$ and $\beta$: $\alpha + \beta = \dfrac{5}{2}$, $\alpha\beta = -\dfrac{3}{2}$
Zeroes of $x^2 + px + q$ are $2\alpha$ and $2\beta$:
Sum: $2\alpha + 2\beta = 5 = -p \Rightarrow p = -5$
Product: $4\alpha\beta = 4 \times \left(-\dfrac{3}{2}\right) = -6 = q$
Therefore, $p = -5$ and $q = -6$.
Zeroes of $x^2 + px + q$ are $2\alpha$ and $2\beta$:
Sum: $2\alpha + 2\beta = 5 = -p \Rightarrow p = -5$
Product: $4\alpha\beta = 4 \times \left(-\dfrac{3}{2}\right) = -6 = q$
Therefore, $p = -5$ and $q = -6$.
If $a$ and $b$ are the zeroes of the polynomial $p(x) = 2x^2 + 5x + k$ satisfying the relation $a^2 + b^2 + ab = \dfrac{21}{4}$, then find the value of $k$.
Answer
From $2x^2 + 5x + k$: $a + b = -\dfrac{5}{2}$ and $ab = \dfrac{k}{2}$
Using $a^2 + b^2 = (a+b)^2 – 2ab$: $$(a+b)^2 – 2ab + ab = \frac{21}{4} \Rightarrow (a+b)^2 – ab = \frac{21}{4}$$ $$\frac{25}{4} – \frac{k}{2} = \frac{21}{4} \Rightarrow \frac{k}{2} = 1 \Rightarrow k = 2$$ The value of $k$ is $\boxed{2}$.
Using $a^2 + b^2 = (a+b)^2 – 2ab$: $$(a+b)^2 – 2ab + ab = \frac{21}{4} \Rightarrow (a+b)^2 – ab = \frac{21}{4}$$ $$\frac{25}{4} – \frac{k}{2} = \frac{21}{4} \Rightarrow \frac{k}{2} = 1 \Rightarrow k = 2$$ The value of $k$ is $\boxed{2}$.
Case Study — Read the following and answer any four questions.
The pictures below show natural examples of parabolic shapes, represented by quadratic polynomials. A parabolic arch is an arch in the shape of a parabola. In structures, the curve of a parabolic arch represents an efficient method of load distribution, which is why it can be found in bridges and architecture.
1. In the standard form of quadratic polynomial $ax^2 + bx + c$, $a$, $b$, and $c$ are:
(A) All are real numbers. (B) All are rational numbers.
(C) ‘$a$’ is a non-zero real number and $b$ and $c$ are any real numbers. (D) All are integers.
2. If $\alpha$ and $\dfrac{1}{\alpha}$ are the zeroes of $2x^2 – x + 8k$, then $k$ is:
(A) 4 (B) $\dfrac{1}{4}$ (C) $-1$ (D) 2
3. The graph of $x^2 + 1 = 0$:
(A) Intersects X-axis at two distinct points. (B) Touches X-axis at a point.
(C) Neither touches nor intersects X-axis. (D) Either touches or intersects X-axis.
4. If the sum of the roots is $-p$ and the product of the roots is $-\dfrac{1}{p}$, then the quadratic polynomial is:
(A) $k\!\left(-px^2 + \dfrac{x}{p} + 1\right)$ (B) $k\!\left(px^2 – \dfrac{x}{p} – 1\right)$
(C) $k\!\left(x^2 + px – \dfrac{1}{p}\right)$ (D) $k\!\left(x^2 + px + \dfrac{1}{p}\right)$
The pictures below show natural examples of parabolic shapes, represented by quadratic polynomials. A parabolic arch is an arch in the shape of a parabola. In structures, the curve of a parabolic arch represents an efficient method of load distribution, which is why it can be found in bridges and architecture.
1. In the standard form of quadratic polynomial $ax^2 + bx + c$, $a$, $b$, and $c$ are:
(A) All are real numbers. (B) All are rational numbers.
(C) ‘$a$’ is a non-zero real number and $b$ and $c$ are any real numbers. (D) All are integers.
2. If $\alpha$ and $\dfrac{1}{\alpha}$ are the zeroes of $2x^2 – x + 8k$, then $k$ is:
(A) 4 (B) $\dfrac{1}{4}$ (C) $-1$ (D) 2
3. The graph of $x^2 + 1 = 0$:
(A) Intersects X-axis at two distinct points. (B) Touches X-axis at a point.
(C) Neither touches nor intersects X-axis. (D) Either touches or intersects X-axis.
4. If the sum of the roots is $-p$ and the product of the roots is $-\dfrac{1}{p}$, then the quadratic polynomial is:
(A) $k\!\left(-px^2 + \dfrac{x}{p} + 1\right)$ (B) $k\!\left(px^2 – \dfrac{x}{p} – 1\right)$
(C) $k\!\left(x^2 + px – \dfrac{1}{p}\right)$ (D) $k\!\left(x^2 + px + \dfrac{1}{p}\right)$
Answer
1. Ans. Option (C) is correct.
In the standard form $ax^2 + bx + c$, ‘$a$’ is a non-zero real number and $b$, $c$ are any real numbers.
2. Ans. Option (B) is correct.
Product of zeroes $= \alpha \times \dfrac{1}{\alpha} = 1$. Also product $= \dfrac{8k}{2} = 4k$. So $4k = 1 \Rightarrow k = \dfrac{1}{4}$.
3. Ans. Option (C) is correct.
Since $x^2 + 1 = 0$ has no real roots (discriminant $< 0$), the graph neither touches nor intersects the X-axis.
4. Ans. Option (C) is correct.
$$P(x) = k\left[x^2 – (-p)x + \left(-\frac{1}{p}\right)\right] = k\left(x^2 + px – \frac{1}{p}\right)$$
In the standard form $ax^2 + bx + c$, ‘$a$’ is a non-zero real number and $b$, $c$ are any real numbers.
2. Ans. Option (B) is correct.
Product of zeroes $= \alpha \times \dfrac{1}{\alpha} = 1$. Also product $= \dfrac{8k}{2} = 4k$. So $4k = 1 \Rightarrow k = \dfrac{1}{4}$.
3. Ans. Option (C) is correct.
Since $x^2 + 1 = 0$ has no real roots (discriminant $< 0$), the graph neither touches nor intersects the X-axis.
4. Ans. Option (C) is correct.
$$P(x) = k\left[x^2 – (-p)x + \left(-\frac{1}{p}\right)\right] = k\left(x^2 + px – \frac{1}{p}\right)$$
If the roots of the quadratic polynomial are equal, where the discriminant $D = b^2 – 4ac$, then:
Solution
Answer: Option (D) is correct.
Explanation: If the roots of the quadratic polynomial are equal, the discriminant is zero: $D = b^2 – 4ac = 0$.
Explanation: If the roots of the quadratic polynomial are equal, the discriminant is zero: $D = b^2 – 4ac = 0$.
Find a quadratic polynomial whose sum and product of zeroes are 0 and $-9$, respectively. Also, find the zeroes of the polynomial so obtained.
Answer
The polynomial: $p(x) = x^2 – 0 \cdot x + (-9) = x^2 – 9$
Factoring: $x^2 – 9 = (x+3)(x-3)$
The zeroes of the polynomial are $x = -3$ and $x = 3$.
Factoring: $x^2 – 9 = (x+3)(x-3)$
The zeroes of the polynomial are $x = -3$ and $x = 3$.
If $\alpha$ and $\beta$ are the zeroes of the polynomial $3x^2 + 6x + k$ such that $\alpha + \beta + \alpha\beta = -\dfrac{2}{3}$, then the value of $k$ is:
Solution
Answer: Option (D) is correct.
Explanation: For $3x^2 + 6x + k$: $\alpha + \beta = -2$ and $\alpha\beta = \dfrac{k}{3}$
Substituting: $-2 + \dfrac{k}{3} = -\dfrac{2}{3} \Rightarrow \dfrac{k}{3} = \dfrac{4}{3} \Rightarrow k = 4$
This corresponds to Option (D).
Explanation: For $3x^2 + 6x + k$: $\alpha + \beta = -2$ and $\alpha\beta = \dfrac{k}{3}$
Substituting: $-2 + \dfrac{k}{3} = -\dfrac{2}{3} \Rightarrow \dfrac{k}{3} = \dfrac{4}{3} \Rightarrow k = 4$
This corresponds to Option (D).
Find the zeroes of the polynomial $p(x) = x^2 + \dfrac{4}{3}x – \dfrac{4}{3}$.
Answer
Working with $\dfrac{1}{3}(3x^2 + 4x – 4)$:
$$= \frac{1}{3}(3x^2 + 6x – 2x – 4) = \frac{1}{3}(3x – 2)(x + 2)$$
Setting to zero: $(3x – 2)(x + 2) = 0$
The zeroes are $x = \dfrac{2}{3}$ and $x = -2$.
The zeroes are $x = \dfrac{2}{3}$ and $x = -2$.
If one of the zeroes of the quadratic polynomial $(a – 1)x^2 + ax + 1$ is $-3$, then the value of $a$ is:
Solution
Answer: Option (C) is correct.
Explanation: Substitute $x = -3$: $$(a-1)(9) + a(-3) + 1 = 0 \Rightarrow 9a – 9 – 3a + 1 = 0 \Rightarrow 6a = 8 \Rightarrow a = \frac{4}{3}$$ This corresponds to Option (C).
Explanation: Substitute $x = -3$: $$(a-1)(9) + a(-3) + 1 = 0 \Rightarrow 9a – 9 – 3a + 1 = 0 \Rightarrow 6a = 8 \Rightarrow a = \frac{4}{3}$$ This corresponds to Option (C).
For what value of $k$, the product of the zeroes of the polynomial $kx^2 – 4x – 7$ is 2?
Solution
Answer: Option (B) is correct.
Explanation: Product of zeroes $= \dfrac{-7}{k} = 2 \Rightarrow k = -\dfrac{7}{2}$
This corresponds to Option (B).
Explanation: Product of zeroes $= \dfrac{-7}{k} = 2 \Rightarrow k = -\dfrac{7}{2}$
This corresponds to Option (B).
Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial $f(x) = ax^2 + bx + c$, where $a \neq 0$, $c \neq 0$.
Answer
Let $\alpha,\ \beta$ be the zeroes of $ax^2 + bx + c$: $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$
For the new polynomial with zeroes $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$:
Sum: $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = -\dfrac{b}{c}$
Product: $\dfrac{1}{\alpha\beta} = \dfrac{a}{c}$
Required polynomial: $p(x) = \dfrac{k}{c}(cx^2 + bx + a)$
For $k = c$: $p(x) = cx^2 + bx + a$
For the new polynomial with zeroes $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$:
Sum: $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = -\dfrac{b}{c}$
Product: $\dfrac{1}{\alpha\beta} = \dfrac{a}{c}$
Required polynomial: $p(x) = \dfrac{k}{c}(cx^2 + bx + a)$
For $k = c$: $p(x) = cx^2 + bx + a$
Find the value of $k$ such that the polynomial $x^2 – (k + 6)x + 2(2k – 1)$ has the sum of its zeroes equal to half of their product.
Answer
Sum of zeroes $= k + 6$ and Product of zeroes $= 2(2k – 1)$
According to the condition: $$k + 6 = \frac{1}{2} \times 2(2k – 1) \Rightarrow k + 6 = 2k – 1 \Rightarrow k = 7$$ The value of $k$ is $7$.
According to the condition: $$k + 6 = \frac{1}{2} \times 2(2k – 1) \Rightarrow k + 6 = 2k – 1 \Rightarrow k = 7$$ The value of $k$ is $7$.
If $\alpha$ and $\beta$ are zeroes of $4x^2 – x – 4$, find the quadratic polynomial whose zeroes are $\dfrac{1}{2\alpha}$ and $\dfrac{1}{2\beta}$.
Answer
From $4x^2 – x – 4$: $\alpha + \beta = \dfrac{1}{4}$ and $\alpha\beta = -1$
Sum: $\dfrac{1}{2\alpha} + \dfrac{1}{2\beta} = \dfrac{\alpha+\beta}{2\alpha\beta} = \dfrac{\frac{1}{4}}{2(-1)} = -\dfrac{1}{8}$
Product: $\dfrac{1}{2\alpha} \cdot \dfrac{1}{2\beta} = \dfrac{1}{4\alpha\beta} = \dfrac{1}{4(-1)} = -\dfrac{1}{4}$
Required polynomial: $k\left(x^2 + \dfrac{1}{8}x – \dfrac{1}{4}\right)$
For $k = 8$: $p(x) = 8x^2 + x – 2$
Sum: $\dfrac{1}{2\alpha} + \dfrac{1}{2\beta} = \dfrac{\alpha+\beta}{2\alpha\beta} = \dfrac{\frac{1}{4}}{2(-1)} = -\dfrac{1}{8}$
Product: $\dfrac{1}{2\alpha} \cdot \dfrac{1}{2\beta} = \dfrac{1}{4\alpha\beta} = \dfrac{1}{4(-1)} = -\dfrac{1}{4}$
Required polynomial: $k\left(x^2 + \dfrac{1}{8}x – \dfrac{1}{4}\right)$
For $k = 8$: $p(x) = 8x^2 + x – 2$
If $\alpha$ and $\beta$ are the zeroes of $p(x) = 6x^2 – 7x + 2$, find the quadratic polynomial whose zeroes are $\dfrac{1}{\alpha}$ and $\dfrac{1}{\beta}$.
Answer
From $6x^2 – 7x + 2$: $\alpha + \beta = \dfrac{7}{6}$ and $\alpha\beta = \dfrac{1}{3}$
Sum: $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{\frac{7}{6}}{\frac{1}{3}} = \dfrac{7}{2}$
Product: $\dfrac{1}{\alpha\beta} = 3$
Required polynomial: $k\left(x^2 – \dfrac{7}{2}x + 3\right)$
For $k = 2$: $p(x) = 2x^2 – 7x + 6$
Sum: $\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha+\beta}{\alpha\beta} = \dfrac{\frac{7}{6}}{\frac{1}{3}} = \dfrac{7}{2}$
Product: $\dfrac{1}{\alpha\beta} = 3$
Required polynomial: $k\left(x^2 – \dfrac{7}{2}x + 3\right)$
For $k = 2$: $p(x) = 2x^2 – 7x + 6$
Find the zeroes of the quadratic polynomial $6x^2 – 3x – 7x$ and verify the relationship between the zeroes and the coefficients of the polynomial.
Answer
Let $f(x) = 6x^2 – 3x – 7x = 6x^2 – 7x – 3$ (combining like terms)
Factoring: $6x^2 – 9x + 2x – 3 = 3x(2x-3) + 1(2x-3) = (3x+1)(2x-3)$
Zeroes: $x = -\dfrac{1}{3}$ or $x = \dfrac{3}{2}$
Verification:
Sum: $-\dfrac{1}{3} + \dfrac{3}{2} = \dfrac{-2+9}{6} = \dfrac{7}{6} = -\dfrac{\text{coeff. of }x}{\text{coeff. of }x^2}$ ✓
Product: $\left(-\dfrac{1}{3}\right)\left(\dfrac{3}{2}\right) = -\dfrac{1}{2} = \dfrac{\text{constant}}{\text{coeff. of }x^2} = \dfrac{-3}{6}$ ✓
The zeroes are $-\dfrac{1}{3}$ and $\dfrac{3}{2}$, and the relationship is verified.
Factoring: $6x^2 – 9x + 2x – 3 = 3x(2x-3) + 1(2x-3) = (3x+1)(2x-3)$
Zeroes: $x = -\dfrac{1}{3}$ or $x = \dfrac{3}{2}$
Verification:
Sum: $-\dfrac{1}{3} + \dfrac{3}{2} = \dfrac{-2+9}{6} = \dfrac{7}{6} = -\dfrac{\text{coeff. of }x}{\text{coeff. of }x^2}$ ✓
Product: $\left(-\dfrac{1}{3}\right)\left(\dfrac{3}{2}\right) = -\dfrac{1}{2} = \dfrac{\text{constant}}{\text{coeff. of }x^2} = \dfrac{-3}{6}$ ✓
The zeroes are $-\dfrac{1}{3}$ and $\dfrac{3}{2}$, and the relationship is verified.
Frequently Asked Questions
What does the Polynomials chapter cover in CBSE Class 10 Maths? ⌄
The Polynomials chapter in CBSE Class 10 covers the geometrical meaning of zeroes of a polynomial, the relationship between zeroes and coefficients (Vieta’s formulas) for quadratic and cubic polynomials, and the division algorithm for polynomials. Students learn to find zeroes, form new polynomials, and verify coefficient relationships — all of which are regularly tested in board exams.
How many marks does the Polynomials chapter carry in the CBSE Class 10 board exam? ⌄
Polynomials is part of the Algebra unit, which carries approximately 20 marks in the CBSE Class 10 Mathematics board paper. Questions from this chapter typically appear as 1-mark MCQs, 2-mark short answers, and occasionally as 3-mark or case-study questions, making it a chapter that contributes across multiple question types.
What are the most important topics students should focus on in Class 10 Polynomials? ⌄
The most important topics are: finding zeroes of a quadratic polynomial using factorisation, applying Vieta’s formulas (sum and product of zeroes), forming a quadratic polynomial when zeroes are given, and understanding the graphical interpretation of zeroes. Questions on finding $k$ when a zero or a condition is given are especially frequent in previous year papers.
What common mistakes do students make when solving Polynomials questions? ⌄
A very common error is confusing the sign in Vieta’s formulas — students often write the sum of zeroes as $+\dfrac{b}{a}$ instead of $-\dfrac{b}{a}$. Another frequent mistake is incorrectly forming the polynomial after finding new zeroes, especially when expressions like $2\alpha + 1$ are involved. Practising a variety of previous year questions helps your child avoid these errors under exam pressure.
How does Angle Belearn help students score well in Polynomials? ⌄
Angle Belearn’s CBSE specialists carefully curate chapter-wise question banks drawn from real board papers, each paired with clear, step-by-step solutions. Students practising on Angle Belearn develop the habit of showing structured working — something that earns full marks in board exams. Regular practice with these verified questions builds both speed and accuracy so your child walks into the exam confident.
