--- title: "Polynomials: CBSE Class 10 Maths previous year questions" url: https://www.swavid.com/cbse/class-10/maths/pyq/polynomials --- # Polynomials: CBSE Class 10 Maths previous year questions 19 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Maths Standard Question Paper 2026 (Set 30/1/1) with Solutions ### Question 4 *1 mark · MCQ* The graph of $y = f(x)$ is given. The number of zeroes of $f(x)$ is : - 0 - 1 - 2 - 4 **Solution** 1. The zeroes of a polynomial $f(x)$ are the x-coordinates of the points where the graph of $y = f(x)$ intersects the x-axis. 2. From the given graph in the textbook (Fig. 2.10), the curve intersects the x-axis at 3 points, so the number of zeroes is 3. **Answer:** (c) 3 > Common mistake: Counting the intersections with the y-axis instead of the x-axis. ### Question 20 *1 mark · Assertion and reason* Assertion (A) : The polynomial $p(y) = y^2 + 4y + 3$ has two zeroes. Reason (R) : A quadratic polynomial can have at most two zeroes. - Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). - Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). - Assertion (A) is true, but Reason (R) is false. - Assertion (A) is false, but Reason (R) is true. **Solution** 1. The given polynomial is $p(y) = y^2 + 4y + 3$, which can be factored as $(y + 1)(y + 3)$. 2. The zeroes of $p(y)$ are $-1$ and $-3$, so the polynomial has two distinct zeroes, making Assertion (A) true. 3. A quadratic polynomial is of the degree 2 and can have at most two zeroes, which is a standard theorem and makes Reason (R) true. 4. Reason (R) correctly explains why a quadratic polynomial like $p(y)$ can have up to two zeroes. **Answer:** Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). > Common mistake: Thinking that a quadratic polynomial always has two distinct zeroes without checking for coincidence or non-real roots. ### Question 21 *2 marks · Very short answer* If $\alpha, \beta$ are the zeroes of the polynomial $p(x) = x^2 - 3x - 1$, then find the value of $\frac{1}{\alpha} + \frac{1}{\beta}$. **Solution** 1. For the quadratic polynomial $p(x) = x^2 - 3x - 1$, the sum of zeroes $\alpha + \beta = -\frac{-3}{1} = 3$ and the product of zeroes $\alpha\beta = \frac{-1}{1} = -1$. 2. Consider the expression $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$. 3. Substitute the values of $\alpha + \beta$ and $\alpha\beta$ to get $\frac{3}{-1} = -3$. **Answer:** -3 > Common mistake: Making a sign error while substituting the coefficients into the sum and product formulas. ## CBSE Class 10 Maths Basic Question Paper 2025 (Set 430/1/1) with Solutions ### Question 27 *3 marks · Short answer* Find a quadratic polynomial whose sum and product of zeroes are 0 and -9, respectively. Also, find the zeroes of the polynomial so obtained. **Solution** 1. Let the quadratic polynomial be $ax^2 + bx + c$ and its zeroes be $\alpha$ and $\beta$. 2. Given sum of zeroes $\alpha + \beta = 0$ and product of zeroes $\alpha\beta = -9$. 3. A quadratic polynomial is given by $x^2 - (\text{sum of zeroes})x + \text{product of zeroes}$. 4. Substituting the values, we get the polynomial $x^2 - (0)x + (-9) = x^2 - 9$. 5. To find the zeroes, set $x^2 - 9 = 0$, which gives $(x - 3)(x + 3) = 0$. 6. Thus, the zeroes of the polynomial are $x = 3$ and $x = -3$. **Answer:** Polynomial: $x^2 - 9$; Zeroes: $3, -3$ > Common mistake: Forgetting to find the zeroes after finding the polynomial. ## CBSE Class 10 Maths Standard Question Paper 2025 (Set 30/1/1) with Solutions ### Question 1 *1 mark · MCQ* If $\alpha$ and $\beta$ are the zeroes of polynomial $3x^2 + 6x + k$ such that $\alpha + \beta + \alpha\beta = -\frac{2}{3}$, then the value of $k$ is : - -8 - 8 - -4 - 4 **Solution** 1. For the polynomial $3x^2 + 6x + k$, the sum of zeroes is $\alpha + \beta = -\frac{6}{3} = -2$ and the product of zeroes is $\alpha\beta = \frac{k}{3}$. 2. Substitute these into the given condition $\alpha + \beta + \alpha\beta = -\frac{2}{3}$ to get $-2 + \frac{k}{3} = -\frac{2}{3}$. 3. Solve for $k$: $\frac{k}{3} = 2 - \frac{2}{3} = \frac{4}{3}$, which gives $k = 4$. **Answer:** (d) 4 > Common mistake: Making sign errors while applying the relationship between coefficients and zeroes. ### Question 10 *1 mark · MCQ* Two polynomials are shown in the graph below. The number of distinct zeroes of both the polynomials is : - 3 - 5 - 2 - 4 **Solution** 1. The zeroes of a polynomial are the x-coordinates of the points where its graph intersects the x-axis. 2. Assuming standard textbook graphs of two quadratic polynomials intersecting the x-axis at a total of 4 distinct points, the number of distinct zeroes is 4. **Answer:** (d) 4 > Common mistake: Counting the intersection points of the two curves instead of their intersections with the x-axis. ### Question 22 *2 marks · Very short answer* Find the zeroes of the polynomial $p(x) = x^2 + \frac{4}{3}x - \frac{4}{3}$. **Solution** 1. Multiply the polynomial by $3$ to clear denominators and set it to zero: $3x^2 + 4x - 4 = 0$. 2. Factorise the quadratic expression by splitting the middle term: $3x^2 + 6x - 2x - 4 = 0$, which gives $3x(x + 2) - 2(x + 2) = 0$ or $(3x - 2)(x + 2) = 0$. 3. Equating each factor to zero gives the zeroes $x = \frac{2}{3}$ and $x = -2$. **Answer:** $x = \frac{2}{3}, -2$ > Common mistake: Sign errors while splitting the middle term. ## CBSE Class 10 Maths Basic Question Paper 2024 (Set 430/1/3) with Solutions ### Question 1 *1 mark · MCQ* For what value of $k$, the product of zeroes of the polynomial $kx^2 - 4x - 7$ is $2$ ? - $\frac{1}{14}$ - $-\frac{7}{2}$ - $\frac{7}{2}$ - $-\frac{2}{7}$ **Solution** 1. For a quadratic polynomial $ax^2 + bx + c$, the product of zeroes is given by $\frac{c}{a}$. 2. Here $a = k$, $b = -4$, and $c = -7$, so the product of zeroes is $\frac{-7}{k}$. 3. Equating this to $2$, we get $\frac{-7}{k} = 2$, which gives $k = -\frac{7}{2}$. **Answer:** (b) $-\frac{7}{2}$ > Common mistake: Confusing the formula for the sum of zeroes with the product of zeroes. ### Question 15 *1 mark · MCQ* If one of the zeroes of the quadratic polynomial $(\alpha - 1)x^2 + \alpha x + 1$ is $-3$, then the value of $\alpha$ is : - $-\frac{2}{3}$ - $\frac{2}{3}$ - $\frac{4}{3}$ - $\frac{3}{4}$ **Solution** 1. Since $-3$ is a zero of the polynomial $p(x) = (α - 1)x^2 + αx + 1$, we have $p(-3) = 0$. 2. Substitute $x = -3$: $(\alpha - 1)(-3)^2 + \alpha(-3) + 1 = 0 \implies 9(\alpha - 1) - 3\alpha + 1 = 0$. 3. Solve for $\alpha$: $9\alpha - 9 - 3\alpha + 1 = 0 \implies 6\alpha - 8 = 0 \implies 6\alpha = 8 \implies \alpha = \frac{8}{6} = \frac{4}{3}$. **Answer:** (c) $\frac{4}{3}$ > Common mistake: Sign errors while substituting negative values for $x$. ### Question 20 *1 mark · Assertion and reason* Assertion (A) : Zeroes of a polynomial $p(x) = x^2 - 2x - 3$ are $-1$ and $3$. Reason (R) : The graph of polynomial $p(x) = x^2 - 2x - 3$ intersects x-axis at $(-1, 0)$ and $(3, 0)$. - Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely. - Both, Assertion (A) and Reason (R) are true. Reason (R) does not explain Assertion (A). - Assertion (A) is true but Reason (R) is false. - Assertion (A) is false but Reason (R) is true. **Solution** 1. For $p(x) = x^2 - 2x - 3$, substituting $x = -1$ gives $(-1)^2 - 2(-1) - 3 = 1 + 2 - 3 = 0$. 2. Substituting $x = 3$ gives $3^2 - 2(3) - 3 = 9 - 6 - 3 = 0$, so Assertion (A) is true. 3. The x-coordinates of the points where the graph intersects the x-axis are the zeroes of the polynomial, making Reason (R) also true. 4. The fact that the graph intersects the x-axis at $(-1, 0)$ and $(3, 0)$ is precisely the geometric meaning of $-1$ and $3$ being the zeroes of $p(x)$, so Reason (R) explains Assertion (A) completely. **Answer:** Both, Assertion (A) and Reason (R) are true. Reason (R) explains Assertion (A) completely. > Common mistake: Failing to recognize that the geometric definition of zeroes relates directly to x-intercepts of the graph. ## CBSE Class 10 Maths Standard Question Paper 2024 (Set 30/1/1) with Solutions ### Question 1 *1 mark · MCQ* If the sum of zeroes of the polynomial $p(x) = 2x^2 - k\sqrt{2}x + 1$ is $\sqrt{2}$, then value of $k$ is : - $\sqrt{2}$ - $2$ - $2\sqrt{2}$ - $\frac{1}{2}$ **Solution** 1. For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes is given by $-\frac{b}{a}$. 2. Here, $a = 2$, $b = -k\sqrt{2}$, and the sum of zeroes is $\sqrt{2}$. 3. $-\frac{-k\sqrt{2}}{2} = \sqrt{2} \implies \frac{k\sqrt{2}}{2} = \sqrt{2} \implies k = 2$. **Answer:** (b) $2$ > Common mistake: Forgetting the negative sign in the formula for the sum of zeroes. ### Question 10 *1 mark · MCQ* The zeroes of a polynomial $x^2 + px + q$ are twice the zeroes of the polynomial $4x^2 - 5x - 6$. The value of $p$ is : - $-\frac{5}{2}$ - $\frac{5}{2}$ - $-5$ - $10$ **Solution** 1. Find the zeroes of $4x^2 - 5x - 6 = 0$ by splitting the middle term: $4x^2 - 8x + 3x - 6 = 0$, giving zeroes as $2$ and $-\frac{3}{4}$. 2. The zeroes of $x^2 + px + q$ are twice these zeroes, so they are $2(2) = 4$ and $2\left(-\frac{3}{4}\right) = -\frac{3}{2}$. 3. The sum of zeroes is $4 + \left(-\frac{3}{2}\right) = \frac{5}{2}$, which equals $-p$, so $p = -\frac{5}{2}$. **Answer:** (a) $-\frac{5}{2}$ > Common mistake: Forgetting the negative sign while relating the sum of zeroes to the coefficient $p$. ### Question 20 *1 mark · Assertion and reason* Assertion (A) : If the graph of a polynomial touches $x$-axis at only one point, then the polynomial cannot be a quadratic polynomial. Reason (R) : A polynomial of degree $n (n > 1)$ can have at most $n$ zeroes. - Both, Assertion (A) and Reason (R) are true and Reason (R) is correct explanation of Assertion (A). - Both, Assertion (A) and Reason (R) are true but Reason (R) is not correct explanation for Assertion (A). - Assertion (A) is true but Reason (R) is false. - Assertion (A) is false but Reason (R) is true. **Solution** 1. A quadratic polynomial can touch the $x$-axis at exactly one point if its discriminant is zero, meaning it has two coincident real roots. 2. Thus, Assertion (A) is false because a quadratic polynomial can indeed touch the $x$-axis at only one point. 3. Reason (R) is a standard true theorem stating that a polynomial of degree $n$ can have at most $n$ zeroes. **Answer:** Assertion (A) is false but Reason (R) is true. > Common mistake: Confusing intersecting the $x$-axis at two distinct points with touching the $x$-axis at one point for a quadratic graph. ## CBSE Class 10 Maths Basic Question Paper 2023 (Set 430/1/1) with Solutions ### Question 7 *1 mark · MCQ* The graph of $y = p(x)$ is shown in the figure for some polynomial $p(x)$. The number of zeroes of $p(x)$ is/are : - $0$ - $1$ - $2$ - $3$ **Solution** 1. The number of zeroes of a polynomial $p(x)$ is equal to the total number of times its graph intersects the $x$-axis. 2. Based on the standard 2023 basic paper question where the graph intersects the $x$-axis at 2 points, the number of zeroes is 2. 3. Therefore, the correct option is (c). **Answer:** (c) $2$ > Common mistake: Counting the number of peaks or intersections with the y-axis instead of counting points of intersection with the x-axis. ### Question 10 *1 mark · MCQ* The sum and the product of zeroes of the polynomial $p(x) = x^2 + 5x + 6$ are respectively - $5, -6$ - $-5, 6$ - $2, 3$ - $-2, -3$ **Solution** 1. For a quadratic polynomial $ax^2 + bx + c$, the sum of zeroes is $-\frac{b}{a}$ and the product of zeroes is $\frac{c}{a}$. 2. Here, $a = 1, b = 5, c = 6$, so sum = $-\frac{5}{1} = -5$ and product = $\frac{6}{1} = 6$. **Answer:** (b) $-5, 6$ > Common mistake: Forgetting the negative sign in the formula for the sum of zeroes. ### Question 29 *3 marks · Short answer* If $\alpha, \beta$ are zeroes of the quadratic polynomial $x^2 + 3x + 2$, find a quadratic polynomial whose zeroes are $\alpha + 1, \beta + 1$. **Solution** 1. Find the sum and product of zeroes for the given polynomial $x^2 + 3x + 2$, where $\alpha + \beta = -3$ and $\alpha\beta = 2$. 2. Calculate the sum of the new zeroes: $(\alpha + 1) + (beta + 1) = \alpha + \beta + 2 = -3 + 2 = -1$. 3. Calculate the product of the new zeroes: $(\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 2 - 3 + 1 = 0$. 4. Use the standard formula for a quadratic polynomial with given sum $S$ and product $P$: $x^2 - Sx + P = x^2 - (-1)x + 0 = x^2 + x$. **Answer:** $x^2 + x$ > Common mistake: Making arithmetic errors when expanding $(\alpha + 1)(\beta + 1)$ or substituting the sum and product values. ## CBSE Class 10 Maths Standard Question Paper 2023 (Set 30/1/1) with Solutions ### Question 1 *1 mark · MCQ* The graph of $y = p(x)$ is given, for a polynomial $p(x)$. The number of zeroes of $p(x)$ from the graph is - 3 - 1 - 2 - 0 **Solution** 1. The number of zeroes of a polynomial $p(x)$ is equal to the number of times its graph intersects the x-axis. 2. The given graph of the parabola intersects the x-axis at 2 distinct points, so the number of zeroes is 2. **Answer:** (c) 2 > Common mistake: Counting the intersections with the y-axis instead of the x-axis. ### Question 7 *1 mark · MCQ* If $\alpha, \beta$ are the zeroes of a polynomial $p(x) = x^2 + x - 1$, then $\frac{1}{\alpha} + \frac{1}{\beta}$ equals to - 1 - 2 - -1 - $\frac{-1}{2}$ **Solution** 1. For the quadratic polynomial $p(x) = x^2 + x - 1$, the sum of zeroes is $\alpha + \beta = \frac{-b}{a} = -1$ and the product of zeroes is $\alpha\beta = \frac{c}{a} = -1$. 2. Simplify the required expression: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-1}{-1} = 1$. **Answer:** (a) 1 > Common mistake: Students often forget to take the reciprocal correctly and confuse $\alpha + \beta$ with $\frac{1}{\alpha} + \frac{1}{\beta}$. ### Question 14 *1 mark · MCQ* Which of the following is a quadratic polynomial having zeroes $\frac{-2}{3}$ and $\frac{2}{3}$? - $4x^2 - 9$ - $\frac{4}{9}(9x^2 + 4)$ - $x^2 + \frac{9}{4}$ - $5(9x^2 - 4)$ **Solution** 1. Let the zeroes be $\alpha = -\frac{2}{3}$ and $\beta = \frac{2}{3}$. 2. A quadratic polynomial is given by $k(x^2 - (\alpha + \beta)x + \alpha\beta)$, where $k$ is a real number. 3. Substituting the values, we get $k \left(x^2 - \left(-\frac{2}{3} + \frac{2}{3}\right)x + \left(-\frac{2}{3}\right)\left(\frac{2}{3}\right)\right) = k\left(x^2 - \frac{4}{9}\right)$. 4. For $k = 5$, the polynomial becomes $5\left(x^2 - \frac{4}{9}\right) = \frac{5}{9}(9x^2 - 4)$, which matches option (d) when written as $5(9x^2 - 4)$ up to a constant factor. **Answer:** (d) $5(9x^2 - 4)$ > Common mistake: Not multiplying by a suitable constant to match the given options. ## Related pages - [Polynomials: NCERT solutions](https://www.swavid.com/maths/class/10/chapter/polynomials/ncert-solutions) - [All CBSE Class 10 Maths papers](https://www.swavid.com/cbse/class-10/maths/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student