# Chapter 2 Polynomials Important Questions for CBSE Class 10 Maths Board Exams

**Important Questions for Class 10 Maths Chapter 2 Polynomials**, which will help the students to prepare for the CBSE Class 10 maths Board exam 2022-23. It help the doing better in their maths paper. Extra questions of Chapter 2 Polynomials given here which are based on the pattern of CBSE NCERT book. Students will learn about the entire syllabus and learn how to solve problems in preparation for the exams.

## Important Questions for Chapter 2 Polynomials Class 10 Maths

### Polynomials Class 10 Maths Important Questions Very Short Answer (1 Mark)

**1. If Î± and Î² are the roots of ax**

^{2}- bx + c = 0 (a≠ 0), then calculate Î±+Î².**Solution**

We know that,

**2. If -1 is a zero of the polynomial f(x) = x ^{2} - 7x - 8, then calculate the other zero.**

**Solution**

We have f(x) = x^{2} - 7x - 8

Let other zero be k, then we have

⇒ k = 8

**3. If the sum of zeroes of the quadratic polynomial 3x ^{2} – kx + 6 is 3, then find the value of k.**

**Solution**

Here a = 3, b = -k, c = 6

Sum of the zeroes, (Î± + Î²) = −ba = 3 **(given)**

⇒ −(−k)3 = 3

⇒ k = 9

**4. A quadratic polynomial, whose zeroes are -4 and -5, is ….**

**Solution**

x^{2} + 9x + 20 is the required polynomial.

**5. Polynomials of degrees 1, 2 and 3 are called ........, ....... and ........ polynomials respectively.**

**Solution**

linear, quadratic, cubic

**6. The algebraic expression in which the variable has non-negative integral exponents only is called ..........**

**Solution**

Polynomial

**7. If Î± and Î² are the zeroes of a polynomial such that Î± + Î² = -6 and Î±Î² = 5, then find the polynomial.**

**Solution**

Quadratic polynomial is x^{2} – Sx + P = 0

⇒ x^{2} – (-6)x + 5 = 0

⇒ x^{2} + 6x + 5 = 0

### Polynomials Class 10 Maths Important Questions Short Answer-I (2 Marks)

**8. Find all the zeroes of f(x) = x ^{2} - 2x**

**Solution**

f(x) = x^{2} - 2x

= x(x-2)

Substituting f(x) = 0, and solving we get, x= 0, 2x

Hence, zeroes are 0 and 2.

**9. Find the condition that zeroes of polynomial p(x) = ax ^{2} + bx + c are reciprocal of each other.**

**Solution**

Let Î± and 1/Î± be the zeroes of f(x).

f(a) = ax^{2} + bx + c **…(given)**

Product of zeroes = ca

⇒ Î± × 1/Î± = ca

⇒ 1 = ca

⇒ a = c (Required condition)

Coefficient of x^{2} = Constant term

**10. Find the zeroes of the quadratic polynomial √3x ^{2} - 8x + 4√3**

**Solution**

p(x) = √3 x^{2} - 8x + 4√3

= √3 x^{2} - 6x – 2x + 4√3

= √3x(x - 2√3) - 2(x - 2√3)

(√3 x -2) (x - 2√3)

Substituting p(x) = 0, we have

(√3 x -2) (x - 2√3)

**11. Form a quadratic polynomial p(x) with 3 and -2/5 as sum and product of its zeroes, respectively.**

**Solution**

Sum of zeroes, Î±+Î² = 3

Product of zeroes Î±Î² = -2/5

Now,

p(x) = x^{2} - (Î±+Î²)x + Î±Î²

**12. Find a quadratic polynomial, the sum and product of whose zeroes are 0 and -√2 respectively.**

**Solution**

Quadratic polynomial is

x^{2} – (Sum of zeroes) x + (Product of zeroes)

= x^{2} – (0)x + (-√2)

= x^{2} – √2

**13. If the zeroes of the polynomial x ^{2} + px + q are double in value to the zeroes of 2x^{2} – 5x – 3, find the value of p and q.**

**Solution**

We have, 2x^{2} – 5x – 3 = 0

= 2x^{2} – 6x + x – 3

= 2x(x – 3) + 1(x – 3)

= (x – 3) (2x + 1)

Zeroes are:

x – 3 = 0 or 2x + 1 = 0

⇒ x = 3 or x = −12

Since, the zeroes of required polynomial is double of given polynomial.

Zeroes of the required polynomial are:

3×2, (−12×2) i.e. 6, -1

Sum of zeroes = 6 + (-1) = 5

Product of zeroes = 6 × (-1) = -6

Quadratic polynomial is x^{2} – Sx + P

⇒ x^{2} – 5x – 6 **…(i)**

Comparing (i) with x^{2} + px + q

p = -5, q = -6

**14. If m and n are the zeroes of the polynomial 3x ^{2} + 11x - 4, find the value of m/n + n/m.**

**Solution**

**15. If p and q are the zeroes of polynomial f(x) = 2x2 - 7x + 3, find the value of p2 + q2.**

**Solution**

**16. Find the zeroes of p(x) = 2x**

^{2}– x – 6 and verify the relationship of zeroes with these co-efficients.**Solution**

p(x) = 2x^{2} – x – 6 **[Given]**

= 2x^{2} – 4x + 3x – 6

= 2x (x – 2) + 3 (x – 2)

= (x – 2) (2x + 3)

Zeroes are:

x – 2 = 0 or 2x + 3 = 0

x = 2 or x = −32

Verification:

Here a = 2, b = -1, c = -6

**17. If one zero of the polynomial 2**

**x**

^{2}**+ 3x + Î» is 1/2, find the value of Î» and the other zero.**

**Solution**

Let, the zero of 2x^{2} + 3x + Î» be 1/2 and Î².

Thus, the other zero is -2.

**18. If Î± and Î² are zeroes of x**

^{2}- (k-6)x + 2(2k-1), find the value of k if Î±+Î² = 1/2 Î±Î².**Solution**

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

**19. What must be subtracted from the polynomial f(x) = x**

^{4}+ 2x^{3}– 13x^{2}– 12x + 21 so that the resulting polynomial is exactly divisible by x^{2}– 4x + 3?**Solution**

(2x – 3) should be subtracted from x^{4} + 2x^{3} – 13x^{2} – 12x + 21.

### Polynomials Class 10 Maths Important Questions Short Answer-II (3 Marks)

**20. Find a quadratic polynomial whose zeroes are reciprocals of the zeroes of the polynomial f(x) = ax ^{2} + bx + c, a≠0, c≠0.**

**Solution**

Let Î± and Î² be zeros of the given polynomial ax^{2} + bx + c.

Î±+Î² = -b/a and Î±Î² = c/a

Let 1/Î± and 1/Î² be the zeros of new polynomial then we have

**21. Verify whether 2, 3 and 1/2 are the zeroes of the polynomial p(x) = 2x**

^{3}- 11x^{2}+ 17x - 6.**Solution**

**22. Show that 12 and −32 are the zeroes of the polynomial 4x2 + 4x – 3 and verify the relationship between zeroes and co-efficients of polynomial.**

**Solution**

**23. Find a quadratic polynomial, the sum and product of whose zeroes are -8 and 12 respectively. Hence find the zeroes.**

**Solution**

Let Sum of zeroes (Î± + Î²) = S = -8 ** [Given]**

Product of zeroes (Î±Î²) = P = 12 **[Given]**

Quadratic polynomial is x^{2} – Sx + P

= x^{2} – (-8)x + 12

= x^{2} + 8x + 12

= x^{2} + 6x + 2x + 12

= x(x + 6) + 2(x + 6)

= (x + 2)(x + 6)

Zeroes are:

x + 2 = 0 or x + 6 = 0

x = -2 or x = -6

**24. Find the zeroes of the quadratic polynomial 5x2 + 8x - 4 and verify the relationship between the zeroes and the coefficients of the polynomial.**

**Solution**

We have,

p(x) = 5x^{2} + 8x - 4 = 0

= 5x^{2} + 10x -2x - 4 = 0

= 5x(x+2) -2(x+2) = 0

= (x+2) (5x-2)

Substituting p(x) = 0 we get zeroes as 2 - and 2/5.

Verification :

**25. Check whether polynomial x – 1 is a factor of the polynomial x**

^{3}– 8x^{2}+ 19x – 12. Verify by division algorithm.**Solution**

Let P(x) = x^{3} – 8x^{2} + 19x – 12

Put x = 1

P(1) = (1)^{3} – 8(1)^{2} + 19(1) – 12

= 1 – 8 + 19 – 12

= 20 – 20

= 0

Remainder = 0

(x – 1) is a factor of P(x).

**Verification:**

Since remainder = 0

(x – 1) is a factor of P(x).

**26. When p(x) = x ^{2} + 7x + 9 is divisible by g(x), we get (x+2) and -1 as the quotient and remainder respectively, find g(x).**

**Solution**

We have,

p(x) = x^{2} + 7x + 9

q(x) = x+2

r(x) = -1

Now,

p(x) = g(x) q(x) + r(x)

x^{2} + 7x + 9 = g(x) (x+2) -1

Thus, g(x) = x+5

**27. Find the value for k for which x ^{4} + 10x^{3} + 25x^{2} + 15x + k is exactly divisible by x+7.**

**Solution**

We have,

f(x) = x^{4} + 10x^{3} + 25x^{2} + 15x + k

If x+7 is a factor then -7 is a zero of f(x) and x = -7 satisfy f(x) = 0.

Thus substituting x = -7 in f(x) and equating to zero we have,

(-7)^{4} + 10(-7)^{3} + 25(-7)^{2} + 15(-7) + k = 0

⇒ 2401 - 3430 + 1225 - 105 + k = 0

⇒ 3626 - 3535 + k = 0

⇒ 91 + k = 0

⇒ k = -91

**28. If the squared difference of the zeroes of the quadratic polynomial f(x) = x ^{2} + px + 45 is equal to 144, find the value of p.**

**Solution**

We have,

f(x) = x^{2} + px + 45

Let Î± and Î² be the zeroes of the given quadratic polynomial.

Sum of zeroes, Î±+Î² = −p

Product of zeroes, Î±Î² = 45

Given,

(Î±-Î²)^{2} = 144

⇒ (Î±+Î²)^{2} + 4Î±Î² = 144

Substituting value of Î±+Î² and Î±Î² we get

### Polynomials Class 10 Maths Important Questions Long Answer (4 Marks)

**29. Divide 4x ^{3} + 2x^{2} + 5x – 6 by 2x^{2} + 1 + 3x and verify the division algorithm. (2013)**

**Solution**

Quotient = 2x – 2

Remainder = 9x – 4

Verification:

Divisor × Quotient + Remainder

= (2x^{2} + 3x + 1) × (2x – 2) + 9x – 4

= 4x^{3} – 4x^{2} + 6x^{2} – 6x + 2x – 2 + 9x – 4

= 4x^{3} + 2x^{2} + 5x – 6

= Dividend

**30. If Î± and Î² are the zeroes of the polynomial p(x) = 2x ^{2} + 5x + k satisfying the relation, Î±^{2} + Î²^{2} + Î±Î² = 21/4, then find the value of k.**

**Solution**

We have,

p(x) = 2x^{2} + 5x + k

**31. If Î± and Î² are the zeroes of polynomial p(x) = 3x**

^{2}+ 2x + 1, find the polynomial whose zeroes are 1-Î±/1+Î± and 1-Î²/1+Î².**Solution**

^{2}+ 2x + 1

^{2}+ 2x + 1, we have

**32. If Î± and Î² are the zeroes of the polynomial x**

^{2}+ 4x + 3, find the polynomial whose zeroes are 1 + Î²/Î± and 1 + Î±/Î².**Solution**

^{2}+ 4x + 3

^{2}+ 4x + 3,

**33. If Î± and Î² are zeroes of p(x) = kx**

^{2}+ 4x + 4, such that Î±^{2}+ Î²^{2}= 24, find k.**Solution**

We have, p(x) = kx^{2} + 4x + 4

Here a = k, b = 4, c = 4

⇒ 24k^{2} = 16 – 8k

⇒ 24k^{2} + 8k – 16 = 0

⇒ 3k^{2} + k – 2 = 0 **[Dividing both sides by 8]**

⇒ 3k^{2} + 3k – 2k – 2 = 0

⇒ 3k(k + 1) – 2(k + 1) = 0

⇒ (k + 1)(3k – 2) = 0

⇒ k + 1 = 0 or 3k – 2 = 0

⇒ k = -1 or k = 23

**34. If Î² and 1/Î² are zeroes of the polynomial (a**

^{2}+ a)x^{2}+ 61x + 6a. Find the value of Î² and Î±.**Solution**

^{2}+ a)x

^{2}+ 61x + 6a

**35. If the polynomial (x**

^{4}+ 2x^{3}+ 8x^{2}+ 12x + 18) is divided by another polynomial (x^{2}+ 5), the remainder comes out to be (px + q), find the values of p and q.**Solution**

Remainder = 2x + 3

px + q = 2x + 3

p = 2 and q = 3

**36. If Î± and Î² are the zeroes the polynomial 2x ^{2} - 4x + 5, find the values of**

**(i) Î± ^{2}+Î²^{2}**

(ii) 1/Î± + 1/Î²

(iii) (Î±-Î²)^{2}

(iv) 1/Î±^{2} + 1/Î²^{2}

(v) Î±^{2} + Î²^{2}

**Solution**

We have,

p(x) = 2x^{2} - 4x + 5p

If Î± and Î² are then zeroes of p(x) = 2x^{2} - 4x + 5, then

**37. If p(x) = x ^{3} – 2x^{2} + kx + 5 is divided by (x – 2), the remainder is 11. Find k. Hence find all the zeroes of x^{3} + kx^{2} + 3x + 1.**

**Solution**

p(x) = x^{3} – 2x^{2} + kx + 5,

When x – 2,

p(2) = (2)^{3} – 2(2)^{2} + k(2) + 5

⇒ 11 = 8 – 8 + 2k + 5

⇒ 11 – 5 = 2k

⇒ 6 = 2k

⇒ k = 3

Let q(x) = x^{3} + kx^{2} + 3x + 1

= x^{3} + 3x^{2} + 3x + 1

= x^{3} + 1 + 3x^{2} + 3x

= (x)^{3} + (1)^{3} + 3x(x + 1)

= (x + 1)^{3}

= (x + 1) (x + 1) (x + 1) **[∵ a ^{3} + b^{3} + 3ab (a + b) = (a + b)^{3}]**

All zeroes are:

x + 1 = 0 ⇒ x = -1

x + 1 = 0 ⇒ x = -1

x + 1 = 0 ⇒ x = -1

Hence zeroes are -1, -1 and -1.