## Important Questions for Class 10 Maths Chapter 2 Polynomials

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

Question 1.

If the sum of zeroes of the quadratic polynomial 3x^{2} – kx + 6 is 3, then find the value of k. (2012)

Solution:

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

Sum of the zeroes, (α + β) = \(\frac { -b }{ a }\) = 3 …..(given)

⇒ \(\frac { -(-k) }{ 3 }\) = 3

⇒ k = 9

Question 2.

If α and β are the zeroes of the polynomial ax^{2} + bx + c, find the value of α^{2} + β^{2}. (2013)

Solution:

Question 3.

If the sum of the zeroes of the polynomial p(x) = (k^{2} – 14) x^{2} – 2x – 12 is 1, then find the value of k. (2017 D)

Solution:

p(x) = (k^{2} – 14) x^{2} – 2x – 12

Here a = k^{2} – 14, b = -2, c = -12

Sum of the zeroes, (α + β) = 1 …[Given]

⇒ \(\frac { -b }{ a }\) = 1

⇒ \(\frac { -\left( -2 \right) }{ { k }^{ 2 }-14 }\) = 1

⇒ k^{2} – 14 = 2

⇒ k^{2} = 16

⇒ k = ±4

Question 4.

If α and β are the zeroes of a polynomial such that α + β = -6 and αβ = 5, then find the polynomial. (2016 D)

Solution:

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

⇒ x^{2} – (-6)x + 5 = 0

⇒ x^{2} + 6x + 5 = 0

Question 5.

A quadratic polynomial, whose zeroes are -4 and -5, is …. (2016 D)

Solution:

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

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

Question 6.

Find the condition that zeroes of polynomial p(x) = ax^{2} + bx + c are reciprocal of each other. (2017 OD)

Solution:

Let α and \(\frac { 1 }{ \alpha }\) be the zeroes of P(x).

P(a) = ax^{2} + bx + c …(given)

Product of zeroes = \(\frac { c }{ a }\)

⇒ α × \(\frac { 1 }{ \alpha }\) = \(\frac { c }{ a }\)

⇒ 1 = \(\frac { c }{ a }\)

⇒ a = c (Required condition)

Coefficient of x^{2} = Constant term

Question 7.

Form a quadratic polynomial whose zeroes are 3 + √2 and 3 – √2. (2012)

Solution:

Sum of zeroes,

S = (3 + √2) + (3 – √2) = 6

Product of zeroes,

P = (3 + √2) x (3 – √2) = (3)^{2} – (√2)^{2} = 9 – 2 = 7

Quadratic polynomial = x^{2} – Sx + P = x^{2} – 6x + 7

Question 8.

Find a quadratic polynomial, the stun and product of whose zeroes are √3 and \(\frac { 1 }{ \surd 3 }\) respectively. (2014)

Solution:

Sum of zeroes, (S) = √3

Product of zeroes, (P) = \(\frac { 1 }{ \surd 3 }\)

Quadratic polynomial = x^{2} – Sx + P

Question 9.

Find a quadratic polynomial, the sum and product of whose zeroes are 0 and -√2 respectively. (2015)

Solution:

Quadratic polynomial is

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

= x^{2} – (0)x + (-√2)

= x^{2} – √2

Question 10.

Find the zeroes of the quadratic polynomial √3 x^{2} – 8x + 4√3. (2013)

Solution:

Question 11.

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. (2012)

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 = \(\frac { -1 }{ 2 }\)

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

Zeroes of the required polynomial are:

3 × 2, (\(\frac { -1 }{ 2 }\) × 2), i.e., 6, -1

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

Product of zeroes, P = 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

Question 12.

Can (x – 2) be the remainder on division of a polynomial p(x) by (2x + 3)? Justify your answer. (2016 OD)

Solution:

In case of division of a polynomial by another polynomial, the degree of the remainder (polynomial) is always less than that of the divisor. (x – 2) can not be the remainder when p(x) is divided by (2x + 3) as the degree is the same.

Question 13.

Find a quadratic polynomial whose zeroes are \(\frac { 3+\surd 5 }{ 5 }\) and \(\frac { 3-\surd 5 }{ 5 }\). (2013)

Solution:

Question 14.

Find the quadratic polynomial whose zeroes are -2 and -5. Verify the relationship between zeroes and coefficients of the polynomial. (2013)

Solution:

Sum of zeroes, S = (-2) + (-5) = -7

Product of zeroes, P = (-2)(-5) = 10

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

= x^{2} – (-7)x + 10

= x^{2} + 7x + 10

Verification:

Here a = 1, b = 7, c = 10

Sum of zeroes = (-2) + (-5) = 7

Question 15.

Find the zeroes of the quadratic polynomial 3x^{2} – 75 and verify the relationship between the zeroes and the coefficients. (2014)

Solution:

We have, 3x^{2} – 75

= 3(x^{2} – 25)

= 3(x^{2} – 52)

= 3(x – 5)(x + 5)

Zeroes are:

x – 5 = 0 or x + 5 = 0

x = 5 or x = -5

Verification:

Here a = 3, b = 0, c = -75

Sum of the zeroes = 5 + (-5) = 0

Question 16.

Find the zeroes of p(x) = 2x^{2} – x – 6 and verify the relationship of zeroes with these co-efficients. (2017 OD)

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 = \(\frac { -3 }{ 2 }\)

Verification:

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

Question 17.

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? (2012, 2017 D)

Solution:

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

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

Question 18.

Verify whether 2, 3 and \(\frac { 1 }{ 2 }\) are the zeroes of the polynomial p(x) = 2x^{3} – 11x^{2} + 17x – 6. (2012, 2017 D)

Solution:

p(x) = 2x^{3} – 11x^{2} + 17x – 6

When x = 2,

p(2) = 2(2)^{3} – 11(2)^{2} + 17(2) – 6 = 16 – 44 + 34 – 6 = 0

When x = 3, p(3) = 2(3)^{3} – 11(3)^{2} + 17(3) – 6 = 54 – 99 + 51 – 6 = 0

Yes, x = 2, 3 and \(\frac { 1 }{ 2 }\) all are the zeroes of the given polynomial.

Question 19.

Show that \(\frac { 1 }{ 2 }\) and \(\frac { -3 }{ 2 }\) are the zeroes of the polynomial 4x^{2} + 4x – 3 and verify the relationship between zeroes and co-efficients of polynomial. (2013)

Solution:

Let P(x) = 4x^{2} + 4x – 3

Question 20.

Find a quadratic polynomial, the sum and product of whose zeroes are -8 and 12 respectively. Hence find the zeroes. (2014)

Solution:

Let Sum of zeroes (α + β) = S = -8 …[Given]

Product of zeroes (αβ) = 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

Question 21.

Find a quadratic polynomial, the sum and product of whose zeroes are 0 and \(\frac { -3 }{ 5 }\) respectively. Hence find the zeroes. (2015)

Solution:

Quadratic polynomial = x^{2} – (Sum)x + Product

Question 22.

Find the zeroes of the quadratic polynomial 6x^{2} – 3 – 7x and verify the relationship between the zeroes and the coefficients of the polynomial. (2015, 2016 OD)

Solution:

We have, 6x^{2} – 3 – 7x

= 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x – 3) (3x + 1)

Question 23.

Find the zeroes of the quadratic polynomial f(x) = x^{2} – 3x – 28 and verify the relationship between the zeroes and the co-efficients of the polynomial. (2012, 2017 D)

Solution:

p(x) = x^{2} – 3x – 28

= x^{2} – 7x + 4x – 28

= x(x – 7) + 4(x – 7)

= (x – 7) (x + 4)

Zeroes are:

x – 7 = 0 or x + 4 = 0

x = 7 or x = -4

Question 24.

If α and β are the zeroes of the polynomial 6y^{2} – 7y + 2, find a quadratic polynomial whose zeroes are \(\frac { 1 }{ \alpha }\) and \(\frac { 1 }{ \beta }\). (2012)

Solution:

Given: 6y^{2} – 7y + 2

Here a = 6, b = -7, c = 2

Question 25.

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

Solution:

Quotient = 3x – 1

Remainder = 1

Verification:

Divisor × Quotient + Remainder

= (x + 2) × (3x – 1) + 1

= 3x^{2 }– x + 6x – 2 + 1

= 3x^{2} + 5x – 1

= Dividend

Question 26.

On dividing 3x^{3} + 4x^{2} + 5x – 13 by a polynomial g(x) the quotient and remainder were 3x +10 and 16x – 43 respectively. Find the polynomial g(x). (2017 OD)

Solution:

Let 3x^{3} + 4x^{2} + 5x – 13 = P(x)

q(x) = 3x + 10, r(x) = 16x – 43 …[Given]

As we know, P(x) = g(x) . q(x) + r(x)

3x^{3} + 4x^{2} + 5x – 13 = g(x) . (3x + 10) + (16x – 43)

3x^{3} + 4x^{2} + 5x – 13 – 16x + 43 = g(x) . (3x + 10)

Question 27.

Check whether polynomial x – 1 is a factor of the polynomial x^{3} – 8x^{2} + 19x – 12. Verify by division algorithm. (2014)

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 facter of P(x).

Verification:

Since remainder = 0

(x – 1) is a factor of P(x).

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

Question 28.

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

Question 29.

Given that x – √5 is a factor of the polynomial x^{3} – 3√5 x^{2} – 5x + 15√5, find all the zeroes of the polynomial. (2012, 2016)

Solution:

Let P(x) = x^{3} – 3√5 x^{2} – 5x + 15√5

x – √5 is a factor of the given polynomial.

Put x = -√5,

Other zero:

x – 3√5 = 0 ⇒ x = 3√5

All the zeroes of P(x) are -√5, √5 and 3√5.

Question 30.

If a polynomial x^{4} + 5x^{3} + 4x^{2} – 10x – 12 has two zeroes as -2 and -3, then find the other zeroes. (2014)

Solution:

Since two zeroes are -2 and -3.

(x + 2)(x + 3) = x^{2} + 3x + 2x + 6 = x^{2} + 5x + 6

Dividing the given equation with x^{2} + 5x + 6, we get

x^{4} + 5x^{3 }+ 4x^{2} – 10x – 12

= (x^{2} + 5x + 6)(x^{2} – 2)

= (x + 2)(x + 3)(x – √2 )(x + √2 )

Other zeroes are:

x – √2 = 0 or x + √2 = 0

x = √2 or x = -√2

Question 31.

Find all the zeroes of the polynomial 8x^{4} + 8x^{3} – 18x^{2} – 20x – 5, if it is given that two of its zeroes are \(\sqrt { \frac { 5 }{ 2 } }\) and \(-\sqrt { \frac { 5 }{ 2 } }\). (2014, 2016 D)

Solution:

Question 32.

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. (2012)

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.

Question 33.

If α and β are zeroes of p(x) = kx^{2} + 4x + 4, such that α^{2} + β^{2} = 24, find k. (2013)

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 = \(\frac { 2 }{ 3 }\)

Question 34.

If α and β are the zeroes of the polynomial p(x) = 2x^{2} + 5x + k, satisfying the relation, α^{2} + β^{2} + αβ = \(\frac { 21 }{ 4 }\) then find the value of k. (2017 OD)

Solution:

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

Here a = 2, b = 5, c = k

Question 35.

What must be subtracted from p(x) = 8x^{4} + 14x^{3} – 2x^{2} + 8x – 12 so that 4x^{2} + 3x – 2 is factor of p(x)? This question was given to group of students for working together. (2015)

Solution:

Polynomial to be subtracted by (15x – 14).

Question 36.

Find the values of a and b so that x^{4} + x^{3} + 8x^{2} +ax – b is divisible by x^{2} + 1. (2015)

Solution:

If x^{4} + x^{3} + 8x^{2} + ax – b is divisible by x^{2} + 1

Remainder = 0

(a – 1)x – b – 7 = 0

(a – 1)x + (-b – 7) = 0 . x + 0

a – 1 = 0, -b – 7 = 0

a = 1, b = -7

a = 1, b = -7

Question 37.

If a polynomial 3x^{4} – 4x^{3} – 16x^{2} + 15x + 14 is divided by another polynomial x^{2} – 4, the remainder comes out to be px + q. Find the value of p and q. (2014)

Solution:

Question 38.

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.