# Polynomials – Exercise 8.1 – Class X

#### Polynomials – Exercise 8.1

1. Find the degree of the following polynomials.

(i) x2 – 9x + 20

(ii) 2x + 3 + 6x2

(iii) x3 + 2x2 – 5x – 6

(iv) x3 + 17x – 21 – x2

(v) √3x3 + 19x + 14

1. If f(x) = 2x3 + 3x2 – 11x + 6, find.

(i) f(0)

(ii)f(1)

(iii)f(-1)

(iv) f(2)

(v)f(-3)

1. Find the values of the following polynomials.

(i) g(x) = 7x2 + 2x + 14 when x = 1

(ii) p(x) = -x3 + x2 – 6x + 5 when x = 2

(iii) p(x) = 2x2 + 1/4 x + 13, when x = -1

(iv) p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 , when x = -2

1. Verify whether the indicated numbers are zeroes of the polynomials in each of the following cases.

(i) f(x) = 3x + 1, x = – 1/3

(ii) p(x) = x2 – 4, x = 2 , x = -2

(iii) g(x) = 5x – 8, x  = 4/5

(iv) p(x) = 3x3 – 5x2 – 11x – 3, x = 3, x = -1 and x = –1/3

1. Find the zeroes of the following quadratic polynomials and verify

(i)x2 + 4x + 4

(ii)x2 – 2x – 15

(iii) 4a2 – 49

(iv) 2a2 – 2√2a + 1

1. If x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + k, find the value of k.
2. For what value of k, -4 is a zero of the polynomial x2 – x – (2k + 2)?

#### Polynomials – Exercise 8.1 – Solution:

1. Find the degree of the following polynomials.

(i) x2 – 9x + 20

Solution:

p(x) = x2 – 9x + 20 is a polynomial in x of degree 2

(ii) 2x + 3 + 6x2

Solution:

p(x) = 6x2 + 2x + 3 is polynomial in x is degree 2

(iii) x3 + 2x2 – 5x – 6

Solution:

p(x) = x3 + 2x2 – 5x – 6 is a polynomial in x is degree 3

(iv) x3 + 17x – 21 – x2

Solution:

p(x) =  x3 – x2 + 17x – 21 is a polynomial in x is degree 3

(v) √3x3 + 19x + 14

Solution:

p(x) = √3x3 + 19x + 14 is a polynomial in x is degree 3

1. If f(x) = 2x3 + 3x2 – 11x + 6, find.

(i) f(0)

(ii)f(1)

(iii)f(-1)

(iv) f(2)

(v)f(-3)

Solution:

f(x) = 2x3 + 3x2 – 11x + 6

(i) f(0) = 2(0)3 + 3(0)2 – 11(0) + 6

= 0 + 0 + 0 + 6

= 6

(ii)f(1) = 2(1)3 + 3(1)2 – 11(1) + 6

= 2 + 3 – 11 + 6

= 0

(iii)f(-1) = 2(-1)3 + 3(-1)2 – 11(-1) + 6

= – 2 + 3 + 11 + 6

= 18

(iv) f(2) = 2(2)3 + 3(2)2 – 11(2) + 6

= 2 x 8 + 3 x 4 – 22 + 6

= 16 + 12 – 22 + 6

= 12

(v)f(-3) = 2(-3)3 + 3(-3)2 – 11(-3) + 6

= 2(-27)  + 3×9 + 33 + 6

= 12

1. Find the values of the following polynomials.

(i) g(x) = 7x2 + 2x + 14 when x = 1

Solution:

g(x) = 7x2 + 2x + 14

if x  = 1, then g(1) = 7(1)2 + 2(1) + 14

= 7 + 2 + 14

= 23

(ii) p(x) = -x3 + x2 – 6x + 5 when x = 2

Solution:

p(x) = -x3 + x2 – 6x + 5

When x = 2, p(2) = -(2)3 + (2)2 – 6(2) + 5

= – 8 + 4 – 12 + 5

= – 11

(iii) p(x) = 2x2 + 1/4 x + 13, when x = -1

Solution:

p(x) = 2x2 + 1/4 x + 13

When x = -1, p(-1) = 2(-1)2 + 1/4 (-1) + 13

= 2 – 1/4 + 13

=59/4

= 14.75

(iv) p(x) = 2x4 – 3x3 – 3x2 + 6x – 2 , when x = -2

Solution:

p(x) = 2x4 – 3x3 – 3x2 + 6x – 2

When x = -2, p(-2) = 2(-2)4 – 3(-2)3 – 3(-2)2 + 6(-2) – 2

= 2(16) – 3(-8) – 3(4) – 12 – 2

=32 + 24 – 12 – 12 – 2

= 30

1. Verify whether the indicated numbers are zeroes of the polynomials in each of the following cases.

(i) f(x) = 3x + 1, x = – 1/3

Solution:

Given,  f(x) = 3x + 1, we have to find out whether at x = ‑1/3 is the zero of the polynomial f(x) = 3x + 1.

If x = – 1/3, f(- 1/3) = 3(- 1/3) + 1

= -1 + 1

= 0

Therefore, x = –1/3 is the zero of the polynomial  f(x) = 3x + 1.

(ii) p(x) = x2 – 4, x = 2 , x = -2

Solution:

Given,  p(x) = x2 – 4, we have to find out whether at x = 2 and x = -2 is the zero of the polynomial p(x) = x2 – 4.

If x = 2, p(2) = (2)2 – 4

= 4 – 4

= 0

Therefore, x = 2 is the zero of the polynomial p(x) = x2 – 4.

If x = -2, p(-2) = (-2)2 – 4

= 4 – 4

= 0

Therefore, x = -2 is the zero of the polynomial p(x) = x2 – 4.
Thus, x = 2 and x = -2 both are zero of the polynomial p(x) = x2 – 4.

(iii) g(x) = 5x – 8, x  = 4/5

Solution:

Given,  g(x) = 5x – 8, we have to find out whether at x = 4/5 is the zero of the polynomial g(x) = 5x – 8.

If x = 4/5, g(4/5) = 5(4/5) – 8

= 4  – 8

= -4

Therefore, x = 4/5 is not a zero of the polynomial g(x) = 5x – 8

(iv) p(x) = 3x3 – 5x2 – 11x – 3, x = 3, x = -1 and x = –1/3

Solution:

Given,  p(x) = 3x3 – 5x2 – 11x – 3, we have to find out whether at x = 3, x = -1 and x = –1/3 is the zero of the polynomial p(x) = 3x3 – 5x2 – 11x – 3.

For x = 3, p(3) = 3(3)3 – 5(3)2 – 11(3) – 3

= 3×27 – 5×9 – 33 – 3

= 81 – 45 – 33 – 3

=0

Therefore, x = 3 is the zero of the polynomial  p(x) = 3x3 – 5x2 – 11x – 3.

For x = -1, p(-1) = 3(-1)3 – 5(-1)2 – 11(-1) – 3

= 3(-1) – 5(1) – 11(-1) – 3

= -3 – 5 + 11 – 3

= 0

Therefore, x = -1 is the zero of the polynomial  p(x) = 3x3 – 5x2 – 11x – 3.

For x = –1/3, p(-1/3) = 3(-1/3)3 – 5(-1/3)2 – 11(-1/3) – 3

= – 3x1/27 – 5x1/9 + 11(1/3) – 3

= –1/95/9 + 11/3 – 3

= 0

Therefore, x = –1/3 is the zero of the polynomial  p(x) = 3x3 – 5x2 – 11x – 3.

1. Find the zeroes of the following quadratic polynomials and verify

(i)x2 + 4x + 4

Solution:

Given, f(x) = x2 + 4x + 4, we have to find the zero of the polynomial.

f(x) = x2 + 4x + 4 = 0

x2 + 2x + 2x + 4 = 0

x(x + 2) +2(x + 2) = 0

(x + 2)(x + 2) = 0

(x + 2)2 = 0

x + 2 = 0

x = -2

Therefore x = -2 is the zero of the polynomial f(x) = x2 + 4x + 4

Verify:

For x = -2, f(-2) = (-2)2 + 4(-2) + 4

= 4 – 8 + 4

= 8 – 8

= 0.

Hence, x = -2 is the zero of the polynomial f(x) = x2 + 4x + 4.

(ii)x2 – 2x – 15

Solution:

Given, f(x) = x2 – 2x – 15, we have to find the zero of the polynomial.

f(x) = x2 – 2x – 15 = 0

x2 + 3x – 5x – 15 = 0

x(x + 3) – 5(x + 3) = 0

(x – 5)(x + 3) = 0

x + 3 = 0 or x – 5 = 0

x = -3 or x = 5

Therefore x = -3 and x = 5 is the zero of the polynomial f(x) = x2 – 2x – 15

Verify:

For x = -3, f(-3) = (-3)2 – 2(-3) – 15

= 9 + 6 – 15

= 15 – 15

= 0.

Hence, x = -3 is the zero of the polynomial f(x) = x2 – 2x – 15

For x = 5, f(5) = (5)2 – 2(5) – 15

= 25 – 10 – 15

= 25 – 25

= 0.

Hence, x = 5 is the zero of the polynomial f(x) = x2 – 2x – 15

(iii) 4a2 – 49

Solution:

Given, f(a) = 4a2 – 49, we have to find the zero of the polynomial.

f(a) = 4a2 – 49 = 0

4a2 – 49 = 0

4a2 = 49

(2a)2 = 72

2a = 7

a = 7/2

Therefore a = 7/2 is the zero of the polynomial f(7/2) = 4a2 – 49

Verify:

For x = 7/2 , f(7/2) = 4(7/2)2 – 49

= 4(49/4) – 49

= 49 – 49

= 0.

Hence, a = 7/2 is the zero of the polynomial f(a) = 4a2 – 49

(iv) 2a2 – 2√2a + 1

Solution:

Given, f(a) = 2a2 – 2√2a + 1, we have to find the zero of the polynomial.

f(a) = 2a2 – 2√2a + 1 = 0

2a2 – 2√2a + 1 = 0

(√2a)2 – 2(√2a)(1) + 12 = 0

a2  – 2ab + b2 = (a – b)2

⇒ (√2a – 1)2 = 0

⇒ (√2a – 1) = 0

⇒√2a – 1 = 0

⇒√2a = 1

⇒a = 1/√2

Verify:

For a = 1/2,

f(a) = 2a2 – 2√2a + 1

= 2(1/√2)2 – 2√2(1/√2) + 1

= 2(1/­2) – 2 + 1

= 1 – 2 + 1

= 0

Hence, a = 1/√2 is the zero of the polynomial f(a) = 2a2 – 2 √2a + 1

1. If x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + k, find the value of k.

Solution:

Given, if x = 1 is a zero of the  polynomial f(x) = x3 – 2x2 + 4x + k. Then we have to find the value of k,

f(1) = (1)3 – 2(1)2 + 4(1) + k = 0

1 – 2 + 4 + k = 0

3 + k = 0

k = -3

1. For what value of k, -4 is a zero of the polynomial x2 – x – (2k + 2)?

Solution:

Given, x = – 4 is the zero of the polynomial f(x) = x2 – x – (2k + 2)

f(-4) = (-4)2 – (-4) – (2k + 2) = 0

⇒ 16 + 4 – (2k + 2) = 0

⇒ 20 – 2(k + 1) = 0

⇒2(10 – (k+1)) = 0

10 – k – 1 = 0

-k + 9 = 0

k = 9

Next  exercise – Polynomials – Exercise 8.2 – Class X