#### Polynomials – Exercise 8.1

- Find the degree of the following polynomials.

(i) x^{2} – 9x + 20

(ii) 2x + 3 + 6x^{2}

(iii) x^{3} + 2x^{2} – 5x – 6

(iv) x^{3} + 17x – 21 – x^{2}

(v) √3x^{3} + 19x + 14

- If f(x) = 2x
^{3}+ 3x^{2}– 11x + 6, find.

(i) f(0)

(ii)f(1)

(iii)f(-1)

(iv) f(2)

(v)f(-3)

- Find the values of the following polynomials.

(i) g(x) = 7x^{2} + 2x + 14 when x = 1

(ii) p(x) = -x^{3} + x^{2} – 6x + 5 when x = 2

(iii) p(x) = 2x^{2} + ^{1}/_{4} x + 13, when x = -1

(iv) p(x) = 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2 , when x = -2

- 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) = x^{2} – 4, x = 2 , x = -2

(iii) g(x) = 5x – 8, x = ^{4}/_{5}

(iv) p(x) = 3x^{3 }– 5x^{2} – 11x – 3, x = 3, x = -1 and x = –^{1}/_{3}

- Find the zeroes of the following quadratic polynomials and verify

(i)x^{2} + 4x + 4

(ii)x^{2} – 2x – 15

(iii) 4a^{2} – 49

(iv) 2a^{2} – 2√2a + 1

- If x = 1 is a zero of the polynomial f(x) = x
^{3}– 2x^{2 }+ 4x + k, find the value of k. - For what value of k, -4 is a zero of the polynomial x
^{2}– x – (2k + 2)?

#### Polynomials – Exercise 8.1 – Solution:

**Find the degree of the following polynomials.**

**(i) x ^{2} – 9x + 20**

Solution:

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

**(ii) 2x + 3 + 6x ^{2}**

Solution:

p(x) = 6x^{2} + 2x + 3 is polynomial in x is degree 2

**(iii) x ^{3} + 2x^{2} – 5x – 6**

Solution:

p(x) = x^{3} + 2x^{2} – 5x – 6 is a polynomial in x is degree 3

**(iv) x ^{3} + 17x – 21 – x^{2}**

Solution:

p(x) = x^{3} – x^{2} + 17x – 21 is a polynomial in x is degree 3

**(v) √3x ^{3} + 19x + 14**

Solution:

p(x) = √3x^{3} + 19x + 14 is a polynomial in x is degree 3

**If f(x) = 2x**^{3}+ 3x^{2}– 11x + 6, find.

**(i) f(0)**

**(ii)f(1)**

**(iii)f(-1)**

**(iv) f(2)**

**(v)f(-3)**

Solution:

f(x) = 2x^{3} + 3x^{2} – 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

**Find the values of the following polynomials.**

**(i) g(x) = 7x ^{2} + 2x + 14 when x = 1**

Solution:

g(x) = 7x^{2} + 2x + 14

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

= 7 + 2 + 14

= 23

**(ii) p(x) = -x ^{3} + x^{2} – 6x + 5 when x = 2**

Solution:

p(x) = -x^{3} + x^{2} – 6x + 5

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

= – 8 + 4 – 12 + 5

= – 11

**(iii) p(x) = 2x ^{2} + ^{1}/_{4} x + 13, when x = -1**

Solution:

p(x) = 2x^{2} + ^{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) = 2x ^{4} – 3x^{3} – 3x^{2} + 6x – 2 , when x = -2**

Solution:

p(x) = 2x^{4} – 3x^{3} – 3x^{2} + 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

**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) = x ^{2} – 4, x = 2 , x = -2**

Solution:

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

If x = 2, p(2) = (2)^{2} – 4

= 4 – 4

= 0

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

If x = -2, p(-2) = (-2)^{2} – 4

= 4 – 4

= 0

Therefore, x = -2 is the zero of the polynomial p(x) = x^{2} – 4.

Thus, x = 2 and x = -2 both are zero of the polynomial p(x) = x^{2} – 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) = 3x ^{3 }– 5x^{2} – 11x – 3, x = 3, x = -1 and x = –^{1}/_{3}**

Solution:

Given, p(x) = 3x^{3 }– 5x^{2} – 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) = 3x^{3 }– 5x^{2} – 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) = 3x^{3 }– 5x^{2} – 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) = 3x^{3 }– 5x^{2} – 11x – 3.

For x = –^{1}/_{3}, p(-^{1}/_{3}) = 3(-^{1}/_{3})^{3 }– 5(-^{1}/_{3})^{2} – 11(-^{1}/_{3}) – 3

= – 3x^{1}/_{27} – 5x^{1}/_{9} + 11(^{1}/_{3}) – 3

= –^{1}/_{9} – ^{5}/_{9} + ^{11}/_{3} – 3

= 0

Therefore, x = –^{1}/_{3} is the zero of the polynomial p(x) = 3x^{3 }– 5x^{2} – 11x – 3.

**Find the zeroes of the following quadratic polynomials and verify**

**(i)x ^{2} + 4x + 4**

Solution:

Given, f(x) = x^{2} + 4x + 4, we have to find the zero of the polynomial.

f(x) = x^{2} + 4x + 4 = 0

x^{2} + 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) = x^{2} + 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) = x^{2} + 4x + 4.

**(ii)x ^{2} – 2x – 15**

Solution:

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

f(x) = x^{2} – 2x – 15 = 0

x^{2} + 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) = x^{2} – 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) = x^{2} – 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) = x^{2} – 2x – 15

** (iii) 4a ^{2} – 49**

Solution:

Given, f(a) = 4a^{2} – 49, we have to find the zero of the polynomial.

f(a) = 4a^{2} – 49 = 0

4a^{2} – 49 = 0

4a^{2} = 49

(2a)^{2} = 7^{2}

2a = 7

a = ^{7}/_{2}

Therefore a = ^{7}/_{2} is the zero of the polynomial f(^{7}/_{2}) = 4a^{2} – 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) = 4a^{2} – 49

**(iv) 2a ^{2} – 2√2a + 1**

Solution:

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

f(a) = 2a^{2} – 2√2a + 1 = 0

2a^{2} – 2√2a + 1 = 0

(√2a)^{2} – 2(√2a)(1) + 1^{2} = 0

a^{2 } – 2ab + b^{2} = (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) = 2a^{2} – 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) = 2a^{2} – 2 √2a + 1

**If x = 1 is a zero of the polynomial f(x) = x**^{3}– 2x^{2 }+ 4x + k, find the value of k.

Solution:

Given, if x = 1 is a zero of the polynomial f(x) = x^{3} – 2x^{2 }+ 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

**For what value of k, -4 is a zero of the polynomial x**^{2}– x – (2k + 2)?

Solution:

Given, x = – 4 is the zero of the polynomial f(x) = x^{2} – 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**