Exercise 2.2

1. Find the value of the polynomial 4x2 - 5x + 3, at:

i. x = 0

Solution:

Substitute x = 0 in 4x2−5x+3:

402−50+3 =

ii. x = -1

Solution:

Substitute x = 1 in 4x2 - 5x + 3:

4−12−5−1+3 = + + 3 =

iii. x = 2

Solution:

Substitute x = 2 in 4x2 - 5x + 3:

422−52+3 = - + 3 =

iv. x = 12

Solution:

Substitute x = 3 in 4x2 - 5x + 3:

4122−512+3 = - 525 + 3 = - 5/2 = 3252

2. Find p(0), p(1) and p(2) for each of the following polynomials.

i. p(x) = x2 - x + 1

Solution:

p(0) = 02−0+1 =

p(1) = 12 - 1 + 1 =

p(2) = 22 - 2 + 1 =

ii. p(y) = 2 + y + 2y2 - y3

Solution:

p(0) = + + - =

p(1) = + + - =

p(2) = + + - =

iii. p(z) = z3

Solution:

p(0) = 03 =

p(1) = 13 =

p(2) = 23 =

iv. p(t) = (t - 1)(t + 1)

Solution:

p(0) = (0 - 1)(0 + 1) =

p(1) = (1 - 1)(1 + 1) =

p(2) = (2 - 1)(2 + 1) =

v. p(x) = x2 - 3x + 2

Solution:

p(0) = 02 - 3(0) + 2 =

p(1) = 12 - 3(1) + 2 =

p(2) = 22 - 3(2) + 2 =

3. Verify whether the values of x given in each case are the zeroes of the polynomial or not?.

i. p(x) = 2x + 1; x = −12

Solution:

Is a zero: YesNo

p−12 = 2−12 + 1

= + 1 =

So, it is a zero.

ii. p(x) = 5x - π; x = 32

Solution:

Is a zero: NoYes

p32 = 532 - π

= 152102 - π ≠= 0.

So, it is not a zero.

iii. p(x) = x2 - 1; x = +1

Solution:

Is a zero: YesNo

p(1) = 12 - 1 =

So, it is a zero.

iv. p(x) = (x - 1)(x + 2); x = -1, -2

Solution:

Is a zero: YesNo

p(-1) = (-1 - 1)(-1 + 2) =

p(-2) = (-2 - 1)(-2 + 2) =

So, -2 is a zero but not -1.

v. p(y) = y2; y = 0

Solution:

Is a zero: YesNo

p(0) = 02 =

So, it is a zero.

vi. p(x) = ax + b; x = −ba

Solution:

Is a zero: YesNo

p−ba = a−ba + b

= + b =

So, it is a zero.

vii. f(x) = 3x2 - 1; x = −13, 13

Solution:

Is a zero: YesNo

f−13 = 3−132 - 1 = 3 × - 1 = - 1 =

f13 = 3132 - 1 = 3 × - 1 = - 1 =

So, −13 and 13 is a zero.

viii. f(x) = 2x - 1; x = 12, −12

Solution:

(i) For x = 12: Is it a zero: YesNo

f12 = 212 - 1 = - 1 =

So, 1/2 is a zero.

Is it a zero: YesNo

(ii) For x = −12

f−12 = 2−12 - 1 = - 1 =

So, -1/2 is not a zero.

4. Find the zero of the polynomial in each of the following cases.

i. f(x) = x + 2

Solution:

x + 2 = 0 => x =

ii. f(x) = x - 2

Solution:

x - 2 = 0 => x =

iii. f(x) = 2x + 3

Solution:

2x + 3 = 0 => 2x = => x = −3232

iv. f(x) = 2x - 3

Solution:

2x - 3 = 0 => 2x = => x = 32−32

v. f(x) = x2

Solution:

x2 = 0 => x =

vi. f(x) = px, p ≠ 0

Solution:

px = 0 => x =

vii. f(x) = px + q, p ≠ 0, p, q are real numbers.

Solution:

px + q = 0 => px = => x = −qpqp

5. If 2 is a zero of the polynomial p(x) = 2x2 - 3x + 7a, then find the value of a.

Solution:

Substitute 2 in the given equation:

p(2) = 2(2242) - 3() + = 0

- + 7a = 0

+ 7a = 0

7a =

a = −2727