AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.2 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.2

Question 1.

Find the value of the polynomial 4x^{2} – 5x + 3, when

(i) x = 0

Solution:

The value at x = 0 is

4(0)^{2} – 5(0) + 3

= 3

(ii) x = – 1

Solution:

The value at x = – 1 is

4 (- 1)^{2} – 5 (- 1) + 3

= 4 + 5 + 3

= 12

iii) x = 2

Solution:

The value at x = 2 is

4(2)^{2} – 5(2) + 3

= 16 -10 + 3

= 9

iv) x = \(\frac{1}{2}\)

Solution:

Question 2.

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

i) p(x) = x^{2} – x + 1

Solution:

p(0) = 0^{2} – 0 + 1 = 1

p(1) = 1^{2} – 1 + 1 = 1

p(2) = 2^{2} – 2 + 1 = 3

ii) P(y) = 2 + y + 2y^{2} – y^{3}

Solution:

p(0) = 2 + 0 + 2(0)^{2} – 0^{3} = 2

p(1) = 2+ 1 + 2(1)^{2} – 1^{3} = 4

p(2) = 2 + 2 + 2(2)^{2} -2^{3} = 4 + 8- 8 = 4

iii) P(z) = z^{3}

Solution:

p(0) = 0^{3} = 0

p(1) = 1^{3} = 1

p(2) = 2^{3} = 8

iv) p(t) = (t – 1)(t + 1) = t^{2} – 1

Solution:

p(0) = (0 – 1) (0 + 1) = – 1

p(1) = t^{2} – 1 = 1^{2} – 1 = 0

p(2) = 2^{2} – 1 = 4 – 1 = 3

v) p(x) = x^{2} – 3x + 2

Solution:

p(0) = 0^{2} – 3(0) + 2 = 2

p(1) = 1^{2} – 3(1) + 2 = 1 – 3 + 2 = 0

p(2) = 2^{2} – 3(2) + 2 = 4- 6 + 2 = 0

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

Solution:

The value of p(x) at x = \(\frac{-1}{2}\) is

\(\mathrm{p}\left(\frac{-1}{2}\right)=2\left(\frac{-1}{2}\right)+1\)

= -1 + 1 = 0

∴ x = \(\frac{-1}{2}\) is a zero of p(x).

(ii) p(x) = 5x – π ; x = \(\frac{-3}{2}\)

Solution:

The value of p(x) at x = \(\frac{-3}{2}\) is

\(\mathrm{p}\left(\frac{-3}{2}\right)=5\left(\frac{-3}{2}\right)-\pi=\frac{-15}{2}-\pi \neq 0\)

∴ x = \(\frac{-3}{2}\) is not a zero of p(x).

iii) p(x) = x^{2} – 1; x = ±1

Solution:

The value of p(x) at x = 1 and – 1 is

p(1) = 1^{2} – 1 = 0

p(-1) = (-1)^{2} -1 = 0

∴ x = ±1 is a zero of p(x).

iv) p(x) = (x – 1) (x + 2); x = – 1, – 2

Solution:

The value of p(x) at x = – 1 is

p(-1) = (-1 – 1) (-1 + 2)

=-2 x 1 =-2 ≠ 0

Hence x = – 1 is not a zero of p(x).

And the value of p(x) at x = – 2 is

p (- 2) = (- 2 – 1) (- 2 + 2) = – 3 x 0 = 0

Hence, x = – 2 is a zero of p(x).

v) p(y) = y^{2}; y = o

Solution:

The value of p(y) at y = 0 is p(0) = 0^{2} = 0

Hence y = 0 is a zero of p(y).

vi) p(x) = ax + b ; x = \(\frac{-\mathbf{b}}{\mathbf{a}}\)

Solution:

The value of p(x) at x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is

\(\mathrm{p}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)=\mathrm{a}\left(\frac{-\mathrm{b}}{\mathrm{a}}\right)+\mathrm{b}\)

= -b + b = 0

∴ x = \(\frac{-\mathbf{b}}{\mathbf{a}}\) is a zero of p(x).

vii) f(x) = 3x^{2} – 1; x = \(\frac{-1}{\sqrt{3}}, \frac{2}{\sqrt{3}}\)

Solution:

viii) f(x) = 2x – 1; x = \(\frac{1}{2} ;-\frac{1}{2}\)

Solution:

Question 4.

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

i) f(x) = x + 2

Solution:

x + 2 = 0

x = – 2

ii) f(x) = x – 2

Solution:

x – 2 = 0

x = 2

iii) f(x) = 2x + 3

Solution:

2x + 3 = 0

2x = – 3

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

iv) f(x) = 2x – 3

Solution:

2x – 3 = 0

2x = 3

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

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

Solution:

x^{2} = 0

x = 0

vi) f(x) = px, p ≠ 0

Solutin:

px = 0

x = 0

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

Solution:

px + q = 0

px = -q

x = \(\frac{-\mathrm{q}}{\mathrm{p}}\)

Question 5.

If 2 is a zero of the polynomial p(x) = 2x^{2} – 3x + 7a, find the value of

a.

Solution:

Given that 2 is a zero of p(x) = 2x^{2} – 3x + 7a

(i.e.) p(2) = 0

⇒ 2(2)^{2} – 3(2) + 7a = 0

⇒ 8 – 6 + 7a = 0

⇒ 2 + 7a = 0

⇒ 7a = – 2

⇒ a = \(\frac{-2}{7}\)

Question 6.

If 0 and 1 are the zeroes of the polynomial f(x) = 2x^{3} – 3x^{2} + ax + b, find the values of a and b.

Solution:

Given that f(0) = 0; f(1) = 0 and

f(x) = 2x^{3} – 3x^{2} + ax + b

∴ f(0) = 2(0)^{3} – 3(0)^{2} + a(0) + b

⇒ 0 = b

Also f(1) = 0

⇒ 2(1)^{3} – 3(1)^{2} + a(1) + 0 = 0

⇒ 2 – 3 + a = 0 .

⇒ a = 1

Hence a = 1; b = 0