**Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.2**

**(1) **Given, p(x) = 4x^{2} – 5x + 3

(i) Now, x = 0

∴ p(o) = 4×0^{2} – 5 x 0 + 3

= 3

(ii) Now, x = -1

∴p(1) = 4x(-1)^{2} – 5(-1) + 3

= 4×1 + 5 + 3

= 12

(iii) No, x = 2

∴ p(2) = 4×2^{2} – 5×2 + 3

= 16 – 10 + 3

= 9

**(2) (i) p(x) = x ^{2} – x + 1**

∴ 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}**

∴ p(0) = 2 + 0 + 2 x 0^{2} – 0^{3}

= 2

P(1) = 2 + 1 + 2 x 1^{2} – 1^{3}

= 3 + 2 – 1

= 4

P(2) = 2 + 2 + 2 x 2^{2} – 2^{3}

= 2 + 2 + 8 – 8

= 4

**(iii) p(z) = z ^{3}**

∴ p(0) = 0^{3}

= 0

P(1) = 1^{3}

= 1

P(2) = 2^{3}

= 8.

**(IV) P(t) = (t – 1) (t + 1) **

∴ p(0) = (0 -1) (0+1)

= -1 x +1

= -1

P(1) = (1 – 1) (1 + 1)

= 0 x 2

= 0

P(2) = (2-1) (2+1)

= 1 x 3

= 3

**(v) p(x) = x ^{2} – 3x + 2**

∴ p(0) = 0^{2} – 3×0 + 2

= 2

P(1) = 1^{2} – 3×1 + 2

= 0

P(2) = 2^{2} – 3×2 + 2

= 4 – 6 + 2

= 0

**(iii) p(x) = x ^{2} – 1; x = ±1**

P(±1) = (±1)^{2} – 1

= 1 – 1

= 0

∴ x = ±1 is the zero of the above given polynomial.

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

P(-1) = (-1 -1) (-1 + 2)

= -2 x 1

= -2 ≠0

P(-2) = (-2 – 1) (-2 + 2)

= -3 x 0

= 0

∴ x = -1 is not the zero of the given polynomial but x = -2 is.

**(iv) p(y) = y ^{2} ; y = 0**

P(0) = 0^{2}

= 0

∴ y = 0 is the zero of the above given polynomial.

**(4) (i) f(x) = x+2**

∴ f(x) = 0

Or, x + 2 = 0

Or, x = -2

∴ x = -2 is the zero of the above given polynomial.

(ii) to given,

f(x) = x – 2

∴ f(x) = 0

Or, x – 2 = 0

Or, x = 2

∴ x = 2 is the zero of the given polynomial.

**(iii) given f(x) = 2x + 3**

∴ f(x) = 0

Or, 2x + 3 = 0

Or, 2x = -3

Or, x = -3/2

∴ x = -3/2 is the zero of the given polynomial.

**(iv) f(x) = 2x – 3**

∴ f(x) = 0

Or, 2x – 3 = 0

Or, x = 3/2

∴ x = 3/2 is the zero of the above given polynomial.

**(v) given, f(x) = x ^{2}**

Or, ∴ f(x) = 0

Or, x^{2} = 0

Or, x = 0

∴ x = 0 is the zero of the above given polynomial.

**(vi) f(x) = px, p ≠ 0**

f(x) = 0

Or, px = 0

∴ x = 0 [∵ p ≠ 0]

∴ x = 0 is the zero of the given polynomial.

**(vii) f(x) = px + q, p ≠ 0, p and q are the real number.**

f(x) = 0

or, px + q = 0

Or, px = -q

Or, x = -q/p

∴ x = -q/p is the zero of the above given polynomial.

**(5) Given, p(x) = 2x ^{2} – 3x + 7a**

Or, p(2) = 2x^{2} -3x + 7a = 0

∴2 x 2^{2} – 3 x 2 + 7a = 0

Or, 2 x 4 – 6 + 7a = 0

Or, 2 + 7a = 0

Or, 7a = -2

Or, a = -2/7

**(6) Given, f(x) = 2x ^{3} – 3x^{2} + ax + b**

Or, f(0) = 2 x 0^{3} – 3 x 0^{2} + 0 x 0 + b

= b

Given, f(0) = 0

∴ b = 0 ….(i)

Now,

f(1) = 2 x 1^{3} – 3 x1^{2} + a x 1 + b

= 2 – 3 + a + b

= a + b -3

Given, f(1) = 0

Or, a + b – 3 = 0

Or, a + b – 3 = 0 [putting the value of b from eqn(i)]

Or, a = 3

∴ a = 3 and b = 0.

The last answer is a=1, and b=0. A is not equal to 3. Please correct it..

Number Please