Telangana SCERT Class 9 Math Solution Chapter 2 Polynomials and Factorisation Exercise 2.2

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

(1) Given, p(x) = 4x² – 5x + 3

(i) Now, x = 0

∴ p(o) = 4×0² – 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) = 4×2² – 5×2 + 3

= 16 – 10 + 3

= 9

(2) (i) p(x) = x² – x + 1

∴ p(0) = 0² – 0 + 1

= 1

P(1) = 1² -1 + 1

= 1

P(2) = 2² – 2 + 1

= 3

(ii) p(y) = 2 + y + 2y² – y³

∴ p(0) = 2 + 0 + 2 x 0² – 0³

= 2

P(1) = 2 + 1 + 2 x 1² – 1³

= 3 + 2 – 1

= 4

P(2) = 2 + 2 + 2 x 2² – 2³

= 2 + 2 + 8 – 8

= 4

(iii) p(z) = z³

∴ p(0) = 0³

= 0

P(1) = 1³

= 1

P(2) = 2³

= 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) = x² – 3x + 2

∴ p(0) = 0² – 3×0 + 2

= 2

P(1) = 1² – 3×1 + 2

= 0

P(2) = 2² – 3×2 + 2

= 4 – 6 + 2

= 0

(iii) p(x) = x² – 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) = y² ; y = 0

P(0) = 0²

= 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) = x²

Or, ∴ f(x) = 0

Or, x² = 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) = 2x² – 3x + 7a

Or, p(2) = 2x² -3x + 7a = 0

∴2 x 2² – 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) = 2x³ – 3x² + ax + b

Or, f(0) = 2 x 0³ – 3 x 0² + 0 x 0 + b

= b

Given, f(0) = 0

∴ b = 0 ….(i)

Now,

f(1) = 2 x 1³ – 3 x1² + 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.