# Telangana SCERT Class 9 Math Solution Chapter 2 Polynomials and Factorisation Exercise 2.3

## Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.3    (2) given f(x) = x3 – px2 + 6x – p ; x – p

∴ using remainder theorem, we can say that,

Remainder of f(x) = f(p)

∴ f(p) = p3 – p x p2 + 6 x p – p

= p3 – p3 + 6p – p

= 5p (5) Given, f(x) 2x3 + ax2 + 3x – 5 ; p(x)x3 + x2 – 4x + a

Divided by x -2

By using remainder theorem we get

∴ f(2) = 2 x 23 + a x 22 + 3 x 2 – 5

= 16 + 4a + 6 – 5

= 4a + 17

P(2) = 23 + 22 – 4×2 + a

= 8 + 4 – 8 + a

= a + 4

Now,

Given that remainder of f(x) and p(x) is same when divided by x – 2

∴ f(2) = p(2)

Or, 4a + 17 = a + 4

Or, 4a – a = 4 – 17

Or, 3a = -13

Or, a = -13/3

(6) given, f(x) = x3 + ax2 + 5 ; p(x)  = x3 -2x2 + a

Divided by x + 2

∴ using remainder theorem,

f(-2) = (-2)3 + a x (-2)2 + 5

= 8 + a x 4 + 5

= 4a – 3

P(-2) = (-2)3 – 2(-2)2 + a

= -8 – 8 + a

= a – 16

Now, given that

f(-2) = p(-2)

∴ 4a – 3 = a – 16

Or, 4a – a = -16 + 3

Or, 3a = -13

Or, a = -13/3 Updated: September 18, 2021 — 4:41 pm

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