**Telangana SCERT Solution Class IX (9) Math Chapter 2 Polynomials and Factorisation Exercise 2.3**

**(2) given f(x) = x ^{3} – px^{2} + 6x – p ; x – p**

∴ using remainder theorem, we can say that,

Remainder of f(x) = f(p)

∴ f(p) = p^{3} – p x p^{2} + 6 x p – p

= p^{3} – p^{3} + 6p – p

= 5p

**(5) Given, f(x) 2x ^{3} + ax^{2} + 3x – 5 ; p(x)x^{3} + x^{2} – 4x + a**

Divided by x -2

By using remainder theorem we get

∴ f(2) = 2 x 2^{3} + a x 2^{2} + 3 x 2 – 5

= 16 + 4a + 6 – 5

= 4a + 17

P(2) = 2^{3} + 2^{2} – 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) = x ^{3} + ax^{2} + 5 ; p(x) = x^{3} -2x^{2} + 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

I 9th bit is missing

Please write these answers in urdu

OP