**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

worst thing some are missing

what are you maintaining

state syllabus answers are difficult to find where cbse is very easy to find

9th bit is missing