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