AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.3 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.3

Question 1.

Find the remainder when

x^{3} + 3x^{2} + 3x + 1 is divided by the following Linear polynomials i) x + 1 Each

Solution:

Let f(x) = x^{3} + 3x^{2} + 3x + 1

By remainder theorem, the remainder is f (- 1)

f (- 1) = (- 1)^{3} + 3(- 1)^{2} + 3(- 1) + 1

= – 1 + 3 – 3 + 1 = 0

ii) \(x-\frac{1}{2}\)

f(x) = x^{3} + 3x^{2} + 3x + 1

By remainder theorem, the remainder is \(\mathrm{f}\left(\frac{1}{2}\right)\)

iii) x

Solution:

f(x) = x^{3} + 3x^{2} + 3x + 1

The remainder is f(0)

∴ f(0) = 0^{3} + 3(0)^{2} + 3(0) + 1 = 1

iv) x + π

Solution:

f(x) = x^{3} + 3x^{2} + 3x + 1

By remainder theorem, the remainder is f(- π)

f(- π) = (- π)^{3} + 3(-π)^{2} + 3 (-π) + 1 .

= – π^{3} + 3π^{2} – 3π + 1

v) 5 + 2x

f(x) = x^{3} + 3x^{2} + 3x + 1

The remainder is \(\mathrm{f}\left(\frac{-5}{2}\right)\)

Question 2.

Find the remainder when x^{3} – px^{2} + 6x – p is divided by x – p.

Solution:

Let f(x) = x^{3} – px^{2} + 6x – p

(x – a) = x – p)

By Remainder theorem, the remainder is f(p)

∴ f(P) = P^{3} – P(P)^{2} + 6p – p

= p^{3} – p^{3} + 5p = 5p

Question 3.

Find the remainder when 2x^{2} – 3x + 5 is divided by 2x – 3. Does it exactly divide the polynomial ? State reason.

Solution:

Let f(x) = 2x^{2} – 3x + 5 and

x – a = 2x – 3 = x – \(\frac{3}{2}\)

By Remainder theorem f(x) when divided by (x – \(\frac{3}{2}\) ) leaves a remainder \(\mathrm{f}\left(\frac{3}{2}\right)\)

As the remainder is 5 we say that (2x – 3) is not a factor of f(x).

Question 4.

Find the remainder when 9x^{3} – 3x^{2} + x – 5 is divided by x – \(\frac{2}{3}\)

Solution:

Let f(x) = 9x^{3} – 3x^{2} + x – 5

x-a = x – \(\frac{2}{3}\)

Remainder theorem the remainder is \(\mathrm{f}\left(\frac{2}{3}\right)\)

Question 5.

If the polynomials 2x^{3} + ax^{2} + 3x – 5 and x^{3} + x^{2} – 4x + a leave the same remainder, when divided by x – 2, find the value of a.

Solution:

Let f(x) = 2x^{3} + ax^{2} + 3x – 5

g(x) = x^{3} + x^{2} – 4x + a

Given that f(x) and g(x) divided x – 2

give same remainder.

i e., f(2) = g(2)

By Remainder theorem.

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

= 2 x 8 + 4a + 6 – 5

= 17 +4a

g(2) = 2^{3} + 2^{2} – 4(2) + a .

= 8 + 4 – 8 + a = 4 + a

i.e., 4 + a = 17 + 4a

∴ a – 4a = 17 – 4

– 3a = 13

a = -13/3

Question 6.

If the polynomials x^{3} + ax^{2 }+ 5 and x^{3} – 2x^{2} + a are divided by (x + 2) leave the same remainder, find the value of a.

Solution:

Let f(x) = x^{3} + ax^{2} + 5

g(x) = x^{3} – 2x^{2} + a

Given that when f(x) and g(x) divided by (x + 2) leaves the same remainder.

i.e.,f(-2) = g(-2)

By Remainder theorem

f(- 2) = (- 2)^{3} + a(- 2)^{2} + 5

= -8 + 4a + 5 = 4a – 3

g(- 2) = (- 2)^{3} – 2(- 2)^{2} + a

= -8 – 8 + a = a – 16

By problem,

4a – 3 = a – 16

4a – a = – 16 + 3

⇒ 3a = – 13 ⇒ a = -13/3

Question 7.

Find the remainder when f(x) = x^{4} – 3x^{2} + 4 is divided by g(x) = x – 2 and verify the result by actual division.

Solution:

Given f(x) = x^{4} – 3x^{2} + 4

g(x) = x – 2

The remainder when f(x) is divided by g(x) is f(2).

f(2) = 2^{4} – 3(2)^{2} + 4

= 16 – 12 + 4

= 8

Actual division

∴ The remainder either by Remainder theorem or by actual division is the same.

Question 8.

Find the remainder when p(x) = x^{3} – 6x^{2} + 14x – 3 is divided by g(x) = 1 – 2x and verify the result by long division method.

Solution:

Given p(x) = x^{3} – 6x^{2} + 14x – 3

g(x) = 1 – 2x

By Remainder theorem when p(x) is divided by g(x) is p(1/2).

Question 9.

When a polynomial 2x^{3} + 3x^{2} + ax + b is divided by (x – 2) leaves remainder 2, and (x + 2) leaves remainder – 2. Find a and b.

Solution:Let f(x) = 2x^{3} + 3x^{2} + ax + b

The remainder when f(x) is divided by (x – 2) is 2.

i.e., f(2) = 2

⇒ 2(2)^{3} + 3(2)^{2}+ a(2) + b = 2

⇒ 16 + 12 + 2a +b = 2

⇒ 2a + b = – 26 …………………..(1)

Also the remainder when f(x) is divided by (x + 2) is – 2.

i.e., f(- 2) = – 2

⇒ 2(- 2)^{3} + 3(- 2)^{2} + a (- 2) + b = – 2

⇒ -16 + 12 – 2a + b = – 2

– 2a + b = 2 ………………..(2)

Solving (1) and (2),

b = – 12

and 2a – 12 = – 26

2a = -26+ 12

a = -14/2 = -7,

a = -7, b = – 12