AP State Syllabus AP Board 9th Class Maths Solutions Chapter 2 Polynomials and Factorisation Ex 2.4 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 2nd Lesson Polynomials and Factorisation Exercise 2.4

Question 1.

Determine which of the following polynomials has (x + 1) as a factor.

i) x^{3} – x^{2} – x + 1

Solution:

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

= -1 – 1 + 1 + 1 = 0

∴ (x + 1) is a factor.

ii) x^{4} -x^{3} +x^{2} – x + 1

Solution:

f(- 1) = (- 1)^{4} – (- 1)^{3} + (- 1)^{2} – (- 1) + 1

= 1 + 1 + 1 + 1 + 1= 5

∴ (x + 1) is not a factor.

iii) x^{4} + 2x^{3} + 2x^{2} + x + 1

Solution:

f(- 1) = (-1)^{4} + 2 (- 1)^{3} + 2 (- 1)^{2} + (-1) + 1

= 1 – 2 + 2 – 1 + 1 = 1

∴ (x + 1) is not a factor.

iv) x^{3} – x^{2} – (3 – √3)x + √3

Solution:

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

= – 1 – 1 + 3 – √3 + √3 = 1

∴ (x + 1) is not a factor.

Question 2.

Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:

i) f(x) = 5x^{3} + x^{2} – 5x – 1; g(x) = x + 1

[Factor theorem : If f(x) is a polynomial; f(a) = 0 then (x – a) is a factor of f(x); a ∈ R]

Solution:

g(x) = x+ 1 = x- a say

∴ a = – 1

f(a) = f(- 1) = 5 (- 1)^{3} + (- 1)^{2} – 5 (- 1) – 1

= -5 + 1 + 5 – 1 = 0

∴ x + 1 is a factor of f(x).

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

Solution:

g(x) = x + 1 = x – a

∴ a = – 1

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

= -1 + 3 – 3 + 1 =0

∴ f(x) is a factor of g(x).

iii) f(x) = x^{3} – 4x^{2} + x + 6;

g(x) = x – 2

Solution:

g(x) = x- 2 = x- a

∴ a = 2

f(a) = f(2) = 2^{3} – 4(2)^{2} + 2 + 6

= 8 – 16 + 2 + 6 = 0

∴ g(x) is a factor of f(x).

iv) f(x) = 3x^{3}+ x^{2} – 20x +12; g(x) = 3x – 2

Solution:

g(x) = 3x – 2 = \(x-\frac{2}{3}\) = x – a

∴ a = 2/3

v) f(x) = 4x^{3}+ 20x^{2}+ 33x + 18; g(x) = 2x + 3

Solution:

g(x) = 2x + 3 = x + \(\frac{3}{2}=\) = x – a

∴ a = -3/2

∴ g(x) is a factor of f(x).

Question 3.

Show that (x – 2), (x + 3) and (x – 4) are factors of x^{3} – 3x^{2} – 10x + 24.

Solution:

Given f(x) = x^{3} – 3x^{2} – 10x + 24

To check whether (x – 2), (x + 3) and (x – 4) are factors of f(x), let f(2), f(- 3) and f(4)

f(2) = 2^{3} – 3(2)^{2} – 10(2) + 24

= 8- 12-20 + 24 = 0

∴ (x – 2) is a factor of f(x).

f(- 3) = (- 3)^{3} – 3(- 3)^{2}– 10(- 3) + 24

= – 27 – 27 + 30 + 24 = 0

∴ (x + 3) is a factor of f(x).

f(4) = (4)^{3} – 3 (4)^{2} – 10 (4) + 24

= 64 – 48 – 40 + 24

= 88 – 88

= 0

∴ (x – 4) is a factor of f(x).

Question 4.

Show that (x + 4), (x – 3) and (x – 7) are factors of x^{3} – 6x^{2} – 19x + 84.

Solution:

Let f(x) = x^{3} – 6x^{2} – 19x + 84

To verify whether (x + 4), (x – 3) and (x – 7) are factors of f(x) we use factor theorem.

Let f(- 4), f(3) and f(7)

f(- 4) = (- 4)^{3} – 6 (- 4)^{2} – 19 (- 4) + 84

= -64 – 96 + 76 + 84

= 0 .

∴ (x + 4) is a factor of f(x).

f(3) = 3^{3} – 6(3)^{2} – 19(3) + 84

= 27 – 54 – 57 + 84

= 0

∴ (x – 3) is a factor of f(x).

f(7) = 7^{3} – 6(7)^{2} – 19(7) + 84

= 343 – 294 – 133 + 84

= 427 – 427

= 0

∴ (x – 7) is a factor of f(x).

Question 5.

If both (x – 2) and \(\left(x-\frac{1}{2}\right)\) of px^{2} + 5x + r, show that p = r.

Solution:

Let f(x) = px^{2}+ 5x + r

As (x – 2) and \(\left(x-\frac{1}{2}\right)\) are factor of f(x), we have f(2) = 0 and f(1/2) = 0

∴ f(2) = p(2)^{2} + 5(2) + r

= 4p + 10 + r = 0

= 4p + r

= – 10 ………………(1)

⇒ p + 10 + 4r = 0

⇒ p + 4r = – 10 ………………. (2)

From (1) and (2);

4p + r = p + 4r

4p – p = 4r – r

3p = 3r

∴ P = r

Question 6.

If (x^{2} – 1) is a factor of ax^{4} + bx^{3} + cx^{2} + dx + e, show that a + c + e = b + d = 0.

Solution:

Let f(x) = ax^{4} + bx^{3} + cx^{2} + dx + e

As (x – 1) is a factor of f(x) we have

x^{2} – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0

f(1) = a + b + c + d + e = 0 ……………. (1)

and f(-1) = a- b + c- d + e = 0

⇒ a + c + e = b + d

Substitute this value in equation (1)

a + c + e + b + d=0

b + d + b + d=0

2 (b + d) = 0

⇒ b + d = 0

∴ a + c + e = b + d = 0

Question 7.

Factorise

i) x^{3} – 2x^{2} – x + 2

Solution:

Let f(x) = x^{3} – 2x^{2} – x + 2

By trial, we find f(l) = 1^{3} – 2(1)^{2} – 1 + 2

= 1 – 2 – 1 + 2

= 0 .

∴ (x – 1) is a factor of f(x).

[by factor theorem]

Now dividing f(x) by (x – 1).

f(x) = (x – 1) (x^{2} – x – 2)

= (x – 1) [x^{2} – 2x + x- 2]

= (x – 1) [x (x – 2) + 1 (x – 2)]

= (x – 1) (x – 2) (x + 1)

ii) x^{3} – 3x^{2} – 9x – 5

Solution:

Let f(x) = x^{3} – 3x^{2} – 9x – 5By trial,

f(- 1) = (- 1)^{3} – 3(- 1)^{2} – 9(- 1) – 5

=-1 – 3 + 9 – 5

=0

∴ (x + 1) is a factor of f(x).

[ ∵ by factor theorem]

Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x^{2} – 4x – 5)

But x^{2}– 4x – 5 = x^{2} – 5x + x – 5

= x (x – 5) + 1 (x – 5)

=(x – 5)(x + 1)

∴ f(x)=(x + 1)(x + 1)(x – 5)

iii) x^{3} + 13x^{2} + 32x + 20

Solution:

Let f(x) = x^{3} + 13x^{2} + 32x + 20

Let f(- 1)

= (- 1)^{3} + 13 (- 1)^{2} + 32 (- 1) + 20

= – 1 + 13 – 32 + 20 = 33 – 33 = 0

∴ (x + 1) is a factor of f(x).

[ ∵ by factor theorem] Now dividing f(x) by (x + 1).

iv) y^{3} + y^{2} – y – 1

Let f(y) = y^{3} + y^{2} – y – 1

f(1) = 1^{3}+ 1^{2}– 1 – 1 = 0

(y – 1) is a factor of f(y).

Now dividing f(y) by (y – 1).

∴ f(x) = (x + 1)(x^{2} + 12x + 20)

But (x^{2} + 12x + 20) = x^{2}+ 10x + 2x + 20

=x(x + 10)+2(x + 10)

=(x + 10)(x + 2)

∴f(x) = (x + 1)(x + 2)(x + 10)

Question 8.

If ax^{2} + bx + c and bx^{2} + ax + c have a common factor x + 1 then show that c = 0 and a = b.

Solution:

Let f(x) = ax^{2} + bx + c and g(x) = bx^{2} + ax + c given that (x + 1) is a common factor for both f(x) and g(x).

∴ f(-1) = g(- 1)

⇒a(- 1)^{2} + b(- 1) + c

= b(- 1)^{2} + a (- 1) + c

⇒ a – b + c = b – a + c

⇒ a + a = b + b

⇒ 2a = 2b

⇒ a = b

Also f(- 1) = a – b + c = 0

⇒ b – b + c = 0

⇒ c = 0

Question 9.

If x^{2} – x – 6 and x^{2} + 3x – 18 have a common factor x – a then find the value of a.

Solution:

Let f(x) = x^{2} – x – 6 and

g(x) = x^{2} + 3x – 18

Given that (x – a) is a factor of both f(x) and g(x).

f(a) = g(a) = 0

⇒ a^{2} – a – 6 = a^{2} + 3a – 18

⇒ – 4a = – 18 + 6

⇒ – 4a = – 12

∴ a = 3

Question 10.

If (y – 3) is a factor of y^{3}– 2y^{2}– 9y + 18, then find the other two factors.

Solution:

Let f(y) = y^{3}– 2y^{2 }– 9y + 18

Given that (y – 3) is a factor of f(y).

Dividing f(y) by (y – 3)

∴ f(y) = (y – 3) (y + y – 6)

But y^{2} + y – 6

= y^{2} + 3y – 2y – 6

= y (y + 3) – 2 (y + 3)

= (y + 3) (y – 2)

∴ f(y) = (y – 2)(y – 3)(y + 3)

The other two factors are (y – 2) and (y + 3).