Gujarat Board GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.4 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.4

Question 1.

Determine which of the following polynomials has x + 1 a factor:

1. x^{3} + x^{2} + x + 1

2. x^{4} + x^{2}+ x^{2} + x + 1

3. x^{4} + 3x^{3} + 3x^{2} + x + 1

4. x^{3} – x^{2} – (2 + \(\sqrt{2}\))x + \(\sqrt{2}\)

Solution:

1.Â Let p(x) = x^{3} + x^{2} + x + 1

x + 1 = 0

x = -1 (If x + 1 is a factor of p(x) then x + 1 will be equal to zero)

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

= -1 + 1 – 1 + 1 = 0

âˆ´ By factor theorem x + 1 is a factor of x^{3} + x^{2} + x + 1.

2. Let p(x) = x^{4} + x^{3} + x^{2} + x + 1

â‡’ x + 1 = 0

x = -1

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

= 1 – 1 + 1 – 1 + 1 = 3 – 2

â‡’ p(-1) = 1

âˆ´ 1 â‰ 0

Hence by factor theorem x + 1 is not a factor of x^{4} + x^{3} + x^{2} + x + 1.

3. Let p(x) = x^{4} + 3x^{3} + 3x^{2} + x + 1

x + 1 = 0

â‡’ x = -1

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

p(-1) = 1 – 3 + 3

p(-1) = 1

âˆ´ 1â‰ 0

Hence by factor theorem x + 1 is not a factor of x^{4} + 3x^{3} + x^{2} + x + 1.

4. Let p(x) = x^{3} – x^{2} – (2 + \(\sqrt {2} \)) x + \(\sqrt {2} \)

x + 1 = 0

â‡’ x = – 1

p(-1) = (-1)^{3} – (-1)^{2} – (2 +\(\sqrt {2} \))(-1) + \(\sqrt {2} \)

p(-1) = -1 – 1 + 2 + \(\sqrt {2} \) + \(\sqrt {2} \)

â‡’ p(-1) = 2\(\sqrt {2} \)

âˆ´ 2\(\sqrt {2} \) = 0

Hence by factor theorem x + 1 is not a factor of x^{3} – x^{2} – (2 + \(\sqrt {2} \))x + \(\sqrt {2} \)

Question 2.

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

1. p(x) = 2x^{3} + x^{2} – 2x – 1, g(x) = x + 1

2. p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

3. p(x) = x^{3} – 4x^{2} + x + 6, g(x) = x – 3

Solution:

1. p(x) = 2 x^{3} + x^{2} – 2x – 1

g(x) = x + 1

g(x) = 0

â‡’ x + 1 = 0

â‡’ x = -1

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

= -2 + 1 + 2 – 1

â‡’ p(-1) = 0

âˆ´ By factor theorem g(x) is a factor of p(x).

2. p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

g(x) = 0

x + 2 = 0

x = -2

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

= p(-2) = – 8 + 3 x 4 – 6 + 1

= p(-2) = – 8 + 12 – 6 + 1

= p(-2) = -1

âˆ´ -1 â‰ 0

Hence by factor theorem g(x) is not a factor of p(x).

3. p(x) = x^{3} – 4x^{2} + x + 6, g(x) = x – 3

g(x) = 0

â‡’ x – 3 = 0

â‡’ x = 3

p(3) = 3^{3} – 4(3)^{2} + 3 + 6

â‡’ p(3) = 27 – 4 x 9 + 9

â‡’ p(3) = 27 – 36 + 9

â‡’ p(3) = 36 – 36

p(3) = 0

Hence by factor theorem g(x) is a factor of p(x).

Question 3.

Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:

1. p(x) = x^{2} + x + k

2. p(x) = 2x^{2} + kx + \(\sqrt {2} \)

3. p(x) = kx^{2} – \(\sqrt {2} \) x + 1

4. p(x) = kx^{2} – 3x + k

Solution:

1. p(x) = x^{2} + x + k

x – 1 will be a factor p(x) therefore p(1) = 0.

By factor theorem

:. p(1) = 1^{2} + 1 + k

0 = 2 + k = k = -2

2. p(x) = 2 x^{2} + kx + \(\sqrt{2}\)

If x – 1 will be a factor of p(x) therefore p(1) = 0.

By factor theorem,

p(1) = 2(1)^{2} + k x 1 + \(\sqrt{2}\)

0 = 2 + k + \(\sqrt{2}\)

k = -2 – \(\sqrt{2}\)

k = – (2 + \(\sqrt{2}\))

3. p (x) = kx^{2} – \(\sqrt{2}\)x + 1

x – 1 will be a factor of p(x), therefore p(1) = 0.

By factor theorem,

p(1) = k(1)^{2} – \(\sqrt{2}\) x 1 + 1

0 = k – \(\sqrt{2}\) + 1

k = \(\sqrt{2}\) – 1

4. p(x) = kx^{2} – 3x + k

x – 1 will be a factor of p(x),

therefore, p(1) = 0.

By factor theorem,

p(1) = k(1)^{2} – 3(1) + k

0 = k – 3 + k

0 = 2k – 3 = k = \(\frac {3}{2}\)

Question 4.

Factorise: 2

1. 12 x^{2} – 7x + 1

2. 2x^{2} + 7x + 3

3. 6x^{2} + 5x – 6

4. 3x^{2} – x – 4

Solution:

1. 12 x^{2} – 7x + 1 (a x c = 12 x 1)

= 12 x^{2} – 4x – 3x + 1 (12 = 4 x 3)

= 4 x (3x -1) -1(3x – 1.)

= (3x – 1)(4x – 1)

2. 2x^{2} + 7x + 3 (a x c = 2 x 3)

-2x^{2} + 6x +x + 3 (6 = 6 x 1)

= 2x (x + 3) + 1(x + 3)

= (x + 3) (2x + 1)

3. 6x^{2} + 5x – 6 (a x c = 6 x 6 = 36)

= 6x^{2} + 9x – 4x – 6 (36 = 9 x 4)

= 3x(2x + 3) -2(2x + 3)

= (2x + 3)(3x – 2)

4. 3x^{2} – x – 4 (a x c = 3 x 4)

= 3x^{2} – 4x + 3x – 4 (12 = 4 x 3)

= x (3x – 4)+ 1(3x – 4)

= (3x – 4) (x + 1)

Question 5.

Factorise:

1. x^{3} – 2x^{2} – x + 2

2. x^{3} – 3x^{2} – 9x – 5

3.Â x^{3}+ 13x^{2} + 32x + 20

5. 2y^{2} + y^{2} – 2y – 1

Solution:

1. Let p(x) = x^{3} – 2x^{2} – x + 2

Here constant form is 2 and its factors can be Â±1, Â±2.

By trial method we find factor of p(x).

Let x – 1 be a factor of p(x).

âˆ´ x – 1 = 0

â‡’ x = 1

p(1) = 1^{3} – 2(1)^{3} – 1 + 2

= 1 – 2 – 1 + 2 = 3 – 3

â‡’ p(1) = 0

âˆ´ By factor theorem x – 1 is a factor of p(x).

Now x^{3} – 2x^{3} – x + 2

= x^{2} – x^{2} + x – 2x + 2

= x(x – 1) -x(x – 1)-2(x – 1)

= (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)

2. Let p(x) = x^{3} – 3x^{2} – 9x – 5

Here constant term is 5 and its factors can be Â±1, Â±5.

By trial method, we find factor of p(x).

Let x – 1 be a factor of p(x).

âˆ´ x – 1= 0

â‡’ x = 1

p(1) = 1^{3} – 3(1)^{2} – 9(1) – 5

= 1 – 3 – 9 – 5 = -16

But -16â‰ 0

âˆ´ x – 1 is not a factor of p(x).

Now let x + 1 be a factor of p(x).

âˆ´ x + 1 = 0

â‡’ x = -1

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

= -1 – 3 + 9 – 5 = -9 + 9

p(-1) = 0

âˆ´ x + 1 is a factor of p(x).

Now x^{3} – 3x^{2} – 9x – 5

= x^{3} + x^{3} – 4x^{3} – 4x – 5x – 5

= x^{2}(x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x^{2} – 4x – 5)

= (x + 1)(x^{2} – 5x + x – 51

= (x + 1) [x(x – 5)+ 1(x – 5)]

= (x + 1)(x – 5)(x + 1)

= (x + 1)(x + 1)(x – 5)

3. Let p(x) = x^{3} + 13x^{2} + 32x + 20 Here constant term is 20 and its factors can be Â±1, Â±2, Â±4, Â±5, Â±10, Â±20

By trial method, we find the factor of p(x).

Let x + 1 be a factor of p(x).

âˆ´ x + 1 = 0

x = – 1

p(-1) = (-1) + 13(-1)2 + 32(-1) + 20

= -1 + 13 – 32 + 20 = 33 – 33

p(-1) = 0

x + 1 is a factor of p(x).

Hence X^{3} + 13x^{2}+ 32x + 20

= x^{3} +x^{2} + 12x^{2} + 12x + 20x + 20

= x^{2}(x + 1) + 12x(x + 1) + 20(x + 1)

= (x + 1)(x^{2} + 12x + 20)

= (x + 1)[x^{2} + 10 x + 2x + 20]

= (x + 1)[x(x + 10)+ 2(x + 10)]

= (x + 1)(x+1o)(x + 2)

= (x + 1)(x + 2)(x + 10)

4. Let p(y) = 2y^{2} + y^{2} – 2y – 1

By trial method, we find the factor of p(y).

Let y – 1 be a factor of p(y).

y – 1 = 0 â‡’ y = 1

p(1) = 2(1)^{3} + 1^{2} – 2 x 1 – 1

= 2 + 1 – 2 – 1 = 3 – 3

p(1) = 0

âˆ´ y – 1 is a factor of p(y).

Hence 2y^{3} + y^{2} – 2y – 1

2y^{3} – 2y^{2} + 3y^{2} – 3y + y – 1

= 2y^{2}(y – 1) + 3y(y – 1) + 1(y – 1)

= (y – 1)[2y^{2} + 3y + 11]

= (y – 1) [2y^{2} + 2y + y + 1]

= (y -1)[2y(y + 1) +1 (y + 1)]

= (y – 1)(y + 1)(2y + 1)