Bihar Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.4 Textbook Questions and Answers.

## BSEB Bihar Board Class 9th Maths Solutions Chapter 2 Polynomials Ex 2.4

Question 1.

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

(i) x³ + x² + x + 1

(ii) x^{4} + x³ + x² + x + 1

(iii) x^{4} + 3x³ + 3x² + x + 1

(iv) x³ – x² – (2 + \(\sqrt{2}\))x + \(\sqrt{2}\)

Solution:

(i) In order to prove that x + 1 is a factor of p(x) = x³ + x² + x + 1, it is sufficient to show that p(- 1) = 0.

Now, p(- 1) = (- 1)³ + (- 1)² + (- 1) + 1

= – 1 + 1 – 1 + 1 = 0

Hence, (x + 1) is a factor of pix) = x³ + x² + x + 1.

(ii) In order to prove that (x + 1) is a factor of

p(x) = x^{4} + x³ + x² + x + 1, it is sufficient to show that p(- 1) = 0.

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

= 1 – 1 + 1 – 1 + 1 = 1 ≠ 0

∴ (x + 1) is not a factor of x^{4} + x³ + x² + x + 1.

(iii) In order to prove that (x + 1) is a factor of

p(x) = x^{4} + 3x³ + 3x² + x + 1, it is sufficient to show that p(- 1) = 0.

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

= 1 – 1 + 1 – 1 + 1 = 1 ≠ 0

∴ (x + 1) is not a factor of x^{4} + x³ + x² + x + 1.

(iv) In order to prove that (x + 1) is a factor of

p(x) = x³ – x² – (2 + \(\sqrt{2}\))x + \(\sqrt{2}\), it is sufficient to show that p(- 1) = 0.

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

= 1 – 1 + 2 + \(\sqrt{2}\) + \(\sqrt{2}\) = 2\(\sqrt{2}\) ≠ 0

∴ (x + 1) is not a factor of x³ – x² (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 :

(i) p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1

(ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2

(iii) p(x) = x³ – 4x² + x + 6, g(x) = x – 3

Solution:

(i) In order to prove that g(x) = x + 1 is a factor of p(x) = 2x³ + x² – 2x – 1, it is sufficient to show that p(- 1) = 0

Now, p(- 1) = 2(- 1)³ + (- 1)² – 2(- 1) – 1

= – 2 + 1 + 2 – 1 = 0

∴ g(x) is a factor of p(x).

(ii) In order to prove that gix) = x + 2 is a factor of p(x) = x³ + 3x² + 3x + 1, it is sufficient to show that p(- 2) = 0

Now, p(- 2) = (- 2)³ + 3(- 2)² – 3(- 2) – 1

= – 8 + 12 – 6 + 1

= – 1 ≠ 0

∴ g(x) is a factor of p(x).

(iii) In order to prove that gix) = x – 3 is a factor of pix) = x³ – 4x² + x – 6, it is sufficient to show that p(+ 3) = 0

Now, p(3) = (3)³ – 4(3)² + 3 – 6

= 27 – 36 + 3 – 6 = 0

= – 12 ≠ 0

∴ g(x) is a factor of p(x).

Question 3.

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

(i) p(x) = x² + x + k

(ii) p(x) = 2x² + kx + \(\sqrt{2}\)

(iii) p(x) = kx² – \(\sqrt{2}\)x + 1

(iv) p(x) = kx² – 3x + k

Solution:

(i) If (x – 1) is a factor of pix) = x² + x + k, then

p(1) = 0

⇒ (1)² + 1 + k = 0

⇒ 1 + 1 + k = 0

⇒ k = – 2

(ii) If (x – 1) is a factor of p(x) = 2x² + kx + \(\sqrt{2}\), then

p(1). = 0

⇒ 2(1)² + k(1) + \(\sqrt{2}\) = 0 ⇒ 2 + k + \(\sqrt{2}\) = 0

⇒ k = – (2 + \(\sqrt{2}\))

(iii) If (x – 1) is a factor of p(x) = kx² – \(\sqrt{2}\)x + 1, then

p(1) =0

⇒ k(1)² – \(\sqrt{2}\) (1) + 1 = 0 ⇒ k – \(\sqrt{2}\) + 1 = 0

⇒ k = \(\sqrt{2}\) – 1

(iv) If (x – 1) is a factor of p(x) = kx² – 3x + k, then

p(1) = 0

⇒ k(1)² – 3(1) + k = 0 = k – 3 + k = 0

2A = 3 ⇒ k = \(\frac { 3 }{ 2 }\)

Question 4.

Factories :

(i) 12x² – 7x + 1

(ii) 2x² + 7x + 3

(iii) 6x² + 5r – 6

(iv) 3x² – x – 4

Solution:

(i) Here p + q = coeff. of x = – 7

pq = coeff.of x² x constant term =

= 12 x 1 = 12

∴ P + q = – 7 = – 4 – 3

and pq = 12 = (- 4)(- 3)

∴ 12x² – 7x + 1 = 12x² – 4x – 3x + 1

= 4x(3x – 1) – 1(3x – 1)

= (3x – 1)(4x – 1)

(ii) Here p + q = coeff. of x = 7

pq = coeff. of x² x constant term

= 2 x 3 = 6

∴ p + q = 7 = 1 + 6

pq = 6 = 1 x 6

∴ 2x² + 7x + 3 = 2x² + x + 6x + 3

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

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

(iii) Here p + q = coeff. of % = 5

pq = coeff. of x² x constant term

= 6 x – 6 = – 36

p + q = 5 = 9 + (- 4)

and pq = – 36 = 9 x (- 4)

∴ 6x² + 5x – 6 = 6x² + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3)

= (2x + 3)(3x – 2)

(iv) Here p + q = coeff. of x = – 1

pq = coeff. of x² x constant term

= 3 x – 4 = – 12

∴ P + q = – 1 = 3 + (- 4)

and pq = – 12 = 3 x (- 4)

∴ 3x² – x – 4 = 3x + 3x – 4x – 4

= 3x(x + 1) – 4(x + 1)

= (x + 1)(3x – 4)

Question 5.

Factorise :

(i) x³ – 2x² – x + 2

(ii) x³ – 3x² – 9x – 5

(iii) x³ + 32x² + 32x + 20

(iv) 2y³ + y² – 2y – 1

Solution:

(i) Let f(x) = x³ – 2x² – x + 2

The constant term in fhc) is + 2 and factors of – 2 are ± 1, ± 2.

Putting x = 1 in f(x), we have

f(1) = (1)³ – 2(1)² – 1 + 2

= 1 – 2 – 1 + 2 = 0

∴ (x – 1) is a factor of fix).

Putting x = – 1 in f(x), we have

f(- 1) = (- 1)³ – 2(- 1)² – ( – 1) + 2

= – 1 – 2 + 1 + 2 = o

∴ x + 1 is a factor of f(x).

Putting x = 2 in f(x), we have

f(2) = (2)³ – 2(2)² – (2) + 2

= 8 – 8 – 2 + 2 = 0

∴ (x + 2) is a factor of f(x)

Putting x = – 2 in fix), we have

f(- 2) = (- 2)³ – 2(- 2)² – (- 2) + 2

= – 8 – 8 + 2 + 2

= – 12 ≠ 0

∴ x + 2 is not a factor of fix).

∴ The factors of f(x) are (x – 1), (x + 1) and (x – 2).

Let f(x) = k(x – 1)(x + 1)(x – 2)

⇒ x³ – 2x² – x + 2 = k(x – 1)(x + 1)(x – 2)

Putting x = 0 on both sides, we have

2 = k(- 1)(1)(- 2) ⇒ k = 1

x³ – 2x² – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) Let p(x) = x³ – 3x² – 9x – 5

We shall look for all factors of – 5. These are ± 1, ± 5.

By trial, we find p(- 1) = – 1 – 3 + 9 – 5 = 0. So, (x + 1) is a factor of p(x).

∴ p(x) = (x + 1)(x² – 4x – 5)

= (x + 1)(x² + x – 5x – 5)

= (x + 1)[x(x + 1) – 5(x + 1)]

= (x + 1)(x + 1)(x – 5)

(iii) Let p(x) = x³ + 13x² + 32x + 20 .

We shall look for all facotrs of + 20, these are + 1, ± 2, ± 4, ± 5, ± 10 and ± 20.

By trial, we find

p(- 2) = – 8 + 52 – 64 + 20 = 0

∴ (x + 2) is a factor of p(x)

Now, divide p(x) by x + 2.

∴ p(x) = (x + 2)(x² + 11x + 10)

= (x + 2)(x² + x + 10x + 10)

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

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

(iv) Let p(y) = 2y³ + y² – 2y – 1

By trial, we find p(1) = 2 + 1 – 2 – 1 = 0.

So, (y – 1) is a factor of p(y).

Now, divide p(y) by y – 1

∴ p(y) = (y – 1)(2y² + 3y + 1)

= (y – 1)(2y² + 2y + y + 1)

= (y – 1)[2y(y + 1) + 1(y + 1)]

= (y – 1)(y + 1)(2y + 1)