Rajasthan Board RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 Textbook Exercise Questions and Answers.

## RBSE Class 9 Maths Solutions Chapter 2 Polynomials Exercise 2.4

Question 1.

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

(i) x^{3} + x^{2} + x + 1

Answer:

To prove, x + 1 is a factor of p(x) = x^{3} + x^{2} + x + 1, it is 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 p(x) = x^{3} + x^{2} + x + 1.

(ii) x^{4} + x^{3} + x^{2} + x + 1

Answer:

To prove, (x + 1) is a factor of p(x) = x^{4} + x^{3} + x^{2} + x + 1, it is to show that p(- 1) = 0.

Now, pi- 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} + x^{3} + x^{4} + x + 1.

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

Answer:

To prove, (x + 1) is a factor of pix) = x^{4} + 3x^{3} + 3x^{2} + x + 1, it is to show that

p(- 1) = 0.

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

= 1 – 3 + 3 – 1 + 1 = 1 ≠ 0

∴ (x + 1) is not a factor of x^{4} + 3x^{4} + 3x^{2} + x + 1.

(iv) x^{3} – x^{2} – (2 + √2 )x + √2

Answer:

To prove, (x + 1) is a factor of p(x) = x^{3} – x^{2} – (2 + √2 )x + √2, it is to show that p(- 1) = 0.

Now,. pi- 1) = (- 1)^{3} – (- 1)^{2} – (2 + √2 )(- 1) + √2

= – 1 – 1 + 2 + √2 + √2 = 2√2 ≠ 0

∴ (x + 1) is not a factor of x^{3} – x^{2} – (2 + √2 )x + √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) = 2x^{3} + x^{2} – 2x – 1, g(x) = x + 1

Answer:

To prove, g(x) = x + 1 is a factor of p(x) = 2x^{3} + x^{2} – 2x – 1, it is 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) p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

Answer:

To prove, g(x) = x + 2 is a factor of pix) = x^{3} + 3x^{2} + 3x + 1, it is 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 not a factor of p(x).

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

Answer:

To prove, g(x) = x – 3 is a factor of p(x) = x^{3} – 4x^{2} + x + 6, it is to show that p(3) = 0.

Now, p(3) = (3)^{3} – 4(3)^{32} + 3 + 6 = 27 – 36 + 3 + 6 = 0

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: .

(i) p(x) = x^{2} + x + k

Answer:

If (x – 1) is a factor of p(x) = x^{2} + x + k, then

p(1) = 0

or (1)^{2} + 1 + k = 0 ⇒ 1 + 1 + k = 0

⇒ k = – 2

(ii) p(x) = 2x^{2} + kx + √2

Answer:

If (x – 1) is a factor of p(x) = 2x^{2} + kx + √2 , then

p(1) = 0

or 2(1)^{2} + k(1) + √2 = 0 ⇒ 2 + k + √2 = 0

⇒ k = – (2 + √2 )

(iii) p(x) = kx^{2} – √2x + 1

Answer:

If (x – 1) is a factor of p(x) = kx^{2} – √2 x + 1, then

p(1) = 0

or k(1)^{2} – √2 (1) + 1 = 0 ⇒ k – √2 + 1 = 0

⇒ k = √2 – 1

(iv) p(x) = kx^{2} – 3x + k

Answer:

If (x – 1) is a factor of p(x) = kx^{2} – 3x + k, then

p(1) = 0

or k(1)^{2} – 3(1) + k = 0 ⇒ k – 3 + k = 0

⇒ 2k = 3

⇒ k = \(\frac{3}{2}\)

Question 4.

Factorise :

(i) 12x^{2} – 7x + 1

Answer:

Here, p + q = coeff. of x = – 7

pq = coeff. of x^{2} × constant term

= 12 × 1 = 12

p + q = -7 = -4-3

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

12x^{2} – 7x + 1 = 12x^{2} – 4x – 3x + 1

= 4x(3x – 1) – 1(3x – 1)

= (3x – 1)(4x – 1)

(ii) 2x^{2} + 7x + 3

Answer:

Here, p + q = coeff. of x = 7

pq = coeff of x^{2} × constant term

= 2 × 3 = 6

∴ p + q = 7 = 1 + 6

and pq = 6 = 1 × 6

∴ 2x^{2} + 7x + 3 = 2x^{2} + x + 6x + 3

= x(2x + 1) + 3(2x + 1) = (2x + 1)(4x + 3)

(iii) 6x^{2} + 5x – 6

Answer:

Here, p + q = coeff. of x = 5

pq = coeff. of x^{2} × constant term

= 6 × (- 6) = – 36

∴ p + q = 5 = 9 + (-4)

and pq = – 36 = 9 × (- 4)

∴ 6x^{2} + 5x – 6 = 6x^{2} + 9x – 4x – 6

= 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2)

(iv) 3x^{2} – x – 4

Answer:

Here, p + q = coeff. of x = – 1

pg = coeff. of x^{2} × constant term

= 3 × (- 4) = – 12

∴ p + q = -1 = 3 + (- 4)

and pq = – 12 = 3 x (- 4)

∴ 3x^{2} – x – 4 = 3x^{2} + 3x – 4x – 4

= 3x(x + 1) – 4(x + 1) = (x + l)(3x – 4)

Question 5.

Factorise :

(i) x^{3} – 2x^{2} – x + 2

Answer:

Let f(x) = x^{3} – 2x^{2} – x + 2

The constant term in f(x) 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 f(x).

Putting x = – 1 in f(x), we have :

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

∴ 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

∴ (x + 2) 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 = -12 ≠ 0

∴ x + 2 is not a factor of f(x).

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

Let f(x) = k(x – 1)(x + 1)(x – 2)

⇒ x^{3} – 2x^{2} – x + 2 = k(x – 1)(x + 1)(x – 2)

Putting x = 0 on both sides, we have :

2 = k(- 1)(1)(- 2)

⇒ k = 1

x^{3} – 2x^{2} – x + 2

= (x – 1)(x + 1)(x – 2)

Aliter: After finding one factor (x – 1), f(x) can be divided by (x – 1) to obtain other 2 factors.

Hence, p(x) = (x – 1) (x^{2} – x – 2)

Further factorising x2 – x – 2

we get x^{2} – x – 2 = x^{2} – 2x + x – 2

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

∴ Factors of p(x) are (x – 1) (x + 1) (x – 2).

(ii) x^{3} – 3x^{2} – 9x – 5

Answer:

Let p(x) = x^{3} – 3x^{2} – 9x – 5

All factors of – 5 are ± 1, ± 5.

By trial, we find p(- 1) = – 1 – 3 + 9 – 5 = 0.

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

Now, we divide x^{3} – 3x^{2} – 9x – 5 by x + 1 as shown below :

p(x) = (x + 1)(x^{2} – 4x – 5) = (x + 1)(x^{2} + x – 5x – 5)

= (x + 1)[x(x + 1) – 5(x + 1)]

= (x + 1)(x + 1)(x – 5)

Aliter : After finding one factor (x + 1), p(x) can be written as follows :

x^{3} + x^{2} – 4x^{2} – 4x – 5x – 5

= x^{2}(x + 1) – 4x(x + 1) – 5(x + 1)

= (x + 1)(x^{2} – 4x – 5) = (x + 1){x^{2} + x – 5x – 5}

= (x + 1){x(x + 1) – 5(x + 1)} = (x + 1)(x+ 1)(x – 5)

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

Answer:

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

All factors of + 20 are ± 1, ± 2, ± 4, ± 5, ± 10 and ± 20.

By trial, we find that :

p(- 2) = – 8 + 52 – 64 + 20 = 0 ’

∴ (x + 2) is a factor of p(x).

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

∴ p(x) = (x + 2)(x^{2} + 11x + 10) = (x + 2)(x^{2} + x + 10x + 10)

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

(iv) 2y^{3} + y^{2} – 2y – 1

Answer:

Let p(y) = 2y^{3} + y^{2} – 2y – 1

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

So, (y – 1) is a factor of p(y).

Now, we divide p(y) by y – 1.

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

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

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