RBSE Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4

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) x³ + x² + x + 1
Answer:
To prove, x + 1 is a factor of p(x) = x³ + x² + 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) = x³ + x² + x + 1.

(ii) x⁴ + x³ + x² + x + 1
Answer:
To prove, (x + 1) is a factor of p(x) = x⁴ + x³ + x² + 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 x⁴ + x³ + x⁴ + x + 1.

(iii) x⁴ + 3x³ + 3x² + x + 1
Answer:
To prove, (x + 1) is a factor of pix) = x⁴ + 3x³ + 3x² + 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 x⁴ + 3x⁴ + 3x² + x + 1.

(iv) x³ – x² – (2 + √2 )x + √2
Answer:
To prove, (x + 1) is a factor of p(x) = x³ – x² – (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 x³ – x² – (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) = 2x³ + x² – 2x – 1, g(x) = x + 1
Answer:
To prove, g(x) = x + 1 is a factor of p(x) = 2x³ + x² – 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) = x³ + 3x² + 3x + 1, g(x) = x + 2
Answer:
To prove, g(x) = x + 2 is a factor of pix) = x³ + 3x² + 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) = x³ – 4x² + x + 6, g(x) = x – 3
Answer:
To prove, g(x) = x – 3 is a factor of p(x) = x³ – 4x² + 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) = x² + x + k
Answer:
If (x – 1) is a factor of p(x) = x² + x + k, then
p(1) = 0
or (1)² + 1 + k = 0 ⇒ 1 + 1 + k = 0
⇒ k = – 2

(ii) p(x) = 2x² + kx + √2
Answer:
If (x – 1) is a factor of p(x) = 2x² + kx + √2 , then
p(1) = 0
or 2(1)² + k(1) + √2 = 0 ⇒ 2 + k + √2 = 0
⇒ k = – (2 + √2 )

(iii) p(x) = kx² – √2x + 1
Answer:
If (x – 1) is a factor of p(x) = kx² – √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) = kx² – 3x + k
Answer:
If (x – 1) is a factor of p(x) = kx² – 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) 12x² – 7x + 1
Answer:
Here, p + q = coeff. of x = – 7
pq = coeff. of x² × constant term
= 12 × 1 = 12
p + q = -7 = -4-3
and pg = 12 = (- 4)(- 3)
12x² – 7x + 1 = 12x² – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (3x – 1)(4x – 1)

(ii) 2x² + 7x + 3
Answer:
Here, p + q = coeff. of x = 7
pq = coeff of x² × constant term
= 2 × 3 = 6
∴ p + q = 7 = 1 + 6
and pq = 6 = 1 × 6
∴ 2x² + 7x + 3 = 2x² + x + 6x + 3
= x(2x + 1) + 3(2x + 1) = (2x + 1)(4x + 3)

(iii) 6x² + 5x – 6
Answer:
Here, p + q = coeff. of x = 5
pq = coeff. of x² × constant term
= 6 × (- 6) = – 36
∴ p + q = 5 = 9 + (-4)
and pq = – 36 = 9 × (- 4)
∴ 6x² + 5x – 6 = 6x² + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3) = (2x + 3)(3x – 2)

(iv) 3x² – x – 4
Answer:
Here, p + q = coeff. of x = – 1
pg = coeff. of x² × constant term
= 3 × (- 4) = – 12
∴ p + q = -1 = 3 + (- 4)
and pq = – 12 = 3 x (- 4)
∴ 3x² – x – 4 = 3x² + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1) = (x + l)(3x – 4)

Question 5.
Factorise :
(i) x³ – 2x² – x + 2
Answer:
Let f(x) = x³ – 2x² – 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)
⇒ 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)

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) (x² – x – 2)
Further factorising x2 – x – 2
we get x² – x – 2 = x² – 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) x³ – 3x² – 9x – 5
Answer:
Let p(x) = x³ – 3x² – 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 x³ – 3x² – 9x – 5 by x + 1 as shown below :

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)

Aliter : After finding one factor (x + 1), p(x) can be written as follows :
x³ + x² – 4x² – 4x – 5x – 5
= x²(x + 1) – 4x(x + 1) – 5(x + 1)
= (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) x³ + 13x² + 32x + 20
Answer:
Let p(x) = x³ + 13x² + 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)(x² + 11x + 10) = (x + 2)(x² + x + 10x + 10)
= (x + 2)[x(x + 1) + 10(x + 1)] = (x + 2)(x + 1)(x + 10)

(iv) 2y³ + y² – 2y – 1
Answer:
Let p(y) = 2y³ + y² – 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)(2y² + 3y + 1) = (y – 1)(2y² + 2y + y + 1)
= (y – 1)[2y(y + 1) + 1(y + 1)] = (y – 1)(y + 1)(2y + 1)