GSEB Solutions Class 9 Maths Chapter 2 Polynomials Ex 2.4

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. x³ + x² + x + 1
2. x⁴ + x²+ x² + x + 1
3. x⁴ + 3x³ + 3x² + x + 1
4. x³ – x² – (2 + \(\sqrt{2}\))x + \(\sqrt{2}\)
Solution:
1.  Let p(x) = x³ + x² + 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
∴ By factor theorem x + 1 is a factor of x³ + x² + x + 1.

2. Let p(x) = x⁴ + x³ + x² + x + 1
⇒ x + 1 = 0
x = -1
p(-1) = (-1)⁴ + (-1)³ + (-l)² + (-l) + 1
= 1 – 1 + 1 – 1 + 1 = 3 – 2
⇒ p(-1) = 1
∴ 1 ≠ 0
Hence by factor theorem x + 1 is not a factor of x⁴ + x³ + x² + x + 1.

3. Let p(x) = x⁴ + 3x³ + 3x² + x + 1
x + 1 = 0
⇒ x = -1
p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
p(-1) = 1 – 3 + 3
p(-1) = 1
∴ 1≠ 0
Hence by factor theorem x + 1 is not a factor of x⁴ + 3x³ + x² + x + 1.

4. Let p(x) = x³ – x² – (2 + \(\sqrt {2} \)) x + \(\sqrt {2} \)
x + 1 = 0
⇒ x = – 1
p(-1) = (-1)³ – (-1)² – (2 +\(\sqrt {2} \))(-1) + \(\sqrt {2} \)
p(-1) = -1 – 1 + 2 + \(\sqrt {2} \) + \(\sqrt {2} \)
⇒ p(-1) = 2\(\sqrt {2} \)
∴ 2\(\sqrt {2} \) = 0
Hence by factor theorem 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:
1. p(x) = 2x³ + x² – 2x – 1, g(x) = x + 1
2. p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2
3. p(x) = x³ – 4x² + x + 6, g(x) = x – 3
Solution:
1. p(x) = 2 x³ + x² – 2x – 1
g(x) = x + 1
g(x) = 0
⇒ x + 1 = 0
⇒ x = -1
p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= -2 + 1 + 2 – 1
⇒ p(-1) = 0
∴ By factor theorem g(x) is a factor of p(x).

2. p(x) = x³ + 3x² + 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
∴ -1 ≠ 0
Hence by factor theorem g(x) is not a factor of p(x).

3. p(x) = x³ – 4x² + x + 6, g(x) = x – 3
g(x) = 0
⇒ x – 3 = 0
⇒ x = 3
p(3) = 3³ – 4(3)² + 3 + 6
⇒ p(3) = 27 – 4 x 9 + 9
⇒ p(3) = 27 – 36 + 9
⇒ 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) = x² + x + k
2. p(x) = 2x² + kx + \(\sqrt {2} \)
3. p(x) = kx² – \(\sqrt {2} \) x + 1
4. p(x) = kx² – 3x + k
Solution:
1. p(x) = x² + x + k
x – 1 will be a factor p(x) therefore p(1) = 0.
By factor theorem
:. p(1) = 1² + 1 + k
0 = 2 + k = k = -2

2. p(x) = 2 x² + 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) = kx² – \(\sqrt{2}\)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) = kx² – 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 x² – 7x + 1
2. 2x² + 7x + 3
3. 6x² + 5x – 6
4. 3x² – x – 4
Solution:
1. 12 x² – 7x + 1 (a x c = 12 x 1)
= 12 x² – 4x – 3x + 1 (12 = 4 x 3)
= 4 x (3x -1) -1(3x – 1.)
= (3x – 1)(4x – 1)

2. 2x² + 7x + 3 (a x c = 2 x 3)
-2x² + 6x +x + 3 (6 = 6 x 1)
= 2x (x + 3) + 1(x + 3)
= (x + 3) (2x + 1)

3. 6x² + 5x – 6 (a x c = 6 x 6 = 36)
= 6x² + 9x – 4x – 6 (36 = 9 x 4)
= 3x(2x + 3) -2(2x + 3)
= (2x + 3)(3x – 2)

4. 3x² – x – 4 (a x c = 3 x 4)
= 3x² – 4x + 3x – 4 (12 = 4 x 3)
= x (3x – 4)+ 1(3x – 4)
= (3x – 4) (x + 1)

Question 5.
Factorise:
1. x³ – 2x² – x + 2
2. x³ – 3x² – 9x – 5
3.  x³+ 13x² + 32x + 20
5. 2y² + y² – 2y – 1
Solution:
1. Let p(x) = x³ – 2x² – x + 2
Here constant form is 2 and its factors can be ±1, ±2.
By trial method we find factor of p(x).
Let x – 1 be a factor of p(x).
∴ x – 1 = 0
⇒ x = 1
p(1) = 1³ – 2(1)³ – 1 + 2
= 1 – 2 – 1 + 2 = 3 – 3
⇒ p(1) = 0
∴ By factor theorem x – 1 is a factor of p(x).
Now x³ – 2x³ – x + 2
= x² – x² + x – 2x + 2
= x(x – 1) -x(x – 1)-2(x – 1)
= (x – 1)(x² – x – 2)
= (x – 1)[x² – 2x + x – 2]
= (x – 1)[x(x – 2) + 1(x – 2)]
= (x – 1) (x – 2) (x + 1)

2. Let p(x) = x³ – 3x² – 9x – 5
Here constant term is 5 and its factors can be ±1, ±5.
By trial method, we find factor of p(x).
Let x – 1 be a factor of p(x).
∴ x – 1= 0
⇒ x = 1
p(1) = 1³ – 3(1)² – 9(1) – 5
= 1 – 3 – 9 – 5 = -16
But -16≠0
∴ x – 1 is not a factor of p(x).
Now let x + 1 be a factor of p(x).
∴ x + 1 = 0
⇒ x = -1
p(-1) = (-1)³ – 3(-1)² – 9(-1) – 5
= -1 – 3 + 9 – 5 = -9 + 9
p(-1) = 0
∴ x + 1 is a factor of p(x).

Now x³ – 3x² – 9x – 5
= x³ + x³ – 4x³ – 4x – 5x – 5
= x²(x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x² – 4x – 5)
= (x + 1)(x² – 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) = x³ + 13x² + 32x + 20 Here constant term is 20 and its factors can be ±1, ±2, ±4, ±5, ±10, ±20
By trial method, we find the factor of p(x).
Let x + 1 be a factor of p(x).
∴ 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 X³ + 13x²+ 32x + 20
= x³ +x² + 12x² + 12x + 20x + 20
= x²(x + 1) + 12x(x + 1) + 20(x + 1)
= (x + 1)(x² + 12x + 20)
= (x + 1)[x² + 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) = 2y² + y² – 2y – 1
By trial method, we find the factor of p(y).
Let y – 1 be a factor of p(y).
y – 1 = 0 ⇒ y = 1
p(1) = 2(1)³ + 1² – 2 x 1 – 1
= 2 + 1 – 2 – 1 = 3 – 3
p(1) = 0
∴ y – 1 is a factor of p(y).
Hence 2y³ + y² – 2y – 1
2y³ – 2y² + 3y² – 3y + y – 1
= 2y²(y – 1) + 3y(y – 1) + 1(y – 1)
= (y – 1)[2y² + 3y + 11]
= (y – 1) [2y² + 2y + y + 1]
= (y -1)[2y(y + 1) +1 (y + 1)]
= (y – 1)(y + 1)(2y + 1)