PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4
PSEB Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4
PSEB 9th Class 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
Answer:
The zero of x + 1 is – 1.
Let, p(x) = x3 + x2 + x + 1.
Then,
p(- 1) = (- 1)4 + (- 1)3 + (- 1)2 + (- 1) + 1
= – 1 + 1 – 1 + 1
≠ 0
So, by the factor theorem, x + 1 is a
factor of x4 + x3 + x2 + x + 1.
(ii) x3+ x3 + x2 + x + 1
Answer:
The zero of x +1 is – 1.
Let, p(x) = x4 + x3 + x2 + x + 1.
Then,
p(- 1) = (- 1)4 + (- 1)3 + (- 1)2 + (- 1) + 1
= 1 – 1 + 1 – 1 + 1
= 1
≠ 0
So, by the factor theorem, x + 1 is not a factor of x4 + x3 + x2 + x + 1.
(iii) x4 + 3x3 + 3x2 + x + 1
Answer:
The zero of x + 1 Is – 1.
Let, p(x )= x4 + 3x3 + 3x2 + x + 1.
Then,
p (-1) = (- 1)4 + 3 (- 1)3 + 3 (- 1)2 + (- 1) + 1
= 1 – 3 + 3 – 1 + 1
= 1
≠ 0
So, by the factor theorem, x + 1 is not a factor of x4 + 3x3 + 3x2 + x + 1.
(iv) x3 – x2 – (2 + √2)x + √2
Answer:
The zero of x + 1 is – 1. .
Let, p(x) = x3 – x2 – (2 + √2)x + √2.
Then.
p(- 1) = (- 1)3 – (- 1)2 – (2 + √2) (- 1) + √2
= – 1 – 1 + 2 + √2 + √2
= 2√2
≠ 0
So, by the factor theorem, 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:
g(x) = 0 gives x + 1 = 0, i.e., x = – 1.
Here, p(x) = 2x3 + x2 – 2x – 1
Then, p(- 1) = 2(- 1)3 + (- 1)2 – 2(- 1) – 1
= – 2 + 1 + 2 – 1
= 0
So, by the factor theorem, g (x) = x + 1 is a factor of p(x) = 2x3 + x2 – 2x – 1.
(ii) p(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2
Answer:
g(x) = 0 gives x + 2 = 0, i.e., x = – 2.
Here, p(x) = x3 + 3x2 + 3x + 1
Then, p(- 2) = (- 2)3 + 3(2)2 + 3(- 2) + 1
= – 8 + 12 – 6 + 1
≠ 0
So, by the factor theorem, g (x) = x + 2 is not a factor of p(x) = x3 + 3x2 + 3x + 1.
(iii) p (x) = x3 – 4x2 + x + 6. g (x) = x – 3
Answer:
g(x) = 0 gives x – 3 = 0, i.e., x = 3.
Here, p (x) = x3 – 4x2 + x + 6
Then, p(3)= (3)3 – 4(3)2 + (3) + 6
= 27 – 36 + 3 + 6
= 0
So, by the factor theorem, g (x) = x – 3 is a factor of p (x) = x3 – 4x2 + x + 6.
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:
Here x – 1 is a factor of p(x) = x2 + x + k.
∴ p(1) = 0
∴ (1)2 + (1) + k = 0
∴ 1 + 1 + k = 0
∴ 2 + k = 0
∴ k = – 2
(ii) p (x) = 2x2 + kx + √2
Answer:
Here, x – 1 is a factor of
p(x) = 2x2 + kx + √2.
∴ p(1) = 0
∴ 2(1)2 + k(1) + √2 = 0
∴ 2 + k + √2 = 0
∴ k = -(2 + √2)
(iii) p (x) = kx2 – √2x + 1
Answer:
Here, x – 1 is a factor of
p(x) = kx2 – √2x + 1.
∴ p (1) = 0
∴ k(1)2 – √2 (1) + 1 = 0
∴ k – √2 + 1 = 0
∴ k = √2 – 1
(iv) p (x) = kx2 – 3x + k
Answer:
Here, x – 1 is a factor of p (x) = kx2 – 3x + k.
∴ p(1) = 0
∴ k(1)2 – 3(1) + k = 0
∴ k – 3 + k = 0
∴ 2k = 3
∴ k = [1atex]\frac{3}{2}[/1atex]
Question 4.
Factorise:
(i) 12x2 – 7x + 1
Answer:
12x2 – 7x + 1 = 12x2 – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (3x – 1)(4x – 1)
(ii) 2x2 + 7x + 3
Answer:
2x2 + 7x + 3 = 2x2 + 6x + x + 3
= 2x(x + 3) + 1(x + 3)
= (x + 3) (2x + 1)
(iii) 6x2 + 5x – 6
Answer:
6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
(iv) 3x2 – x – 4
Answer:
3x2 – x – 4 = 3x2 + 3x – 4x – 4
= 3x(x + 1) – 4(x + 1)
= (x + 1) (3x – 4)
Question 5.
Factorise:
(i) x3 – 2x2 – x + 2
Answer:
Let, p(x) = x3 – 2x2 – x + 2
All the factors of 2 are ± 1 and ± 2.
By trial, we find that p (1) = 0.
∴ x – 1 is a factor of p (x).
x3 – 2x2 – x + 2
= x3 – x2 – x2 + x – 2x + 2
= x2(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)
(ii) x3 – 3x2 – 9x – 5
Answer:
Let, p(x) = x3 – 3x2 – 9x – 5
All the factors of – 5 are ± 1 and ±5.
By trial, we find that p (- 1) = 0.
∴ x + 1 is a factor of p (x).
x3 – 3x2 – 9x – 5
= 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 the factors of 20 are ± 1, ± 2, ± 4, ± 5, ± 10 and ± 20.
By trial, we find that p (- 1) = 0.
∴ x + 1 is a factor of p (x).
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 + 2x + 10x + 20)
= (x + 1) {x(x + 2) + 10(x + 2)}
= (x + 1) (x + 2) (x + 10)
(iv) 2y3 + y2 – 2y – 1
Answer:
Let, p (y) = 2y3 + y2 – 2y – 1
All the factors of – 1 are ± 1.
By trial, we find that p (- 1) = 0.
∴ y + 1 is a factor of p (y).
2y3 + y2 – 2y – 1
= 2y3 + 2y2 – y2 – y – y – 1
= 2y2(y + 1) – y(y + 1) – 1 (y + 1)
= (y + 1) (2y2 – y – 1)
= (y + 1) (2y2 – 2y + y – 1)
= (y + 1) {2y (y – 1) + 1(y – 1)}
= (y + 1) (y – 1)(2y + 1)
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