Class IX Math NCERT Solutions For Polynomials Exercise– 2.4
Exercise– 2.4
1. Determine which of the following polynomials has a factor (x + 1):
(i) x3 + x2 + x + 1
(ii) x4 + x + x2 + x + 1
(iii) x4 + 3x3 + 3x2 + x +1
(iv) 
Ans. For x + 1 = 0, we have x = –1.
(i) p(x) = x3 + x2 + x + 1
∴ p(–1) = (–1)3 + (–1)2 + (–1) + 1
= –1 + 1 – 1 + 1 = 0
i.e. when p(x) is divided by (x + 1), then the remainder is zero.
∴ (x + 1) is a factor of x3 + x2 + x + 1.
(ii) p(x) = x4 + x3 + x2 +x + 1
∴ p(–1) = (–1)4 + (–1)3 + (–1)2 + (–1) + 1
= 1 – 1 + 1 – 1 + 1
= 3 – 2 = 1
∵ p(x) is not divisible by x + 1.
i.e. (x + 1) is not a factor of
x4 + x3 + x2 + x + 1.
(iii) ∵ p(x) = x4 + x3 + x2 +x + 1
∴ p(–1) = (–1)4 + 3(–1)3 + 3(–1)2 + (–1) + 1
= (1) + 3(– 1) + 3(1) + (– 1) + 1
= 1 – 3 + 3 – 1 + 1
= 1 ≠ 0
∵ f(–1) ≠ 0
∴ (x + 1) is not a factor of x4 + 3x3 + 3x2 +x +1.
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
Ans. (i) We have p(x) = 2x3 + x2 – 2x – 1 and g(x) = x + 1
∴ p(–1) = 2(–1)3 + (–1)2 – 2(–1) – 1
= 2(–1)3 + (–1)2 – 2(–1) – 1
= 2(–1) + 1 + 2 – 1
= –2 + 1 + 2 – 1
= –3 + 3 = 0
∵ p(–1) = 0
∴ g(x) is a factor of p(x).
(ii) We have p(x) = x3 + 3x2 + 3x + 1 and g(x) = x + 2
∴ p(–2) = (–2)3 + 3(–2)2 + 3(–2) + 1
= –8 + 3(4) + (–6) + 1
= –8 + 12 – 6 + 1
= –8 – 6 + 12 + 1
= –14 + 13 = –1
∵ p(–2) ≠ 0
Thus, g(x) is not a facot of p(x).
(iii) We have p(x) = x3 – 4x2 + x + 6 and g(x) = x – 3
∴ p(3) = (3)3 – 4(3)2 + (3) + 6
= 27– 4(9) + 3 + 6
= 27– 36 + 3 + 6 = 0
Since g(x) = 0
∴ g(x) is a factor of p(x).
3. Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
Ans. Here p(x) = x3 + x + k
For x – 1 be a factor of p(x), p(1) should be equal to 0.
We have p(1) = (1)3 + 1 + k
or p(1) = 1 + 1 + k = k + 2
∴ k + 2 = 0
⇒ k = –2
(ii) Here, p(x) = 2x2 + kx + 2
For x – 1, be a factor of p(x), p(1) = 0
Since, 
∵ p(1) must be equal to 0.
⇒ k = –2 – 2
or k = –(2 + 2 ).
(iii) 
∴ For (x – 1) be a factor of p(x), p(1) should be equal to 0.
(iv) Here p(x) = kx2 – 3x + k and g(x) = x – 1
For g(x) be a factor of p(x), p(1) should be equal to 0.
Since p(1) = k(1)2 – 3(1) + k
= k – 3 + k
= 2k – 3
∴ 2k – 3 = 0
4. Factorise:
(i) 12x2 – 7x + 1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4
Ans. (i) 12x2 – 7x + 1
Here co-efficient of x2 = 12
Co-efficient of x = –7
and constant term = 1
∴ a = 12, b = –7, c = 1
Now, l + m = –7 and lm = ac = 12 × 1
∴ We have l = (–4) and m = (–3)
i.e. b = –7 + (–4 –3).
Now, 12x2 –7x + 1 = 12x2 – 4x – 3x + 1
= 4x(3x – 1) – 1(3x – 1)
= (3x – 1)(3x – 1)
Thus, 12x2 – 7x + 1 = (3x – 1)(3x – 1)
(ii) 2x2 + 7x + 3
Here, a = 12, b = –7, c = 1
∴ l + m = 7 and lm = 2 × 3 = 6
i.e. l + 6 = 7 and 1 × 6 = 6∴ l = 1 and m = 6
We have
2x2 + 7x + 3 = 2x2 + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 1)
Thus, 2x2 + 7x + 3 = (2x + 1)(x + 1)
(iii) 6x2 + 5x – 6
We have a = 6, b = 5 and c = –6
∴ l + m = 5 and lm = ac = 6 × (–6) = –36
∴ l + m = 9 + (–4)
∴ 6x2 + 5x – 6 = 6x2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2)
(iv) 3x2 – x – 4
We have a = 3, b = –1 and c = –4
∴ l + m = –1 and lm = 3 × (–4) = –12
∴ l = – 4 and m = 3
Now, 3x2 – x – 4 = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
Thus, 3x2 – x – 4 = (3x – 4)(x + 1)
5. Factorise:
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
Ans. (i) x3 – 2x2 – x + 2
Rearranging the terms, we have
x3 – 2x2 – x + 2 = x3 – x – 2x2 + 2
= x(x2 – 1) – 2(x2 – 1)
= (x2 – 1) (x – 2)
= [(x2 – (1)2][x – 2]
= (x – 1) (x + 1)(x – 2)
[∴ a2 – b2 = (a + b)(a – b)]
Thus,
x3 – 2x2 – x + 2 = (x – 1) (x + 1)(x – 2)
(ii) x3 – 3x2 – 9x – 5
We have p(x) = x3 – 3x2 – 9x – 5
By trial, let us find:
p(1) = (1)3 – 3(1)2 – 9(1) –5
= 3 – 3 – 9 – 5
= –14 ≠ 0
Now p(–1) = (–1)3 – 3(–1)2 – 9(–1) –5
= –1 – 3(1) + 9 – 5
= –1 – 3 + 9 – 5 = 0
∴ By factor theorem, [x – (–1)] is a factor of p(x).
Now, 
∴ x3 – 3x2 – 9x – 5 = (x + 1)(x2 – 4x – 5)
= (x + 1)[x2 – 5x + x – 5]
[Splitting –4 into –5 and +1]
= (x + 1)[x(x – 5) +1(x – 5]
= (x + 1)[(x – 5) (x + 1)]
= (x + 1)(x – 5)(x + 1)
(iii) x3 + 13x2 + 32x + 20
We have p(x) = x3 + 13x2 + 32x + 20
By trial, let us find:
p(1) = (1)3 + 13(1)2 + 32(1) + 20
= 1 + 13 + 32 + 20
= 66 ≠ 0
Now
p(–1) = (–1)3 + 13(–1)2 + 32(–1) + 20
= –1 + 13 – 32 + 20
= 0
∴ By factor theorem, [x – (–1)], i.e. (x + 1) is a factor p(x).
or x3 + 13x2 + 32x + 20
= (x + 1) (x2 + 12x + 20)
= (x + 1) (x2 + 2x + 12x + 20)
[Splitting the middle term]
= (x + 1)[x(x + 2) + 10(x + 2)]
= (x + 1)[(x + 2) (x + 10)]
= (x + 1)(x + 2) (x + 10)
(iv) 2y3 + y2 – 2y – 1
We have p(y) = 2y3 + y2 – 2y – 1
By trial, we have
p(1) = 2(1)3 + (1)2 – 2(1) – 1
= 2(1) + 1 – 2 – 1
= 2 + 1 – 2 – 1 = 0
∴ By factor theorem, (y – 1) is a factor of p(y).
∴ 2y3 – y2 – 2y – 1 = (y – 1)(2y2 + 3y + 1)
= (y – 1)[2y2 + 2y + y + 1)
[Splitting the middle term]
= (y – 1)[2y(y + 1) + 1(y + 1)]
= (y – 1)[(y + 1) (2y + 1)]
= (y – 1)(y + 1) (2y + 1)
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