Ex 2.4 NCERT Solutions - Polynomials Class 9 Notes | EduRev

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Class 9 : Ex 2.4 NCERT Solutions - Polynomials Class 9 Notes | EduRev

The document Ex 2.4 NCERT Solutions - Polynomials Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
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FACTOR THEOREM
If p(x) is any polynomial of degree 1 or greater than 1 and ‘a’ be any real number such that: (i) (x – a) is a factor of p(x), then p(a) = 0; and (ii) p(a) = 0, then (x – a) is a factor of p(x).

FACTORIZATION OF POLYNOMIALS:
[Using the “splitting the middle-term” method] Let ax2 + bx + c be a quadratic polynomial such that a, b and c are constants and a ≠ 0.

To factorise it, we split the co-efficient of x into two parts ‘l’ and ‘m’ such that l + m = b
l x m = a x c

Ques 1: Determine which of the following polynomials has a factor (x + 1): 
(i) x3 + x2 + x + 1 
(ii) x+ x+ x2 + x + 1 
(iii) x4 + 3x3 + 3x+ x + 1 
(iv) x3 – x2 – Ex 2.4 NCERT Solutions - Polynomials Class 9 Notes | EduRev
Ans: For x + 1 = 0, we have x = –1.

∴ The zero of x + 1 is –1.
(i) p(x) = x3 + x2 + x + 1

∴ p(–1) = (–1)+ (–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 x+ 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

∵ f(–1) ≠ 0

∴ 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 + 3x3 + 3x2 + 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 + 3x+ x + 1.

(iv) ∵ p(x) = x3 – x2 – (2 + √2) x + √2)

∴ p(–1) = (–1)3 – (–1)2 – (2 + 2) (–1) + 2 = –1 – 1 – (–1) (2 + √2) + 2
= –1 – 1 + 1 (2 + √2) + 2
= –1 – 1 + 2 + √2 + √2)
= –2 + 2 + 2√2
= 2√2 ≠ 0
Since p(–1) ≠ 0.

∴ (x + 1) is not a factor of x4 + 3x3 + 3x+ x + 1.

Ques 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)+ (–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 factor of p(x).

(iii) We have p(x) = x– 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).

Ques 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 
(ii) p(x) = 2x2 + kx + √2 
(iii) p(x) = kx2 – √2 x + 1 
(iv) p(x) = kx2 – 3x + k
Ans: (i) Here p(x) = x2 + x + k
For x – 1 be a factor of p(x), p(1) should be equal to 0.
We have p(1) = (1)2 + 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) = 2(1)+ k(1) + √2
= 2 + k + √2

∵ p(1) must be equal to 0.
∴ k + 2 + √2 = 0
⇒ k = –2 – √2
or k = – (2 + √2) .

(iii) Here p(x) = kx2 – √2 x + 1 and g(x) = x – 1
∴ For (x – 1) be a factor of p(x), p(1) should be equal to 0.
Since p(1) = k(1)2 – √2 (1) + 1
or p(1) = k – √2 + 1
or p(1) = k – √2 + 1
∴ k – √2 +1 = 0 ⇒ k= √2 – 1

(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 ⇒ k = 3/2

Ques 4: Factorise: 
(i) 12x– 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 x 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)(4x – 1)
Thus, 12x2 – 7x + 1 = (3x – 1)(4x – 1)

(ii) 2x2 + 7x + 3 Here, a = 2, b = 7 and c = 3
∴ l + m = 7 and lm = 2 x 3 = 6
i.e. 1 + 6 = 7 and 1 x 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 + 3)
Thus, 2x2 + 7x + 3 = (2x + 1)(x + 3)

(iii) 6x2 + 5x – 6

We have a = 6, b = 5 and c = –6

∴ l + m = 5 and lm = ac = 6 x (–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 x (–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)

Ques 5: Factorise: 
(i) x– 2x2 – x + 2 
(ii) x– 3x2 – 9x – 5 
(iii) x3 + 13x+ 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)

= [(x)2 – (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) x– 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).

Ex 2.4 NCERT Solutions - Polynomials Class 9 Notes | EduRev

∴ x– 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)+ 13(1)+ 32(1) + 20

= 1 + 13 + 32 + 20

= 66 ≠ 0

Now p(–1) = (–1)3 + 13(–1)+ 32(–1) + 20

= –1 + 13 – 32 + 20
= 0
∴ By factor theorem, [x – (–1)], i.e. (x + 1) is a factor p(x).

Ex 2.4 NCERT Solutions - Polynomials Class 9 Notes | EduRev

or x3 + 13x+ 32x + 20

= (x + 1)(x2 + 12x + 20)

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

Ex 2.4 NCERT Solutions - Polynomials Class 9 Notes | EduRev

∴ 2y3 – y– 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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