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)