Question

Factorise : (i) x3 – 2x 2 – x + 2 (ii) x 3 – 3x 2 – 9x – 5 (iii) x 3 + 13x 2 + 32x + 20 (iv) 2y 3 + y 2 – 2y – 1.

Answer 1

(i) Let p(x) = x3 − 2x2 − x + 2

All the factors of 2 have to be considered.

These are ± 1, ± 2. By trial method,

p(2) = (2)3 − 2(2)2 − 2 + 2

= 8 − 8 − 2 + 2 = 0

Therefore, (x − 2) is factor of polynomial p(x).

Let us find the quotient on dividing

x3 − 2x2 − x + 2 by x − 2.

By long division,

x + 1 ÷ x3 - 2x2 - x + 2 = x3 + x2 - - - / 3x2 - x + 2 - 3x2 - 3x + + / 2x + 2 2x + 2 - - / 0 = x2 - 3x + 2

It is known that, Dividend = Divisor × Quotient + Remainder

x 3 − 2x2 − x + 2 = (x + 1) (x2 − 3x + 2) + 0

= (x + 1) [x2 − 2x − x + 2]

= (x + 1) [x (x − 2) − 1 (x − 2)]

= (x + 1) (x − 1) (x − 2)

= (x − 2) (x − 1) (x + 1)

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(ii) Let p(x) = x3 − 3x2 − 9x − 5 All the factors of 5 have to be considered.

These are ±1, ± 5. By trial method,

p(−1) = (−1)3 − 3(−1)2 − 9(−1) − 5

= − 1 − 3 + 9 − 5 = 0

Therefore, x + 1 is a factor of this polynomial.

Let us find the quotient on dividing

x3 + 3x2 − 9x − 5 by x + 1.

By long division, x + 1 ÷ x3 - 3x2 - 9x - 5

= x3 + x2 - - / -4x2 - 9x - 5 -4x2 - 4x + + / -5x - 5 -5x - 5 + + / 0 It is known that,

Dividend = Divisor × Quotient + Remainder

x 3 − 3x2 − 9x − 5 = (x + 1) (x2 − 4x − 5) + 0

= (x + 1) (x2 − 5x + x − 5)

= (x + 1) [(x (x − 5) +1 (x − 5)]

= (x + 1) (x − 5) (x + 1)

= (x − 5) (x + 1) (x + 1)

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(iii) Let p(x) = x3 + 13x2 + 32x + 20 All the factors of 20 have to be considered.

Some of them are ±1, ± 2, ± 4, ± 5 …… By trial method,

p(−1) = (−1)3 + 13(−1)2 + 32(−1) + 20

= − 1 +13 − 32 + 20

= 33 − 33 = 0 As p(−1) is zero,

therefore, x + 1 is a factor of this polynomial p(x).

Let us find the quotient on dividing

x3 + 13x2 + 32x + 20 by (x + 1). x + 1 ÷ x3 + 13x2 + 32x + 20

= x3 + x2 - - / 12x2 + 32x 12x2 + 12x - - / 20x + 20 20x + 20 - - / 0

= x2 + 12x + 20 It is known that,

Dividend = Divisor × Quotient + Remainder

x3 + 13x2 + 32x + 20 = (x + 1) (x2 + 12x + 20) + 0

= (x + 1) (x2 + 10x + 2x + 20)

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

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

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

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(iv) Let p(y) = 2y3 + y2 − 2y − 1 By trial method,

p(1) = 2 ( 1)3 + (1)2 -2 (1) - 1

= 2 + 1 - 2 -1 = 0

Therefore, y − 1 is a factor of this polynomial.

Let us find the quotient on dividing

2y3 + y2 − 2y − 1 by y − 1. y - 1 ÷ 2y3 + y2 - 2y - 1

= 2y3 - 2y2 - + / 3y2 - 2y - 1 3y2 - 3y - + / y - 1 y - 1 / 0

= p(y) = 2y3 + y2 − 2y − 1 = (y − 1) (2y2 +3y + 1)

= (y − 1) (2y2 +2y + y +1)

= (y − 1) [2y (y + 1) + 1 (y + 1)]

= (y − 1) (y + 1) (2y + 1)