# Factorise: (i) x 3 − 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

(i) Let p(x) = x3 − 2x2x + 2

All the factors of 2 have to be considered. These are ± 1, ± 2.

By trial method,

p(−1) = (−1)3 − 2(−1)2 − (−1) + 2

= −1 − 2 + 1 + 2 = 0

Therefore, (x +1 ) is factor of polynomial p(x).

Let us find the quotient on dividing x3 − 2x2x + 2 by x + 1.

By long division, It is known that,

Dividend = Divisor × Quotient + Remainder

x3 − 2x2x + 2 = (x + 1) (x2 − 3x + 2) + 0

= (x + 1) [x2 − 2xx + 2]

= (x + 1) [x (x − 2) − 1 (x − 2)]

= (x + 1) (x − 1) (x − 2)

= (x − 2) (x − 1) (x + 1)

(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, It is known that,

Dividend = Divisor × Quotient + Remainder

x3 − 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)

(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).

By long division, 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)

(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. 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)

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