Factorise:

(i) *x* ^{3} − 2*x* ^{2} − *x*
+ 2 (ii) *x* ^{3} + 3*x* ^{2} −9*x *−
5

(iii) *x* ^{3} + 13*x* ^{2} + 32*x* +
20 (iv) 2*y* ^{3} + *y* ^{2} − 2*y*
− 1

(i) Let *p*(*x*) = *x*^{3} − 2*x*^{2} − *x* + 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 *x*^{3} − 2*x*^{2} − *x* + 2 by *x* + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ *x*^{3} − 2*x*^{2} − *x* + 2 = (*x* + 1) (*x*^{2} − 3*x* + 2) + 0

= (*x* + 1) [*x*^{2} − 2*x* − *x* + 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*) = *x*^{3} − 3*x*^{2} − 9*x *− 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 *x*^{3} + 3*x*^{2} − 9*x *− 5 by *x* + 1.

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

∴ *x*^{3} − 3*x*^{2} − 9*x *− 5 = (*x *+ 1) (*x*^{2} − 4*x* − 5) + 0

= (*x *+ 1) (*x*^{2} − 5*x* + *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*) = *x*^{3} + 13*x*^{2} + 32*x* + 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 *x*^{3} + 13*x*^{2} + 32*x* + 20 by (*x* + 1).

By long division,

It is known that,

Dividend = Divisor × Quotient + Remainder

*x*^{3} + 13*x*^{2} + 32*x* + 20 = (*x *+ 1) (*x*^{2} + 12*x* + 20) + 0

= (*x *+ 1) (*x*^{2} + 10*x* + 2*x *+ 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*) = 2*y*^{3} + *y*^{2} − 2*y* − 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 2*y*^{3} + *y*^{2} − 2*y* − 1 by *y* − 1.

*p*(*y*) = 2*y*^{3} + *y*^{2} − 2*y* − 1

= (*y *− 1) (2*y*^{2} +3y + 1)

= (*y *− 1) (2*y*^{2} +2y + *y* +1)

= (*y *− 1) [2*y *(*y *+ 1) + 1 (*y *+ 1)]

= (*y *− 1) (*y *+ 1) (2*y *+ 1)

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