**Q.5.** Factorise:

(i) |
(ii) |

(iii) |
(iv) 2 |

**Solution:**

Using trial and error with factor theorem.

(i) Let *p*(*x*) = *x*^{3} - 2*x*^{2} - *x* + 2

Factors of 2 are ±1 and ± 2

By trial and error method, we find that,

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

= -1 -2 + 1 + 2 = 0

Therefore, (*x*+1) is a factor of *p*(*x*)

Now, we can write, using long division,

*x*^{3} - 2*x*^{2} - *x* + 2 =(*x*+1) (*x*^{2} - 3*x* + 2)

= (*x*+1) (*x*^{2} - *x* - 2*x* + 2) [by midterm splitting]

= (*x*+1) {*x*(*x*-1) -2(*x*-1)}

= (*x*+1) (*x*-1) (*x*+2)

(ii) Let *p*(*x*) = *x*^{3} - 3*x*^{2} - 9*x* - 5

Factors of 5 are ±1 and ±5

By trial and error method, we find that,

*p*(5) = (5)^{3} - 3(5)^{2} - 9(5) - 5

= 125 - 75 - 45 - 5 = 0

Therefore, (*x*-5) is the factor of *p*(*x*)

Now, we can write, using long division,

*x*^{3} - 3*x*^{2} - 9*x* - 5 = (*x *- 5) (*x*^{2} + 2*x* + 1)

= (*x*-5) (*x*^{2} + *x* + *x* + 1) [by midterm splitting]

= (*x*-5) {*x*(*x *+ 1) +1(*x *+ 1)}

= (*x *- 5) (*x*+1) (*x*+1)

(iii) Let *p*(*x*) = *x*^{3} + 13*x*^{2} + 32*x* + 20

Factors of 20 are ±1, ±2, ±4, ±5, ±10 and ±20

By trial and error method, we find that,

p(-1) = (-1)^{3} + 13(-1)^{2} + 32(-1) + 20

= -1 + 13 - 32 + 20 = 0

Therefore, (*x*+1) is a factor of *p*(*x*)

Now, we can write, using long division,

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

= (*x*+1) (*x*^{2} + 2*x* + 10*x* + 20) [by midterm splitting]

= (*x*-5) {*x*(*x*+2) +10(*x*+2)}

= (*x*-5) (*x+*2) (x+10)

(iv) Let *p*(*y*) = 2*y*^{3} + *y*^{2} - 2*y* - 1

Factors of ab = 2×(-1) = -2 are ±1 and ±2

By trial and error method, we find that,

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

= 2 +1 - 2 - 1 = 0

Therefore, (*y*-1) is a factor of *p*(*y*)

Now, we can write, using long division,

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

= (*y*-1) (2*y*^{2} + 2*y* + *y* + 1) [by midterm splitting]

= (*y*-1) {2*y*(*y*+1) +1(*y*+1)}

= (*y*-1) (2*y*+1) (*y*+1)