Q.5. Factorise:

 (i) x3 - 2x2 - x + 2 (ii) x3 - 3x2 - 9x - 5 (iii) x3 + 13x2 + 32x + 20 (iv) 2y3 + y2 - 2y - 1

Solution:

Using trial and error with factor theorem.

(i) Let p(x) = x3 - 2x2 - 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,

x3 - 2x2 - x + 2 =(x+1) (x2 - 3x + 2)

= (x+1) (x2 - x - 2x + 2) [by midterm splitting]

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

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

(ii) Let p(x) = x3 - 3x2 - 9x - 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,

x3 - 3x2 - 9x - 5 = (x - 5) (x2 + 2x + 1)

= (x-5) (x2 + x + x + 1) [by midterm splitting]

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

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

(iii) Let p(x) = x3 + 13x2 + 32x + 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,

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

= (x+1) (x2 + 2x + 10x + 20) [by midterm splitting]

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

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

(iv) Let p(y) = 2y3 + y2 - 2y - 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,

2y3 + y2 - 2y - 1 = (y-1) (2y2 + 3y + 1)

= (y-1) (2y2 + 2y + y + 1) [by midterm splitting]

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

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