#### Question

If (x + 1) and (x – 2) are factors of x^{3} + (a + 1)x^{2} – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.

#### Solution

Let f(x) = x^{3} + (a + 1)x^{2} – (b – 2)x – 6

Since, (x + 1) is a factor of f(x).

Remainder = f(-1) = 0

(-1)^{3} + (a + 1)(-1)^{2} – (b – 2) (-1) – 6 = 0

-1 + (a + 1) + (b – 2) – 6 = 0

a + b – 8 = 0 …(i)

Since, (x – 2) is a factor of f(x).

Remainder = f(2) = 0

(2)^{3} + (a + 1) (2)^{2} – (b – 2) (2) – 6 = 0

8 + 4a + 4 – 2b + 4 – 6 = 0

4a – 2b + 10 = 0

2a – b + 5 = 0 …(ii)

Adding (i) and (ii), we get,

3a – 3 = 0

a = 1

Substituting the value of a in (i), we get,

1 + b – 8 = 0

b = 7

∴ f(x) = x^{3} + 2x^{2} – 5x – 6

Now, (x + 1) and (x – 2) are factors of f(x). Hence, (x + 1) (x – 2) = x^{2} – x – 2 is a factor of f(x).

f(x) = x^{3} + 2x^{2} – 5x – 6 = (x + 1) (x – 2) (x + 3)