Question

If (x + 1) and (x – 2) are factors of x³ + (a + 1)x² – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.

Answer 1

Let f(x) = x³ + (a + 1)x² – (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) = x³ + 2x² – 5x – 6

Now, (x + 1) and (x – 2) are factors of f(x).

Hence, (x + 1)(x – 2) = x² – x – 2 is a factor of f(x). 

                      x + 3
`x^2 - x - 2")"overline(x^3 + 2x^2 - 5x - 6)`
                    x³ – x² – 2x             
                         3x² – 3x – 6
                         3x² – 3x – 6        
                                0
∴ f(x) = x³ + 2x² – 5x – 6 = (x + 1)(x – 2)(x + 3)