Question

Factorise:

x³ – 6x² + 11x – 6

Answer 1

Let p(x) = x³ – 6x² + 11x – 6

Constant term of p(x) = – 6

Factors of – 6 are ±1, ±2, ±3, ±6.

By trial, we find that p(1) = 0, so (x – 1) is a factor of p(x)  ...[∵ (1)³ – 6(1)² + 11(1) – 6 = 1 – 6 + 11 – 6 = 0]

Now, we see that x³ – 6x² + 11x = 6

= x³ – x² – 5x² + 5x + 6x – 6

= x²(x – 1) – 5x(x – 1) + 6(x – 1)

= (x – 1)(x² – 5x + 6)  ...[Taking (x – 1) common factor]

Now, (x² – 5x + 6) = x² – 3x – 2x + 6  ...[By splitting the middle term]

= x(x – 3) – 2(x – 2)

= (x – 3)(x – 2)

∴ x³ – 6x² + 11x – 6 = (x – 1)(x – 2)(x – 3)