Question

Factorise:-
x^3 - 6x^2 + 11x - 6

Answer 1

(x - 1)(x - 2)(x - 3).

Explanation:

f(x) = x^3 - 6x^2 + 11x - 6

f(1) = 1 - 6 + 11 - 6 = 12 - 12

= 0 so by the Factor Theorem (x - 1) is a factor.

Also, by the Rational Root theorem, as the last term is -6 and the leading coefficient is 1 some of -1, 2,-2, 3 -3, 6,-6 might be zeroes of the function.

f(-1) = -1 + 6 - 11 - 6 = -12 so -1 is not a zero and (x+ 1) is not a factor.

f(2) = 8 - 24 +22 - 6

= 30 - 30 = 0 so (x - 2) is also a factor)

Since the last term is -6 the last factor must be (x - 3)

=Checking: f(3) = 27 -6(9) + 33 - 6

= 27 - 54 + 33 - 6

= 60-60 = 0.

So the factors are (x - 1)(x - 2)(x - 3).

Answer 2

(x - 1)(x - 2)(x - 3)

Explanation:

Note the sum of the coefficients

1 - 6 + 11 - 6 = 0

hence x = 1 is a root and (x - 1) is a factor

dividing x³ - 6x² + 11x - 6 by (x - 1) gives

(x - 1)(x² - 5x + 6)

To factor the quadratic

Consider the factors of + 6 which sum to give - 5

The factors are - 2 and - 3, since

- 2 × - 3 = 6 and - 2 - 3 = - 5, hence

x² - 5x + 6 = (x - 2)(x - 3) and

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