Question

Factorise the expression $x^3-6x^2+11x-6$.

• A. $(x+1)(x+2)(x+3)$
• B. $(x-1)(x+2)(x+3)$
• C. $(x+1)(x+2)(x-3)$
• D. $(x-1)(x-2)(x-3)$

Answer 1

The correct option is D $(x-1)(x-2)(x-3)$

Let us assume that all factors are integers. Product of the factors is equal to the constant term (-6). Possible roots are the factors of -6. Hence, possible factors are:
$\pm 1,\pm 2,\pm 3,\pm 6$

$\text{For x = 1, the value of the given}\text{expression becomes:}p\left(1\right)=1^3-6\times 1^2+11\times 1-6=0$

$\text{By factor theorem,}(x+1)\text{is a factor}\text{of}{x}^3-6x^2+11x-6.$

Performing long division as shown below:

$x^2-5x+6x-1)\overline{x^3-6x^2+11x-6}{x}^3-x^2-+\overline{-5x^2+11x-6}-5x^2+5x+-\overline{6x-6}\underset{–}{6x-6}\underset{–}{0}$

We get the quotient:
$q\left(x\right)=x^2-5x+6=x^2-2x-3x+6=x(x-2)-3(x-2)=(x-3)(x-2)$

$\text{Polynomial p(x) can be written as:}p\left(x\right)=(x-1)q\left(x\right)p\left(x\right)=(x-1)(x-2)(x-3)$