Question

using the factor theorem factorise the following
$x^3-6x^2+11x-6$

Answer 1

Zeroes of $x^3-6x^2+11x-6$

By using rational theorem, the roots can be among the factors of $\frac{6}{1}=6$

Let us try $x=1$

$\Rightarrow {\left(1\right)}^3-6{\left(1\right)}^2+11\left(1\right)-6=0$

$\therefore (x-1)$ is a factor of $x^3-6x^2+11x-6.$

Now, using synthetic division method :

So, the quotient $=x^2-5x+6$

Now, using common factor theorem,

$\Rightarrow x^2-5x+6=x^2-2x-3x+6$

$=x(x-2)-3(x-2)$

$=(x-2)(x-3)$

$\therefore$ Zeroes of the polynomial $=1,2,3$

So, factor of the polynomial $=(x-1)(x-2)(x-3).$

Hence, the answer is $(x-1)(x-2)(x-3).$