# How do you factor the expression x^3 - 2x^2 + 3x - 6?

Feb 16, 2016

Factors are hence $\left(x - 2\right) \left({x}^{2} + 3\right)$

#### Explanation:

To factorize the expression x^3−2x^2+3x−6, we first identify a value of $x$ for which the value of function is $0$. This could be a factor of last term i.e. $- 6$ i.e. among $\left(1 , 2 , 3 , 6 , - 1 , - 2 , - 3 , - 6\right)$.

It is seen that for $x - 2$ function is zero. Hence $\left(x - 2\right)$ is a factor of x^3−2x^2+3x−6. Dividing latter by former, we get the factors of the function as

$\left(x - 2\right) \left({x}^{2} + 3\right)$

Now as the determinant (${b}^{2} - 4 a c$ if the function is $a {x}^{2} + b x + c$) of ${x}^{2} + 3$ is

${0}^{2} - 4.1 .3$ = $- 12$. a negative number, this cannot be factorized into rational factors (assuming that to be a condition).

The factors are hence $\left(x - 2\right) \left({x}^{2} + 3\right)$