# How do you factor x^3 – 4x^2 +x + 6?

Mar 29, 2017

$\left(x - 2\right) \left(x - 3\right) \left(x + 1\right)$

#### Explanation:

It is usually really, really hard to factorize a cubic function. However, for this polynomial, we can factor by grouping. We try values for splitting the term $- 4 {x}^{2}$.

For example, we split it into $- 2 {x}^{2} - 2 {x}^{2}$.

The equation becomes this: $\left({x}^{3} - 2 {x}^{2}\right) - \left(2 {x}^{2} - x - 6\right)$. We can factorize each of the expressions in the parentheses: ${x}^{2} \left(x - 2\right) - \left(x - 2\right) \left(2 x + 3\right)$. There is a common factor $\left(x - 2\right)$.

Factoring the common factor out, we get $\left(x - 2\right) \left({x}^{2} - 2 x - 3\right)$. We then factorize ${x}^{2} - 2 x - 3$ to $\left(x - 3\right) \left(x + 1\right)$.

The fully factored form is then $\left(x - 2\right) \left(x - 3\right) \left(x + 1\right)$.

Mar 29, 2017

${x}^{3} - 4 {x}^{2} + x + 6 = \textcolor{m a \ge n t a}{\left(x - 2\right) \left(x - 3\right) \left(x + 1\right)}$

#### Explanation:

Provided the expression has rational roots, we can use the rational root theorem.

For the expression $\textcolor{g r e e n}{{x}^{3} - 4 {x}^{2} + x + 6}$
according to the rational root theorem, possible rational roots are:
$\textcolor{w h i t e}{\text{XXX}} \left\{\pm 1 , \pm 2 , \pm 3 , \pm 6\right\}$

With the use of a spreadsheet these values can be easily checked (it can also be done with a calculator or even manually with a bit more effort).

From this, we see that $x = 2$, $x = 3$, and $x = - 1$ all are zeros for the given expression.
This implies that $\left(x - 2\right)$, $\left(x - 3\right)$, and $\left(x + 1\right)$ are factors of the given expression.

Since ${x}^{3} - 4 {x}^{2} + x + 6$ is of degree $3$, it has a maximum of $3$ factors and we have found them all.