# How to factor cubic trinomials? x^3-7x-6

Jul 25, 2015

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

#### Explanation:

You could solve this by plotting the equation and inspecting where the roots are:
graph{x^3-7x-6 [-5, 5, -15, 5]}

We can see there appear to be roots in the areas of $x = - 2 , - 1 , 3$, if we try these we see this is indeed a factorisation of the equation:

$\left(x - 3\right) \left(x + 1\right) \left(x + 2\right) = \left(x - 3\right) \left({x}^{2} + 3 x + 2\right) = {x}^{3} - 7 x - 6$

Jul 25, 2015

Use the rational roots theorem to find possible roots, try each to find roots $x = - 1$ and $x = - 2$ hence factors $\left(x + 1\right)$ and $\left(x + 2\right)$ then divide by these to find $\left(x - 3\right)$

${x}^{3} - 7 x - 6 = \left(x + 1\right) \left(x + 2\right) \left(x - 3\right)$

#### Explanation:

Find roots of ${x}^{3} - 7 x - 6 = 0$ and hence factors of ${x}^{3} - 7 x - 6$.

Any rational root of a polynomial equation in standard form is of the form $\frac{p}{q}$, where $p$, $q$ are integers, $q \ne 0$, $p$ a factor of the constant term and $q$ a factor of the coefficient of the term of highest degree.

In our case $p$ must be a factor of $6$ and $q$ a factor of $1$.

So the only possible rational roots are: $\pm 1$, $\pm 2$, $\pm 3$ and $\pm 6$.

Let $f \left(x\right) = {x}^{3} - 7 x - 6$

$f \left(1\right) = 1 - 7 - 6 = - 12$
$f \left(- 1\right) = - 1 + 7 - 6 = 0$
$f \left(2\right) = 8 - 14 - 6 = - 12$
$f \left(- 2\right) = - 8 + 14 - 6 = 0$

So $x = - 1$ is a root of $f \left(x\right) = 0$ and $\left(x + 1\right)$ a factor of $f \left(x\right)$.
$x = - 2$ is a root of $f \left(x\right) = 0$ and $\left(x + 2\right)$ a factor of $f \left(x\right)$.

$\left(x + 1\right) \left(x + 2\right) = {x}^{2} + 3 x + 2$

Divide $f \left(x\right)$ by the factors we've found so far to find:

${x}^{3} - 7 x - 6 = \left({x}^{2} + 3 x + 2\right) \left(x - 3\right)$

Actually you can deduce the $x$ and the $- 3$ simply by looking at what you need to multiply ${x}^{2}$ and $2$ by to get ${x}^{3}$ and $- 6$.

So the complete factorisation is:

${x}^{3} - 7 x - 6 = \left(x + 1\right) \left(x + 2\right) \left(x - 3\right)$