# How do you factor x^3 + 14x^2 + 60x + 72 = 0?

Jul 24, 2018

$\left(x + 2\right) {\left(x + 6\right)}^{2} = 0$

#### Explanation:

Given:

${x}^{3} + 14 {x}^{2} + 60 x + 72 = 0$

By the rational roots theorem, any rational zeros of the given cubic are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $72$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

$\pm 1 , \pm 2 , \pm 3 , \pm 4 , \pm 6 , \pm 8 , \pm 9 , \pm 12 , \pm 18 , \pm 24 , \pm 36 , \pm 72$

In addition, note that all of the coefficients are positive and the constant term is non-zero. As a result, any real zero (rational or otherwise) of this cubic must be negative.

So that leaves rational possibilities:

$- 1 , - 2 , - 3 , - 4 , - 6 , - 8 , - 9 , - 12 , - 18 , - 24 , - 36 , - 72$

We find:

${\left(\textcolor{b l u e}{- 2}\right)}^{3} + 14 {\left(\textcolor{b l u e}{- 2}\right)}^{2} + 60 \left(\textcolor{b l u e}{- 2}\right) + 72 = - 8 + 56 - 120 + 72 = 0$

So $x = - 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{3} + 14 {x}^{2} + 60 + 72 = \left(x + 2\right) \left({x}^{2} + 12 x + 36\right)$

Without trying any more of our "possible" zeros, we can recognise the remaining quadratic factor as a perfect square trinomial:

${x}^{2} + 12 x + 36 = {x}^{2} + 2 \left(x\right) \left(6\right) + {6}^{2} = {\left(x + 6\right)}^{2}$

So the factored form of the given cubic equation can be written:

$\left(x + 2\right) {\left(x + 6\right)}^{2} = 0$