# Can y=x^3+5x^2-12x-36  be factored? If so what are the factors ?

Sep 19, 2016

${x}^{3} + 5 {x}^{2} - 12 x - 36 = \left(x - 3\right) \left(x + 2\right) \left(x + 6\right)$

#### Explanation:

$f \left(x\right) = {x}^{3} + 5 {x}^{2} - 12 x - 36$

By the rational roots theorem, any rational zeros of $f \left(x\right)$ are expressible in the form $\frac{p}{q}$ for integers $p , q$ with $p$ a divisor of the constant term $- 36$ and $q$ a divisor of the coefficient $1$ of the leading term.

That means that the only possible rational zeros are:

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

In addition note that the signs of the coefficients are in the pattern: $+ + - -$. By Descartes' rule of signs we can deduce that $f \left(x\right)$ has exactly $1$ positive Real zero. It has $0$ or $2$ negative Real zeros.

So let's try the positive possibilities first:

$f \left(1\right) = 1 + 5 - 12 - 36 = - 42$

$f \left(2\right) = 8 + 20 - 24 - 36 = - 32$

$f \left(3\right) = 27 + 45 - 36 - 36 = 0$

So $x = 3$ is a zero and $\left(x - 3\right)$ a factor:

${x}^{3} + 5 {x}^{2} - 12 x - 36 = \left(x - 3\right) \left({x}^{2} + 8 x + 12\right)$

To factor the remaining quadratic note that $2 + 6 = 8$ and $2 \cdot 6 = 12$

Hence:

${x}^{2} + 8 x + 12 = \left(x + 2\right) \left(x + 6\right)$

So our complete factorisation is:

${x}^{3} + 5 {x}^{2} - 12 x - 36 = \left(x - 3\right) \left(x + 2\right) \left(x + 6\right)$