# How do you write P(x) = x^3 + 6x^2 − 15x − 100 in factored form?

Sep 13, 2015

${x}^{3} + 6 {x}^{2} - 15 x - 100 = \left(x - 4\right) \left(x + 5\right) \left(x + 5\right)$

#### Explanation:

By the rational root theorem, any rational roots of $P \left(x\right) = 0$ must be expressible in the form $\frac{p}{q}$, where $p , q \in \mathbb{Z}$, $q > 0$, $\text{hcf} \left(p , q\right) = 1$, $p$ a divisor of the constant term $- 100$ and $q$ a divisor of the coefficient $1$ of the leading term.

So the only possible rational roots of $P \left(x\right) = 0$ are:

$\pm 1$, $\pm 2$, $\pm 4$, $\pm 5$, $\pm 10$, $\pm 25$, $\pm 50$, $\pm 100$

Let's try some:

$P \left(1\right) = 1 + 6 - 15 - 100 = - 108$
$P \left(- 1\right) = - 1 + 6 + 15 - 100 = - 80$
$P \left(2\right) = 8 + 24 - 30 - 100 = - 102$
$P \left(- 2\right) = - 8 + 24 + 30 - 100 = - 54$
$P \left(4\right) = 64 + 96 - 60 - 100 = 0$

So $\left(x - 4\right)$ is a factor of $P \left(x\right)$.

Use synthetic division to divide $P \left(x\right)$ by $\left(x - 4\right)$...

So ${x}^{3} + 6 {x}^{2} - 15 x - 100 = \left(x - 4\right) \left({x}^{2} + 10 x + 25\right)$

Now we can recognise ${x}^{2} + 10 x + 25$ as a perfect square trinomial. It is of the form ${a}^{2} + 2 a b + {b}^{2} = {\left(a + b\right)}^{2}$, with $a = x$ and $b = 5$.

So ${x}^{2} + 10 x + 25 = {\left(x + 5\right)}^{2}$

Putting this together:

${x}^{3} + 6 {x}^{2} - 15 x - 100 = \left(x - 4\right) \left(x + 5\right) \left(x + 5\right)$