# How do you solve  (x+5)(x^2-7x+12)=0?

Sep 28, 2015

The solutions are $- 5$, $3$ and $4$.

#### Explanation:

A multiplications gives zero as a result if and only if at least one of his factors equals zero. So that's what you need to impose.

The first factor is quite simple: $x + 5$ equals zero if and only if $x = - 5$.

The second factor is a parabola, whose zeroes you may find through the classical formula {-b \pm \sqrt{b^2-4ac}/2a, but in simple cases as this, I prefer this simplier one: if the coefficient of ${x}^{2}$ is one, then you can read your equation like this:

${x}^{2} - s x + p = 0$, where $s$ is the sum of the roots, and $p$ is their product. So, you have $- s = - 7$, and $p = 12$. This means that we are looking for two numbers $a$ and $b$ such that $a + b = 7$, and $a \cdot b = 12$. It's easy to see, with barely no calculations, that these numbers are $3$ and $4$.

So, your equation is solved by three numbers, namely $- 5$, $3$ and $4$.