Question

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

Answer 1

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-sx+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*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`.