# How do you find all the zeros of f(x)=x^3+3x^2-4x-12?

May 31, 2016

Factor by grouping to find zeros: $2 , - 2 , - 3$

#### Explanation:

Notice that the ratio between the first and second terms is the same as that between the third and fourth terms.

So this cubic factors by grouping:

${x}^{3} + 3 {x}^{2} - 4 x - 12$

$= \left({x}^{3} + 3 {x}^{2}\right) - \left(4 x + 12\right)$

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

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

$= \left(x - 2\right) \left(x + 2\right) \left(x + 3\right)$

Hence zeros: $2 , - 2 , - 3$

Jul 15, 2018

$\left(x + 2\right) \left(x - 2\right) \left(x + 3\right)$

#### Explanation:

We essentially have two parts to our function:

$\textcolor{s t e e l b l u e}{{x}^{3} + 3 {x}^{2}} \textcolor{p u r p \le}{- 4 x - 12}$

Let's factor a ${x}^{2}$ out of the blue term, and a $- 4$ out of the purple term. We now have

$\textcolor{s t e e l b l u e}{{x}^{2} \left(x + 3\right)} \textcolor{p u r p \le}{- 4 \left(x + 3\right)}$

Since both terms have an $x + 3$ in common, we can factor that out. We now have

$\textcolor{red}{\left({x}^{2} - 4\right)} \left(x + 3\right)$

You might immediately recognize that what I have in red is a difference of squares, which can be factored as $\left(x + 2\right) \left(x - 2\right)$.

We now have

$\left(x + 2\right) \left(x - 2\right) \left(x + 3\right)$

Hope this helps!