# How do you factor y=x^3 - 3x^2 + 4x -12 ?

Dec 8, 2015

You can use the grouping method to factor $y = {x}^{3} - 3 {x}^{2} + 4 x - 12$ to get $\left({x}^{2} + 4\right) \left(x - 3\right)$.

#### Explanation:

We can look at this equation in two parts, indicated by the parentheses:

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

There, we just grouped the terms! (:
Now, do you notice anything about the groups?
You'll find that we can factor out an ${x}^{2}$ from the first group and we can factor out a $4$ from the second:

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

Hey look at that--what do you notice about our "leftover" terms? They're both $\left(x - 3\right)$. We can merge these together into one common $\left(x - 3\right)$ term. And then we're going to do something that might be new for you; we're going to take the two terms that we just factored out (${x}^{2}$ and $4$), and add them to get one term to multiply the "leftovers" with:

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

And there you have it--factored form!
As a side note, if you wanted to solve this equation, all you would need to do from here is set the equation equal to zero ("plug in" 0 for $y$) and solve for $x$ to get $x = + 2 i , - 2 i , \mathmr{and} 3$.