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

1 Answer
Dec 8, 2015

Answer:

You can use the grouping method to factor #y = x^3 - 3x^2 + 4x - 12# to get #(x^2 + 4)(x-3)#.

Explanation:

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

#y = (x^3 - 3x^2) + (4x - 12)#

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(x - 3) + 4(x-3)#

Hey look at that--what do you notice about our "leftover" terms? They're both #(x - 3)#. We can merge these together into one common #(x - 3)# 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 = (x^2 + 4)(x - 3)#

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 = +2i, -2i, and 3#.