University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Factoring a complicated difference of squares, so what I have behind me is actually a difference of squares. I have something squared minus something else squared, now what we want to do is factor this. The first thing that a lot of students see is squares, let’s FOIL it out, combine like terms and then factor it out. While that’s perfectly acceptable, I do want to show you a little bit of a short cut.
We know how to factor a² minus b²; this is just going to be a minus b times a plus b. This problem is really no different. If we call 3x minus 1, a, and x plus 3, b, what we actually have is something of this form, so let’s just write that down so you can see it.
So if a is equal to 3x minus 1 and b is equal to x plus 3, what we actually have then is a² minus b², easy enough hopefully you see that. So really all we have to do then is plug these two things for a and b, so a minus b turns into 3x minus 1, minus x plus 3 and then we have our in parenthesis there, our 3x minus 1 plus x plus 3. And the reason I throw in all these parenthesis is we do have to remember to distribute this negative sign through, a lot of times people forget that. Also a lot of people just want to leave their answer like this. This is fairly easy phrase to simplify, so we really don’t have any excuse not simplify it up.
Simplifying terms combining terms on this side we have 3x minus x, so that is going to be 2x, minus 1 minus 3, so that's just going to be -4. 3x plus x is 4x and then -1 plus 3 is +2. What I see here now is that I actually have a term I can factor out of both of these things. I can factor out a 2.
So I’m going to come back way this way. What we have here is actually 2, x minus 2 and then we can factor out 2 from over there, 2 times 2x plus 1. We can combine our 2s and we’re left with 4x minus 2, 2x plus 1.
Okay so factoring a difference of squares where you’re not actually dealing with single terms, you’re dealing with binomials. It’s really not that hard, just make a simple substitution for each of your terms, put it in your difference of squares formula and then just simplify. In this case both of our simplification terms had a common factor we could take that out more step, but by making a substitution we’re able to solve this without having to FOIL everything out.
