###### Carl Horowitz

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!

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Review of the Methods of Factoring - Problem 2

# Review of the Methods of Factoring - Problem 1

Carl Horowitz
###### Carl Horowitz

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!

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I now like to take a look at this fairly ugly expression behind me and try to factor that out. So the first thing that you see is that there is a lot going on. We have squared terms and we have binomial squared and all sorts of stuff. And a lot of students first reaction to this is to FOIL everything up, get rid of your parenthesis and make it sort of one long string of a polynomial.

You could do that and it’s perfectly fine, but it’s going to create a lot more work and what I do want to point out is that every single term has this x plus 1² which means it’s a common factor and we can take it out okay? Just like if every term had an even coefficient we could take out a list of two, maybe even greater.

So whenever you see every term have something in common just take it out, it makes your life a little bit easier. So let’s take out x plus 1² and then I like to put brackets in there just to distinguish what we factored out from what’s inside, solve parenthesis will work just fine.

So then we have a 2x², a minus 7x and a plus 6 left over. So now we have this x plus 1² on the outside which is fine and then we want to factor this inside portion as well, so we can bring down the x plus 1² and then we want to write this as the product of two binomials.

A lot of students have different tricks up their sleeve, you could use a square, you could use a diamond, there’s all sorts of different ways to think about the factors of this term, but basically figure out a way that works for you and stick with it. The way I do is I tend to just think about logically and go through some products in my head in order to best figure out.

So what I see is I see a negative sign in the middle and a positive sign on the end. The only way to get this to be negative is by having this negative as well, so I know that I have to be subtracting in the middle. A negative times a negative is positive which will take care of my positive 6. I have a 2x² which means I have to break it down to a 2x and an x, it’s the only factors and because they’re both negative where I put the 2x and where I put the x is going to matter. So I know I have to have an x in one spot and a 2x in the other one.

Now we have to figure out how to get this 6 and the 6 is going to be the last one to multiply together. The factors of 6 are 1 and 6, or 2 and 3. Let’s think about 1 and 6. If I put the 1 and the 6 in; if I put the 1 here this will give me -2, I put the 6 here that will give me -6. -2 and -6 is -8 that’s not going to work. If I switch the 1 and the 6, have the 6 here that’s going to give me -12 plus another negative number there’s no way to make -7, so I know I’m not dealing with 1 and 6 which leaves me with 2 and 3.

If I put 3 here, -3 times 2, -6 to leave the 2 here give me -8, so the only way left then is I have to pair the 2 with the 2 to give me -4 and the 3 with the 1 to give me -7. So that’s logically how I think about it and in general, obviously I don’t talk it loud, but that’s my thought process as to how I figure out what term is going to go where. It’s a lot harder when we have a leading coefficient of a term other than just the one, but it’s still doable.

So what we’ve done is factored out our common factor and then we were left with the trinomial which we just factored, leading term was a number other than 1 which makes it a little bit difficult but still difficult and we have our answer in factored form.