###### 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 6

# Review of the Methods of Factoring - Problem 5

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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Factoring a trinomial when your leading coefficient is a number other than 1. So factoring something like this is probably one of the hardest things to factor because we’re left with a number other than 1 upfront. It makes it a little bit harder to deal with in terms of what our factors are going to have to be, but I know that I have 3 terms, so I know it’s going to have to be two binomials that we’re multiplying together and I also know that by looking at this I don’t have any common factors I can take out to make my life easier.

If I could take out a factor, it’d make life a little bit easier. Both 10 and 5 are multiples of 5, but this 23 in the middle kind of throws a wrench in that, so I have to leave it as is. Okay so what I see here is that I know that I need a x in either first term and I know that I need a 5y in one of them because the only way to get 5y² is the 5y and the y. So I can assign that to either of those spots and then work around them.

So what I need to do now is figure out a number of things. My last sign is negative which tells me that I need one positive and one negative sign when I factor it out. My middle term is positive which means that my positive portion has to be greater. So when I multiply the two positives, that has to be bigger than the positive times a negative and I’m looking at 10 which I know I’ll need to factor. The factors of 10 are 1 and 10 or 2 and 5 that’s all we’re dealing with.

So if I throw in a 1 and a 10, I’m multiplying it by 1 and 5, so the only way numbers I’m going to be able to get are multiples of 5. 10 times 5, 10 times 1 they’re all multiples of 5 or 10. There is no way this is going to be a 10, so we’re going to have to break it up as 2 times 5, the only other option.

So what I’m left with is trying to figure out what gets paired with the 5 and what gets paired with the 2. I need 23 which is the fairly good sized number. 5 times 2 isn’t going to get me that, so I’m going to have to take this 5 and multiply it by the +5 as well; so I know that this has to be 5 and then leaving this as 2.

The last thing we have to do is deal with the sign. This is 25 and this is 2. I want the 25 to be positive so therefore this 5y actually is positive leaving this y as negative. Always check your work make sure your numbers work out. Here is my 10x², here is my -5y² those are the easy ones. Work on the middle term we end up with +25xy and -2xy adding those together we end up with 23xy.

So just by sort of thinking our way through it, we were able to factor this trinomial down into the product of two binomials.