# Greatest Common Factors - Problem 2

Here I’m given polynomials and I’m going to look for monomial that’s the greatest common factor. First thing I’m going to look at is the numbers or the constants or the coefficients and then I’m going to be looking at the variable parts. So here we go.

5 and 10, the number that multiplies into both 5 and 10 is 5. So that’s going to be part of my greatest common factor. X to the third and x, the number of Xs that goes into both of those is just 1x. So if I were to be un-distributing I need to think 5x multiplied by what gives me 5x to the third. 5x times x² is what gives me 5x to the third.

Here I have my greatest common factor and now I’m kind of thinking backwards. I’m like un-distributing. 5x times what gives me 10x? Just 2. There we go, that’s the factor form of this binomial. I took this binomial and wrote it as the product of 2 different other polynomials.

Let’s look at this guy. I want to figure out what number goes into 6, -12 and -60. Well the number that goes into all of those is 6.that’s going to be part of my greatest common factor. Then I want to look at the ps. How many ps go into each of these? It’s got to be one p. Next I want to think 6p times what gives me 6p to the third? There it is. 6p times -2 would give me that, sorry, 6p times -2gives me that -12 part. I still need one more p. Last but not least I want to do 6p times what gives me -60p. That’s going to be -10. I think that’s my factored form, but that’s kind of confusing. I want to show you guys how to check your work and that’s by distributing. So let me just check.

By distributing 6p times p² I have 6p to the third take away 12p² take away 60p. Good, that’s how I know I did it correctly because I went back and distributed. So you guys you can absolutely do these problems. It’s just important that you start by finding the common, greatest common factor and from there you go through the un-distributing process. Figure out what do you multiply that greatest common factor by in order to get each of the original common terms.

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