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CCSS.Math: , ,

- Whereas to factor the
polynomial below as the product of two binomials and we have n times n minus one plus 3 times n minus one. So I encourage you to pause this video and see if you can figure this out. Well, the key is to realizing
that both of these terms have n minus one as a factor. Let me just rewrite the whole thing so we can work on it down here. So this is n times n minus one plus 3 times n minus one. And notice both of them
have an n minus one, have an n minus one as a factor. So what we can do is factor out the n minus one or you can view it as undistributing the n minus one and if we do that, we're gonna factor out the n minus one and what are we going to have left over? Well if you take out the n minus one here if you undistribute it out, you're just going to be left with that n. So you're going to have an n there and then for this second term, you factor this n minus one out. You're just going to be left with this positive three plus three. And just like that we are done. We have factored the polynomial below as a product of two binomials. So this is the same thing as n minus one times n plus three. And once again you can check this. You can take this n minus one and distribute it. n minus one times n is
this term right over here n times n minus one and then n minus one times three is this term right over here. n minus one times three or three times n minus one.