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# Multiplying Complicated Polynomials - Concept

###### 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!

When multiplying polynomials, sometimes we come across **complicated polynomials** that we can use substitution and multiplication to solve. When we have a trinomial where a binomial follows the (a-b)(a+b) format, simply subsitute the entire binomial in for one of the variables and simplify.

Multiplying more complicated polynomials, so we know that when we multiply a minus b times a plus b we can foil this up. This a got distributed to both items over here, this negative b got distributed by the both items and we end up with another polynomial. So in doing that we know we end up with the a times the a which is a squared, distributing this a to the b and the a to the negative b those 2 are equal and opposite so those cancel out to nothing and we have the b times the negative b ending up with negative b squared. So the standard foil operations. If we want to multiply out these 2 polynomials things get a little bit more complicated, we could do the exact same approach as we did before which would be take this 3x multiply it by the 3x the 1 and the 3y, take the 1 add it by each of these 3 and the negative 3 add to each 3 and then combine like terms.

That seems like a pretty long arduous process, so what I want to do is take a step back and look at this and see if there's anything that we can do to make our life easier okay. So what I see is there's a 3x plus 1 and a 3x plus 1 in both okay. So I'm just sort of group those off to the side and sort of distinguish them for the rest of the problem. We also then have a minus 3y and a plus 3y okay. So we have one item minus something else and that same item plus something else. There's actually a really close relationship between this problem and this one up here. So what we can do is if we say let a equal 3x plus 1 and let b equal 3y what we've actually done is turn this equation into a minus b and this equation into a plus b.

We just did this calculation up here so we know that this is going to be a squared minus b squared okay. More specifically we know that a is a 3x plus 1 so this turns into 3x plus 1 squared and our b turns into 3y squared okay. Now let's say we're dealing with having to multiply our 3 terms we just have to square something a lot easier, so we just would have to foil this out becomes 9x squared plus we're going to have the 2 of the 1 times the 3x that becomes 6x plus 1 and 3y squared, square goes to both thing so this ends up minus 9 squared. So using a little bit of a shortcut started seeing some similarities in these 2 equations we can make a substitution and make our life a lot easier than if we had to take each of these elements and distribute it through.

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###### Carl Horowitz

B.S. in Mathematics University of Michigan

He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his step-by-step explanations are easy to follow.

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## Rex · 10 months, 2 weeks ago

you should add a search bar to find the right video