Multiplying Complicated Polynomials - Concept 7,283 views

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.