Polynomial special products: difference of squares | Algebra 2 | Khan Academy

Earlier in our mathematical adventures, we had expanded things like ( x + y \times x - y ). Just as a bit of review, this is going to be equal to ( x \times x ), which is ( x^2 ), plus ( x \times \text{negative } y ), which is negative ( xy ), plus ( y \times x ), which is plus ( xy ), and then minus ( y \times y ) or you could say ( y \times \text{negative } y ), so it's going to be minus ( y^2 ). Negative ( xy ) plus ( xy ) means this is just going to simplify to ( x^2 - y^2 ).

This is all review; we covered it. When we thought about factoring things that are differences of squares, we thought about this when we were first learning to multiply binomials. What we're going to do now is essentially just do the same thing but do it with slightly more complicated expressions.

So, another way of expressing what we just did is we could also write something like ( a + b \times a - b ) is going to be equal to what? Well, it's going to be equal to ( a^2 - b^2 ). The only difference between what I did up here and what I did over here is instead of an ( x ), I wrote an ( a ); instead of a ( y ), I wrote a ( b ).

Given that, let's see if we can expand and then combine like terms. If I'm multiplying these two expressions, say I'm multiplying ( 3 + 5x^4 ) times ( 3 - 5x^4 ), pause this video and see if you can work this out.

All right, well, there's two ways to approach it. You could just approach it exactly the way that I approached it up here, but we already know that when we have this pattern where we have something plus something times that same original something minus the other something, well, that's going to be of the form of this thing squared minus this thing squared.

Remember, the only reason why I'm applying that is I have a ( 3 ) right over here and here. So the ( 3 ) is playing the role of the ( a ). So, let me write that down. That is our ( a ), and then the role of the ( b ) is being played by ( 5x^4 ), so that is our ( b ) right over there.

This is going to be equal to ( a^2 - b^2 ), but our ( a ) is ( 3 ), so it's going to be equal to ( 3^2 - ) and then our ( b ) is ( 5x^4 ) minus ( 5x^4 ) squared. Now, what does all of this simplify to? Well, this is going to be equal to ( 3^2 ), which is ( 9 ), and then minus ( 5x^4 ) squared.

Let’s see, ( 5^2 ) is ( 25 ), and then ( x^4 ) squared, well that is just going to be ( x^{4 \times 4} ), which is just ( x^8 ). Another way to think about it: our exponent properties say this is the same thing as ( 5^2 \times x^{4 , \text{squared}} ). If I raise them to an exponent and then raise that to another exponent, I multiply the exponents, and there you have it.

Let's do another example. Let's say that I were to ask you: what is ( 3y^2 + 2y^5 \times 3y^2 - 2y^5 )? Pause this video and see if you can work that out.

Well, we're going to do it the same way. You can, of course, always just try to expand it out the way we did originally, but we could recognize here that, hey, I have an ( a + b ) times the ( a - b ), so that's going to be equal to our ( a^2 ).

So, what's ( 3y^2 )? Well, that's going to be ( 9y^4 ) minus our ( b^2 ). Well, what's ( 2y^5 ) squared? Well, ( 2^2 ) is ( 4 ), and ( y^5 ) squared is ( y^{5 \times 2} ) or ( y^{10} ).

And there's no further simplification that I could do here; I can't combine any like terms, and so we are done here as well.