### Video Transcript

Difference of Two Squares

So if we start off with a square, that is, dimensions π by π, we know the area of this square will be π squared. This is the difference of two squares. So letβs take away another square from this square. So, say that we want to find the area of the white shape. We can see that weβre taking away the area of the blue shape. So weβve got π squared and weβre taking away π squared. Or we know that by cutting down here, that gives us different dimensions on the sides for different parts. This part is of course π minus π, because the whole length is π and we can see just that part at the bottom is π, so that part is π minus π. And then, thatβs of course is π. But if we take that shape and move it up on the side, we know that we can because our dotted length is as well π minus π, we get a different shape. Weβll end up with this rectangle. So we can see the side length, again, is π minus π and the top has a length of π. And then weβre adding on the length of π. So itβs π minus π is one dimension, and π plus π is the other. We know that the area of our original shape was π squared minus π squared, because it was the big square take away the smaller square. So our new shape would be π minus π all multiplied by π plus π, as itβs a rectangle. And putting these two together, gives us the difference of two squares.

Letβs have a look at it algebraically as well. So letβs expand these two brackets using FOIL. Weβll use the first terms multiplied together, which will be π multiplied by π, gives us π squared. Then π multiplied by minus π gives us minus ππ. π multiplied by π gives us plus ππ. And finally, π multiplied by minus π gives us minus π squared. So we can see our middle terms cancel out, and weβre left with π squared minus π squared. So weβve seen it geometrically and also algebraically. Letβs actually apply this to a problem.

So, we need to find what π is. So what is being squared to give us sixteen π₯ squared. Well we know that four squared is sixteen, and π₯ squared is π₯ squared. So thatβll become four π₯ all squared. So four π₯ is the π value. Now looking at our π. Well what squared give us nine, three. So thatβll become three π¦ all squared is nine π¦ squared. And that is π. So substituting it in for the relation, weβve got π plus π all multiplied by π minus π. So we have four π₯ plus three π¦ all multiplied by four π₯ minus three π¦. Letβs look at one more example.

So looking at this example, again, you can say what squared gives us thirty-six π₯ squared. Well thatβll be six π₯ all squared. And then, what squared gives us thirty-six π¦ squared. Again, thatβll be six π¦ all squared. So we can just apply the formula putting π as six π₯ and π as six π¦, and weβre done. But in this case, thereβs actually another way we couldβve done it. If we look back to the first part of the question, thirty-six π₯ squared minus thirty-six π¦ squared, we can see that theyβve both got a common factor, the greatest common factor, of thirty-six. So we can factor the thirty-six first. And that wouldβve given us this. Well then in this case, our π is π₯ and our π is π¦. So we could substitute that into the relation, and we would get this. Now these are exactly the same. So the answer we got using the first method and the second method are exactly the same. Because if in the first method we factored six out of each individual bracket, we would then have to do six multiplied by six, which we of course know is thirty-six. So we get the same thing either way but this second method is a fully factored method. So in summary, the only thing we really need to remember, in the difference of two squares, is the relation, and just working out how to spot it when weβre given an option. So when we see a square number subtracted by another square number, obviously with variables that squared as well, then we can just go straightaway and use this relation.