Perfect square factorization intro | Mathematics II | High School Math | Khan Academy

We're going to learn to recognize and factor perfect square polynomials in this video. So, for example, let's say I have the polynomial x² + 6x + 9. If someone asks you, "Hey, can you factor this into two binomials?" Well, using techniques we learned in other videos, you say, "Okay, I need to find two numbers whose product is nine and whose sum is six."

So, I encourage you to think of, to pause this video and say, "Well, what two numbers can I add up to six, and if I take their product, I get nine?" Well, 9 only has so many factors, really: 1, 3, and 9. And 1 + 9 does not equal 6.

But 3 × 3 equals 9, and 3 + 3 does equal 6. So we can factor this as (x + 3)(x + 3), which is of course the same thing as (x + 3)².

Now, what was it about this expression that made us recognize, or maybe now we will start to recognize it, as being a perfect square? Well, I have, of course, some variable that is being squared, which we need. I have some perfect square as the constant, and that whatever is being squared there—I have two times that as the coefficient on this first-degree term here.

Here, let's see if that is generally true, and I'll switch up the variables just to show that we can. So, let's say that I have a² + 14a + 49. A few interesting things are happening here.

Alright, I have my variable squared. I have a perfect square constant term that is 7² right over here. And my coefficient on my first-degree term here is 2 times the thing that’s being squared; that is, 2 × 7. Or you could say it's 7 + 7.

So, you can immediately say, "Okay, if I want to factor this, this is going to be (a + 7)²." You can verify that by multiplying out by figuring out what (a + 7)² is. Sometimes when you're first learning, it's like, "Hey, isn't that just a² + 7a?"

No! Remember, this is the same thing as (a + 7)(a + 7). You can calculate this by using the FOIL technique. I don't like that so much because you're not thinking mathematically about what's happening, really. You just have to do the distributive property twice here.

First, you can multiply (a + 7)(a) so (a + 7)(a) and then multiply (a + 7)(7) so plus (a + 7)(7). This is going to be a² + 7a plus now we distribute this 7, plus 7a plus 49.

So now you see where that 14a came from—it's from the 7a plus the 7a. You see where the a² came from, and you see where the 49 came from. You can speak of this in more general terms if I wanted to.

If I wanted to just take the expression (A + B)², that's just (A + B)(A + B). We do exactly what we did just here, but here I'm just doing it in very general terms with A or B. You can think of A as either a constant number or even a variable.

So this is going to be, if we distribute this, it's going to be A + B times A plus A + B times B. This is going to be A². Now I'm just doing the distributive property again.

A² + 2Ab + B². So this is A² + 2Ab + B². This is going to be the general form. So if A is the variable (which was X or a in this case), then it's just going to be whatever squared in the constant term is going to be 2 times that times the variable.

I want to show that there's some variation that you can entertain here. So, if you were to see 25 + 10x + x² and someone wanted you to factor that, you could say, "Look, this right here is a perfect square. It's 5². I have the variable squared right over here, and then this coefficient on our first-degree term is 2 × 5."

So you might immediately recognize this as (5 + x)². Now, of course, you could just rewrite this polynomial as x² + 10x + 25, in which case you might say, "Okay, variable squared, some number squared, 5²; two times that number is the coefficient here."

So that's going to be (x + 5)². And that's good because these two things are absolutely equivalent.