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Video transcript

Factor t squared plus 8t plus 15. So let's just think about what happens if we multiply two binomials t plus a times t plus b. And I'm using t here because t is the variable in the polynomial that we need to factor. So if you multiply this out, applying the distributive property twice or using FOIL, you get t times t, which is t squared. Plus t times b, which is bt. Plus a times t, which is at. Plus a times b, which is ab. We essentially multiplied every term here by every term over there. And then we have two t terms, I guess you could call them, this bt plus at. So we could combine those. And we get t squared plus a plus bt-- I could have written this as b plus at as well-- plus ab. Now, if we compare this to this right over here, we see that we have a similar pattern. Our coefficient on the second degree term is 1. Our coefficient on the second degree term here is 1. We didn't have to write it. Then a plus b is the coefficient on t. So 8 right over here, this 8 could be a plus b. And then finally, our constant term, ab, that could be 15. So if we wanted to factor this out, we just have to find and a and a b where their product is 15 and their sum is 8 in general. In general, if you ever see-- and I'll write it in the more traditional with an x variable-- if you see anything of the form x squared plus bx plus c, the coefficient here is 1. Then you just have to find two numbers whose sum is equal to this thing right here and whose product is equal to that thing right there. Whose sum is equal to 8 and whose product is equal to 15. So what are two numbers that add up to 8 whose products are 15? So if you just factor 15, it could be 1 and 15. Those don't add up to 8 in any way. 3 and 5, those do add up to 8. So a and b could be 3 and 5. So a and b, this could be 3 times 5, and then 8 is 3 plus 5. Now we could just go straight and factor this and say, hey, this is t plus 3 times t plus 5, since we already figured out a and b are. But what I want to do is kind of factor this by grouping. So I'm essentially going to go in reverse steps from what I just showed you. So this polynomial right here, I'm going to write it as t squared plus-- instead of writing 8t I'm going to write it as the sum of at plus bt, or as the sum of 3t plus 5t. So plus 3t plus 5t. So essentially I've started here. And then I'm going to this step where I break up that middle term into the coefficients that add up to the 8. And then finally plus 15. And now I can factor it by grouping. These two guys right over here have a common factor t. And then these two guys over here have a common factor 5. So let's factor out the t in this first expression right over here, or this part of the expression. So it's t times t plus 3. Plus-- and then over here, if you factor out a 5 you get a 5-- times t plus 3. 5t divided by 5 is t. 15 divided by 5 is 3. And now you can factor out a t plus 3. You have a t plus 3 being multiplied times both of these terms. So let's factor that out. So it becomes t plus 3 times t plus 5. Let me write that plus a little bit neater. Times t plus 5. And we are done. But we actually didn't even have to do this grouping step, although hopefully you see that it does work. We could have just said, look, from this pattern over here I have two numbers that add up to 8 and their product is 15. So this must be t plus 3 times t plus 5. Or t plus 5 times t plus 3, either way.