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# Introduction to grouping

Sal factors t^2+8t+15 as (t+3)(t+5), first using the "sum-product" method, then using the grouping method. Created by Sal Khan and Monterey Institute for Technology and Education.

Video transcript

Factor t squared plus 8t plus 15. So let's just think about what happen if we multiply 2 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 de distributive property twice or using F.O.I.L, 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 essentialy multiplied every term here by every term over there. And then we have 2 terms, 2 t terms I guess we could call them. This bt plus at. So we can combine those. So we get t squared plus a plus b t, I can write this is b plus a t as well. plus ab. If we compare this to this right over here. we see that we have a similar pattern. Our coeffient, Our coeffient on the second degree term is one Our coeffient on the second degree term here is one. We didn't have to write it. Then, a plus b is a 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. That could be 15. So if we want to factor this out. We just have to find an a and a b where their product is 15 and their sum is 8. In general, in general if you ever see I'll write it in kind of the more traditional with the 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 2 numbers whose sum is equal to this thing right here and his whose product is equal that thing right there. Whose sum is equal to 8 and whose product is equal to 15 So what are 2 numbers that add up to 8 and whose product are 15. So if we just factor 15, We have 1 and 15. Those don't add up to 8 in anyway. 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 can just go straight and factor this and say, hey, This is t plus 3 time t plus 5. Since we already figured out what 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 step. >From what I just showed you. So this first, this polynomial right here, I'm going to write it as t squared plus, instead of 8t i'm going to write it as a sum of at plus bt. Or as a sum of 3 t plus 5 t. So plus 3t plus 5t. So, I'm essentially, I starded here and 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 by grouping. These 2 guys right over here have a common factor t. And these 2 guys over here have a common factor 5. So let's factor out the t in this first expression 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 multplying times both of these terms So, let's factor, let's factor that out. So it becomes t plus 3 times, times, t plus 5 times, I'll write that plus a little bit neater, times t plus 5. And we are done. We didn't actually have to do this grouping step altough 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 is t plus 3 times t plus 5. or t plus 5 times t plus 3 either way