Current time:0:00Total duration:4:12

0 energy points

Studying for a test? Prepare with these 5 lessons on Equation du second degré.

See 5 lessons

# Introduction to grouping

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.