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# Factorizing a polynomial by splitting the middle term

Video transcript

- [Voiceover] So we have a
quadratic expression here. X squared minus three x minus 10. And what I'd like to do in this video is I'd like to factor it as
the product of two binomials. Or to put it another way, I want to write it as
the product x plus a, that's one binomial, times x plus b, where we need to figure out
what a and b are going to be. So I encourage you to pause the video and see if you can figure
out what a and b need to be. Can we rewrite this
expression as the product of two binomials where
we know what a and b are? So let's work through this together now. I'll highlight a and
b in different colors. I'll put a in yellow and
I'll put b in magenta. So one way to think about
it is let's just multiply these two binomials using a and b, and we've done this in previous videos. You might want to review
multiplying binomials if any of this looks strange to you. But if you were to multiply what we have on the right-hand side
out it would be equal to, you're going to have the x times the x which is going to be x squared. Then you're going to
have the a times the x, which is ax. And then you're going to
have the b times the x, which is bx. Actually just let me, I'm
not gonna skip any steps here just to see it this time. This is all review, or should be review. So then we have, so we did
x times x to get x squared. Then we have a times x
to get ax, to get a x. And then we're gonna have x times b, so we're multiplying each
term times every other term. So then we have x times b to get bx. So plus bx. b x. And then finally we have
plus the a times the b, which is of course going to be ab. And now we can simplify this, and you might have been
able to go straight to this if you are familiar with
multiplying binomials. This would be x squared plus, we can add these two coefficients because they're both on
the first degree terms, they're both multiplied by x. If I have ax's and I add bx's to that I'm going to have a plus b x's. So let me write that down. a plus b x's, and then
finally I have the plus, I'll do that blue color,
finally I have it. I have plus ab. Plus ab. And now we can use this to think about what a and b need to be. If we do a little bit of pattern matching, we see we have an x squared there, we have an x squared there. We have something times x, in this case it's a
negative three times x. And here we have something times x. So one way to think about it is that a plus b needs to be
equal to negative three. They need to add up to
be this coefficient. So let me write that down. So we have a plus b needs to
be equal to negative three. And we're not done yet. We finally look at this last
term, we have a times b. Well a times b needs to
be equal to negative 10. So let's write that down. So we have a times b needs
to be equal to negative 10. And in general, whenever
you're factoring something, a quadratic expression that has
a one on second degree term, so it has a one coefficient
on the x squared, you don't even see it but
it's implicitly there. You could write this as one x squared. A way to factor it is to
come up with two numbers that add up to the coefficient
on the first degree term, so two numbers that add
up to negative three. And if I multiply those same two numbers, I'm going to get negative 10. So two numbers that add
up to negative three, to add up to the coefficient here. And now when I multiply it,
I get the constant term. I get this right over here. Two numbers when I
multiply I get negative 10. Well what could those numbers be? Well since when you multiply
them we get a negative number, we know that they're going
to have different signs. And so let's see how we
could think about it. And since when we add them
we get a negative number, we know that the negative
number must be the larger one. So if I were to just factor 10, 10 you could do that as one times ten, or two times five. And two and five are interesting because if one of them are negative, their difference is three. So if one is negative... So let's see if we're
talking about negative 10, you could say negative two times five. And when you multiply them
you do get negative 10. But if you add these two, you're going to get positive three. But what if you went positive
two times negative five. Now this is interesting because still when you multiply them
you get negative 10. And when you add them,
two plus negative five is going to be negative three. So we have just figured
out our two numbers. We could say that a is two or
we could say that b is two, but I'll just say that a is equal to two and b is equal to negative five. And so our original expression, we can rewrite as, so we can rewrite x squared
minus three x minus 10. We can say that that
is going to be equal to x plus two times x, instead
of saying plus negative five which we could say, we could just say, actually let me write that down. I could write just plus
negative five right over there because that's our b. I could just write x minus
five, and we're done. We've just factored it as
a product of two binomials. Now, I did it fairly involved mainly so you see where all this came from. But in the future whenever you
see a quadratic expression, and you have a one coefficient on the second degree term right over here, you could say alright well I
need to figure out two numbers that add up to the coefficient
on the first degree term, on the x term, and those same two numbers when I multiply them need to
be equal to this constant term, need to be equal to negative 10. You say okay well let's see, they're gonna be different signs because when I multiply them
I get a negative number. The negative one is
gonna be the larger one, since when I add them I
got a negative number. So let's see, let's say five
and two seem interesting. Well negative five and positive two, when you add them you're
gonna get negative three, when you multiply them
you get negative 10.