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## Dividing polynomials by x

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# Dividing polynomials by x (no remainders)

CCSS Math: HSA.APR.D.6, HSA.APR.D

## Video transcript

- [Instructor] What I'd
like to do in this video is try to figure out what x to the fourth minus two x to the third plus five x divided by x is equal to. So pause this video and see
if you can have a go at that before we work through this together. All right, so if we're saying
what is this top expression divided by this bottom expression, another way to think about it is, what do I have to multiply,
so I'm going to multiply something, I'll put that in parentheses. If I multiply that something times x, I should get x to the fourth minus two x to the third plus five x. Now how do I approach that? Well there's two ways
that I could tackle it. One way is I could just rewrite
this expression as being, and I will just make this x in yellow so I can keep track of it. I could just rewrite
this as one over x times, times x to the fourth minus two x to the third plus five x. And then I can distribute the one over x, and so what is that going to be equal to? Well it's going to be
equal to x to the fourth. Let me do this, x to the fourth over x minus two x to the third over x plus five x, plus five x over x. And so what are each of
these going to be equal to? X to the fourth divided by x, if I have four x's that
I'm multiplying together and then I divide by x,
that's going to be equivalent to x to the third power. So this right over here is
equal to x to the third. You could also get there from
your exponent properties. In the denominator, you have
an x to the first power, and so you would subtract the exponents. You have the same base here,
so that's x to the third. And then, in this part right over here, what would that equal to? Well it's going to be minus two x to the third divided by x to the first. Well by the same property,
that's going to be x squared. And then last but not
least, if you take five x's and then you divide by x, you are just going to be left with five. And you can verify that this, indeed, if I were to multiply it by x, I'm gonna get x to the fourth minus two x to the third plus five x. Let me do that. If I put x to the third
minus two x squared plus five times x, what I
could do is distribute the x. X times x to the third is x to the fourth. X times negative two x squared is negative two x to the third. X times five is five x. Now I mentioned there's two
ways that I could do it. Another way that I could
try to tackle it is I could look at this numerator and try to factor an x out. I would try to factor out
whatever I see in the denominator. So if I do that, actually let
me just rewrite the numerator. So I can rewrite x to the fourth as x times x to the third. And then I can rewrite the
minus two x to the third as, let me write it this way, as plus x times negative two x squared. And then I could write this five x as being equal to plus x times five. And then I'm gonna divide everything by x, divide everything by x. I just rewrote the numerator here, but for each of those
terms, I factored out an x. And now I can factor out
x out of the whole thing. So I sometimes think of factoring out an x out of the whole thing as
reverse distributive property. So if I factor out this
x out of every term, what am I left with? I'm left with an x times x to the third minus two x squared plus five; I ended up doing
that in the same color, but hopefully you're following, plus five, and then all of that is divided by x. And as long as x does not equal zero, x divided by x is going
to be equal to one, and we're left with what
we had to begin with, or the answer that we had to begin with. So these are two different approaches. Nothing super sophisticated here. When you're dividing by
x, you're just like hey, that's the same thing as
multiplying every term by one over x, or you can factor out an x out of the numerator,
and then they cancel out.