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## Algebra (all content)

### Unit 10: Lesson 23

Practice dividing polynomials with remainders- Divide polynomials by x (with remainders)
- Divide polynomials by monomials (with remainders)
- Divide polynomials by monomials (with remainders)
- Dividing polynomials with remainders
- Divide polynomials by linear expressions
- Divide polynomials with remainders

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# Dividing polynomials with remainders

CCSS.Math:

Sal divides (x^3+5x-4) by (x^2-x+1) using long division. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

We divide x to the
third plus 5x minus 4 by x minus x--
actually, I think this is supposed to be an x squared. Let me correct it. So we're going to divide x
to the third plus 5x minus 4 by x squared minus x plus 1. And just to see the different
ways we can rewrite this. We could rewrite this
as x to the third plus 5x minus 4 divided by x
squared minus x plus 1. Or maybe the best
way to write it in this circumstance,
since we're going to do algebraic
long division, is to write it as x
squared minus x plus 1, divided into x to
the third plus-- and actually, I'm going to
leave some blank space here. We don't have an x
squared term here. But I'm going to leave
some space for it, just so that we can align
everything in the proper place when we actually
do the division. So x to the third
plus nothing to the x squared power plus 5x minus 4. So we have a place for the
third power, the second power, the first power,
and the 0-th power. So now let's just do a little
bit of algebraic long division. Let's look at the
highest-degree term. x squared goes into x to
the third how many times? Well, it goes into it x times. x to the third
divided by x squared is equal to x to the 3 minus 2,
which is equal to x to the 1, which is equal to x. So it goes x times. I'll write the x
right over here. And we multiply x times
this entire thing. x times x squared
is x to the third. x times negative x is
negative x squared. x times 1 is positive x. And now we want to subtract
this whole expression from that whole expression. And that's the same thing
as adding the opposite, or multiplying each of
these terms by negative 1, and then adding
it to these terms. So let's do that. So we have negative
x to the third. Negative 1 times negative x
squared is positive x squared. And then positive x times
negative 1 is negative x. And so let's now add everything. x to the third minus x to
third-- those cancel out. 0 plus x squared
gives us an x squared. 5x minus x gives us a plus 4x. And then we bring
down this minus 4. We're not adding
anything to it there. You could view there's a 0 here. So let's bring down the minus 4. And now let's look at
the highest-degree terms. x squared goes into x
squared exactly one time. It's the same thing,
so we put a plus 1. And then you have 1 times
x squared is x squared. 1 times negative
x is negative x. 1 times 1 is 1. And now we want to
subtract this from that, or we want to add the opposite. And to add the
opposite, we can just multiply each of these
terms by negative 1. x squared becomes
negative x squared. Negative x times
negative 1 is positive x. And then positive 1 times
negative 1 is negative 1. Now let's do the
addition. x squared minus x squared--
they cancel out. 4x plus x is 5x. And then we have negative
4 minus 1 is negative 5. Now, you might be
tempted to keep dividing, but you can't any more. This term right here, the
highest-degree term here, is now higher than the
highest-degree term that you're going to
try to divide into. So we have a remainder. So the answer to this is--
this expression right over here is equal to x plus 1
plus the remainder, plus 5x minus 5-- whatever
the remainder is-- divided by x squared minus x plus 1. If this was divisible,
we could keep dividing, but we're saying it's not. It's now a lower degree
than this down here. So we could say it's
x plus 1 plus whatever this remainder is, divided
by this thing over here. So our answer-- I'm going
to write it one more time. It's x plus 1 plus 5x minus 5,
over x squared minus x plus 1. And we can check
that this works. If we take this thing over
here, and we multiply it by this thing over
here, we should get the x to the
third plus 5x minus 4. So let's do that. Let's multiply this thing
by x squared minus x plus 1. And to do that,
let's just distribute this whole trinomial
times each of these terms. When we do the
first term, we have x squared minus
x plus 1 times x. So that's going to
be x times x squared, which is x to the third;
x times negative x, which is negative x squared; x
times 1, which is plus x. Then we can multiply
this whole thing times 1. So it's going to be plus
x squared minus x plus 1. I'm just multiplying
all of these times 1. And then we can multiply this
whole thing times this thing. Now, this is the same
as the denominator here. So it'll cancel out. This will cancel with that. And we're just going to be left
with the numerator over here, so plus 5x minus 5. And now we can try
to simplify it. We only have one third-degree
term, the x to the third. So we have x to the third here. Second-degree terms-- we
have a negative x squared. And then we also have
a positive x squared. So they cancel out
with each other. First-degree terms-- let's see. We have a positive
x and a negative x. Those cancel out
with each other. So we're just going to
have that 5x over here. So we're just going
to have this 5x. So then we have plus 5x. And then we have the 0-th degree
terms, or the constant terms. We have a positive
1 and a negative 5. Add them together. You get negative 4. So you get x to the
third plus 5x minus 4, which is exactly what
we had over here.