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### Course: 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

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.

## Want to join the conversation?

- OK, so my teacher gave us a problem just like this. But one of the terms comes out as a zero! what do i do??? :((41 votes)
- You can write the 0x^...If you want to but you need to remember you're either adding or subtracting a number from 0, which is the just opposite of whatever number you have...(6 votes)

- Can't you factor a 5 out of 5x - 5 at4:26? Also, you could factor x^2 - x +1 to get (x - 1)(x - 1). So, you would have 5(x - 1) divided by (x - 1)(x - 1). This would result in 5 divided by x - 1. So, shouldn't the final answer be x + 1 + 5/x-1?(13 votes)
- How do you handle a polynomial division when the initial terms don't divide evenly? For example, 2x - 5 divided into 3x^3 -9x^2 +15. Thanks!(6 votes)
- Actually, you simply do the division as normal. 3x^3/2x = 3(x^2)/2. See? No issue here. The coefficients might get sort of messy, but division works normally.(9 votes)

- Is it possible to use synthetic division on a problem like this?(6 votes)
- No, it's not possible, because the divisor has an exponent higher than 1.

In order to divide polynomials using synthetic division, the denominator (the number(s) on the bottom of the fraction) must satisfy two rules:

1 - Be a linear expression, in other words, each term must either be a constant or the product of a constant and a single variable to the power of 1.

2 - The leading coefficient (first number) must be a 1.

For example, you can use synthetic division to divide a polynomial by (x + 2) or (x – 6), but you cannot use synthetic division to divide by 6x, or (2x + 3) or (3x^2 – x + 3).(5 votes)

- Is synthetic division the same as polynomial long division?(4 votes)
- Yes, but different technique and it only works on special case, the divisor needs to be linear factor.

https://www.khanacademy.org/math/algebra2/arithmetic-with-polynomials/synthetic-division-of-polynomials/v/synthetic-division(4 votes)

- i have a question that I am struggling with it asks... Using polynominal long division to determine the quotient when 3x^3 - 5x^2 + 10x + 4 is divided by 3x + 1

Any help with this would be greatly appreciated!

Thank you.(4 votes)- well since the x values are already in descending order, you can start and you always use the first term in the divisor to determine what to multiply it by to get the dividend, so to get from 3x to 3x^3 you multiply by x^2 and then you have both terms of the divisor multiplied by that to get 3x^3 +x^2 which you subtract from 3x^3-5x^2 and you get -6x^2 and you bring down the 10x and to get from 3x to -6x^2 you multiply by -2x and you do that to both terms and you get -6x^2-2x which you subtract from -6x^2+10x to get 12x and you bring down the +4 and to get from 3x to 12x you multiply by 4 and 3x+1 times 4 is 12x+4 so when you subtract you get 0 and the answer is x^2-2x+4 (remainder 0)(2 votes)

- What would you do if the x term for the divisor is bigger than the dividend? Help would be much appreciated(3 votes)
- Then you have a proper fraction. It's like have 3/4. All you can do is factor the numerator and denominator to see if you can reduce the fraction.(3 votes)

- at5:36, Sal canceled the denominator and is only left with the numerator. Why he doesn't distribute the x^2-x+1 into 5x-5 as well? Thanks!(3 votes)
- The entire fraction is one term, so he is distributing x^2-x+1 into the entire fraction, not just the denominator.

Let's say 5x-5 = a and x^2-x+1 = b. Distributing into the last term would leave you with (a/b) * b. Simplifying this would just leave you with a, or 5x-5.

This question was asked last year, so my answer is a bit late - you may have already figured it out yourself - but maybe it'll help with someone else who has this question, since there are no other answers here.(2 votes)

- Can you use long division for polynomials where the divisor is a higher degree then the dividend?(3 votes)
- You could, but you would have negative exponents in your answer. Answers are typically expressed without negative exponents whenever possible, so you won't see this done often.(1 vote)

- Which property is used to express f(x)=5x^4-3x^3+x^2+1 as a polynomial in power of (x-1) by using synthetic division plzzz tell the property not the method of solving question....(3 votes)

## 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.