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### Course: Algebra 2 > Unit 4

Lesson 3: Dividing polynomials by linear factors- Dividing polynomials by linear expressions
- Dividing polynomials by linear expressions: missing term
- Divide polynomials by linear expressions
- Factoring using polynomial division
- Factoring using polynomial division: missing term
- Factor using polynomial division

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# Dividing polynomials by linear expressions

Learn how to divide polynomials by linear factors! This tutorial guides you through the process of algebraic long division, focusing on highest degree terms first. You'll master the art of subtracting polynomials, handling remainders, and rewriting expressions in different forms.

## Want to join the conversation?

- Is dividing the polynomials by algebraic long division the only way to solve this problem?(19 votes)
- Well, there is a shortcut called synthetic division that can be used in special cases.(24 votes)

- in1:32why does 2x^2 become 2x(6 votes)
- The steps of polynomial long division are as follows.

1) find the term you have to multiply the leading term of the divisor (denominator) you have to multiply by to get the first term of the dividend (numerator.) In this case the denominator is x+2 and the numerator is 3x^3 + 4x^2 -3x +7. We want what we have to multiply x in x+2 by to get 3x^3. so x times what is 3x^3? the answer is 3x^2. This is the first term of your answer (the quotient.) And just to make sure other terms are explicitly stated, the dividend is 3x^3 + 4x^2 -3x +7 and divisor is x + 2

2) take the term of the quotient you just found and multiply it by the divisor. so the term found was 3x^2 and the divisor is x + 2, so ultiply those to get 3x^3 + 6x^2. You should notice 3x^3 is the same as the first term in the dividend.

3) subtract this new expression from the dividend. so (3x^3 + 4x^2 -3x +7) - (3x^3 + 6x^2) = -2x^2 - 3x + 7. THIS is your new divisor, or in other words you can think of the division problem now as (-2x^2 - 3x + 7) / (x + 2).

4) Repeat steps 1-3 to find the next term of your quotient, the expression you subtract from the current dividend and then the new dividend.

so step 1, what do we multiply the first term of the divisor by to get the first term of the dividend? -2x (this is where the -2x you asked about came from.) So the second term of the quotient is -2x, Then you multiply by the dividend -2x(x + 2) = -2x^2 - 4x. now subtract this from the current dividend (-2x^2 - 3x + 7) - (-2x^2 - 4x) = x + 7.

Now just keep going until the dividend has a smaller degree than the divisor. Then this is your remainder.

Let me know if that didn't help. just to make sure your question specifically is answered, the -2x is the number that multiplies the first term of the divisor to get the first term of the new dividend, which is part of the process of solving these problems.(20 votes)

- What happened to the synthetic division video and practice?(10 votes)
- They were changes on this site so I saw, so I guess they changed a few things.(3 votes)

- Do we still have to put a condition on the domain that 'x cannot equal -2', even though the remainder of 'x + 2' clearly shows that a value of -2 would create an undefined denominator?

Asking for a friend...(6 votes)- There are many ways to write the domain, and it really depends on your teacher. For the easiest, you may write (just going off you're example), (-inf,inf) x ≠ -2.

More formally, which would be nice, is to use the "union term." More specifically, instead of (-inf,inf) x ≠ -2, it would be (-inf,-2]**U**[-2,inf). Why union? It's the term for the combination of two unconnected intervals. Hopefully this helps.(3 votes)

- I have a question. When Khan does 4 - (-6) he gets 2. However, in practice, I have to make the 6 positive and get 10. Why doesn't Khan fix the problem?(4 votes)
- Sal is not actually doing 4 - (-6) but is actually doing 4 - 6, which becomes negative 2.(4 votes)

- What are some real-life applications for theses formulas and equations?(3 votes)
- In Hungary, we do not learn (algebraic) long division, therefore I do not understand what Sal is doing in the video. As far as I am concerned, dividing polynomials by linear expressions is needed for some integration problems, so could you recommend me a video in connection with the topic, or suggest another method?(2 votes)
- If you look under the Arithmetic section in Khan Academy, then go to Multiplication and Division, there are videos on how to perform long division.(3 votes)

- in the past video we would stop and end it as a remainder in2:12,, why are we continuing? or the question is when do I know to stop(2 votes)
- At2:12, you still have an x on the bottom, so you have not reached the remainder part yet. He finally gets the remainder of 5 at2:30. This is also clearly seen when he rewrites at2:49.(2 votes)

- i no understand(2 votes)
- Why do we always focus on the highest degree term? Can't we start dividing by the constant?(2 votes)
- We can. Try it. Dividing by the highest degree term is simply convention.(1 vote)

## Video transcript

- We're told, divide the polynomials. The form of your answer should either be just a clean polynomial, or some polynomial plus some
constant over x plus two, where p of x is a polynomial
and k is an integer. Fair enough, and if we were
doing this on Khan Academy, this is a screen shot from Khan Academy, we would have to type this in, but we're just going to do it by hand. And, like always, pause this video and try to do it on your own before we work through it together. All right, now let's
work through it together. And what we're trying to
do is divide x plus two, into three x to the third power plus four x squared
minus three x plus seven. And so, like always, we focus on the highest
degree terms first. x goes into three x to the
third power how many times? Three x squared times. We'd want to put that in
the second degree column. Three x squared. Three x squared times
two is six x squared. Three x squared times x is
three x to the third power. There's something very meditative about algebraic long division. Anyway, we'd want to
subtract what we just wrote from what we have up here. So let's subtract. And these characters cancel out. And then four x squared
minus six x squared is negative two x squared. Bring down that negative three x. And now we would wanna say, hey, how many times does x go
into negative two x squared? Well, it would go negative two x times. Put that in our first degree column. Negative two x times
two is negative four x. Negative two x times x is
negative two x squared. Now we wanna subtract what
we have here in orange from what we have up here in teal. So we either put a negative
around the whole thing or we distribute that negative
and that becomes a positive, that becomes a positive. And so this is equal to, the
x squared terms cancel out. Negative three x plus four x is just going to be a straight up x. Bring down that seven, x plus seven. How many times does x go into x? Well, one time. Actually, let me do a,
use a new color here. So, how many times does x go into x? It goes one time. Put that in the constant column. One times two is two. One times x is x. We wanna subtract these characters. And we're left with
seven minus two is five. And so, we can rewrite
this whole thing as, we deserve I guess a
little bit of a drum roll. Three x squared minus two x plus one, plus the remainder, five, over x plus two. One way to think about it is, hey, I have this remainder, I'd have to keep dividing it by x plus two if I really wanted to figure
out exactly what this is. Now if I wanted these expressions
to be completely identical I would put a condition on the domain that x cannot be equal to negative two because if x was equal to negative two, we'd be dividing by zero here. But for the purposes of this exercise, you just have to input
this part right over here, you'd have to type it in, which I guess isn't the easiest
thing to do in the world. But it's worth doing. All right, see you in the next video.