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### Course: Get ready for AP® Calculus>Unit 6

Lesson 2: Dividing quadratics by linear factors

# Dividing quadratics by linear expressions with remainders: missing x-term

An interesting case in polynomial division is when one of the terms is missing. The video explains how to divide a quadratic expression, like (x²+1), by a linear one, such as (x+2). It shows two methods: re-expressing the numerator and using algebraic long division. Both methods lead to the same answer.

## Want to join the conversation?

• how does this help me in real life problems
• Knowing how to divide polynomials is essential in calculus. A lot of unsolvable problems become solvable just because it's possible to divide and reduce them
• I have a question. Where did you get the 5 from in the first example in the video? (At )
• The closest expression we can get to x^2+1 that is also a multiple of x+2 is x^2-4 since x^2-4 has a degree of 2 and doesn't have an "x" term. However, x^2-4 is not exactly the same as x^2+1, so we have to do something to make them the same. We can add 5 to x^2-4 so it is equal to x^2+1. That way, we can manipulate the expression (x^2-1)/(x+2) to look like (x^2-4+5)/(x+2). We can then factor x^2-4 in the numerator as (x+2)(x-2). Now the expression should look something like this:
((x+2)(x-2)+5)/ x+2
You can rewrite this expression as ((x+2)(x-2))/(x-2) + 5/(x+2)
In the first term, the x+2 cancels out so you are just left with x-2
You can't really go any further with the second term, so the answer should be
x-2 + 5/(x+2)

Hope this helped! ^~^
• How is Sal so popular on the street?
• He helps a lot of people (like us) learn math, that's why XD.
• Surely x-2+5/(x+2) is identical to (x^2+1)/(x+2) without declaring it not defined at x=-2, as the former also contains an expression with (x+2) as the denominator.
I understand that you'd want to restrict the domain in the case there was no remainder, but here it's just redundant, or am I mistaken here?
• Good catch! Yes, in that case it would be redundant.

He included it here because it's a good habit to get into. If you always restrict the domain, you don't need to think about whether you have to or not. I'm with you though, I wouldn't have included it because the remainder already excludes "x = -2" from being a possibility.
• At why isn't the x+2 also included in the remainder? Shouldn't the remainder be 5/ x+2 ?
• Sal is placing emphasis on what is the number that was left over from the division, which is — your remainder.
If you were writing it down however as an expression overall , yes you are right , you are writing it as 5/(x+2).

Hopefully that helps !
• Wait, in the answer of algebraic long division, where do you put the remainder?
(1 vote)
• If you try to divide 28/5, you get 5 with a remainder of 3, so you would either get a decimal (5.6) or like in the video, you get a fraction 5 3/5. So remainder goes on top and what you divided by goes on bottom. Of you are dividing by x + 2 and get a remainder of 5, the last part would be 5/(x+2). Is this what you are asking?
• How does the -2x-4 become 2x+4 at ? Is it because of the double negatives?
• Yup ! When doing long division, you subtract to go to the next step.

-(-2x-4) = 2x+4 through distributing. hopefylly tht helps !
• Why do people keep walking up!? Where did they come from!?
• they saw this video and they decided to be a part of the video, so they went back in time, and walked up to you and asked you the question.

anyway, it is just an excuse to present the problem
(1 vote)
• What would you do if instead of x+2 it was x-2