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### Course: Integrated math 3>Unit 3

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 !