Dividing Polynomials and the Remainder Theorem Part 2

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This lesson shows how to divide a polynomial by a binomial using both long division and synthetic division. The lesson also discusses the Remainder Theorem and shows how to use it to find remainders in algebraic divisions. This is the second part of a three part lesson. This video was created for the MHF4U Advanced Functions course in the province of Ontario, Canada.

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This is one of the best tutorials on synthetic divisions. Thank you very much.

pinklady

This is the second tutorial I have watched and again he covers basics. He will not assume his viewers recall basic mathematics. I have been tutoring for over 30 years and in most cases I have to go over the basics before we can launch the actual problems.

revincentiii

Superb explaination ....not a single video is as awsm as ur on synthetic division

godsgrace

This is awesome!!! I'm using a college algebra math book that gave us a similar problem but never went through the trouble explaining how to solve this type of synthetic division. Very helpful! Thank you!!

luisdfernandez

THANK YOU ALL OF THE VIDEOS I COULD FIND HAD 1X IT WAS RIDICULOUS

honkernoodles

this tutorial really help us from our homework and quizzes the best tutorial ever thanks... :)

dentalclinic

Since in the division I used 4/3 then I'm actually dividing by (x - 4/3). When you divide by (x - a) you place an "a" at the left of the Synthetic Division statement. In the first example the polynomial was divided by (x - 3) so "3" was used at the left of the Synthetic Division statement.

AlRichards

@kiiish I was shown how to use Synthetic Division when dividing by a power larger than 1, but that was about 15 years ago and I forget how it went. I'd use the Long Division, that's much more the same. I searched on the Internet during the past year to try and find how Synthetic Division works when dividing by a quadratic, but I was unable to find anything.

AlRichards

Thank you very much for the well research information

desmondrose

oh god i thought i was gonna fail this quiz thank you man

TehShikamaruStory

@AlRichards314 Hey can you use this method, if you're divisor has a power of x, like x^2 + 1.

kiiish

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