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Using Remainder Theorem to find remainder

  1. Jan 13, 2017 #1
    1. The problem statement, all variables and given/known data
    (y4 - 5y2 + 2y - 15) / (3y - √(2))
    The answer says (2√(2)/3)-(1301/81)
    https://lh3.googleusercontent.com/t...DKJeR9_wevbfnEStla1ntJRTakXqP8dG=w449-h311-no

    2. Relevant equations


    3. The attempt at a solution
    Using synthetic division
    3y - √(2) = 0
    3y = √(2)
    y = √(2)/3
    View attachment 111547

    My final answer that I keep getting is (2√(2)/3)-21. I can't seem to get (2√(2)/3)-(1301/81). I was just wondering if someone could show me where I'm going wrong? Sorry writing this out was very difficult so I hope you can still understand it.
     
    Last edited: Jan 13, 2017
  2. jcsd
  3. Jan 13, 2017 #2

    fresh_42

    Staff: Mentor

    I don't actually can follow your calculations. If I compare my calculation with your result, then I think you simply lost some denominators.
    I get ##\frac{4}{81}-\frac{10}{9}-15+\frac{2\sqrt{2}}{3}## whereas you seem to have added only the nominators ##4-10-15+\frac{2\sqrt{2}}{3}##.

    I once gave an example on PF here:
    https://www.physicsforums.com/threa...r-a-polynomial-over-z-z3.889140/#post-5595083

    Maybe it helps.
     
  4. Jan 13, 2017 #3
    upload_2017-1-13_19-24-50.png Sorry, I should have just posted this to begin with.
     
  5. Jan 13, 2017 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Your long-division process is all wrong. Let ##p(y) = y^4 - 5y^2 + 2y - 15## and ##d(y) = 3y - \sqrt{2}##. You want to find a "quotient" ##q(y)## and a number ##r## (the "remainder") that give you ##p(y) = q(y) d(y) + r.## The question is asking you to find ##r##.

    Step 1: see how many times the leading term of ##d(y)## will go into the leading term of ##p(y)##; that is the number of times ##3y## goes into ##y^4##. The answer in this case is ##y^4/(3y) = (1/3)y^3##, so that is the first term in your quotient ##q(y)##. Now we have
    $$ p(y) - \frac{1}{3} y^3 d(y) = \frac{\sqrt{2}}{3} y^3 -5 y^2 + 2y - 15 \equiv p_1(y).$$

    However, all that would be doing it the hard way. The easy way would be to go ahead and use the "remainder theorem".

    Step 2: see how many times the leading term of ##d(y)## goes into the leading term of ##p_1(y)##. The answer is ##(\sqrt{2}/3)/(3 y) = (\sqrt{2}/9) y^2##, so that is the next term in your quotient ##q(y)##. We have
    $$p_2(y) \equiv p_1(y) - \frac{\sqrt{2}}{9} y^2 d(y) = -\frac{43}{9} y^2 + 2y - 15.$$

    So, up to now we have
    $$p(y) = \left( \frac{1}{3} y^3 + \frac{\sqrt{2}}{9} y^2 \right) d(y) + p_2(y).$$

    Keep going like that; the next term in the quotient ##q(y)## will be the number of times the leading term of ##d(y)## goes into the leading term of ##p_2(y)##, so is the number of times ##3y## goes into ##-(43/9) y^2##, etc., etc.

    The remainder will be what you have left when the process comes to an end.

    However, all that would be doing it the hard way; I included the material just because you attempted to do it, but did it all wrong. The easy way would be to just use the so-called "remainder theorem".
     
  6. Jan 14, 2017 #5
    I was using synthetic division which worked on every other question but I'll try it the way you're suggesting it.
     
  7. Jan 14, 2017 #6
    I got the same result as the answer. Thanks for your help!
     
  8. Jan 14, 2017 #7
    It looks to me like you simply made a couple multiplication errors when working with fractions. I worked the problem with synthetic division and got the right answer.
     
    Last edited: Jan 14, 2017
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