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Algebraic Division Help

  1. Nov 23, 2004 #1
    I didn't really know where to put this, but I thought I'd post it in General Math. Any way, here's the problem, I'd like it solved step by step, please.

    a4 (a to the fourth power) + Oa3 (a to the third power) + 9a2 (a squared) + Oa

    divided by:

    a2 (a squared) - 3a + 9

    Thanks in advance!

  2. jcsd
  3. Nov 23, 2004 #2

    matt grime

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    what is O?
  4. Nov 23, 2004 #3


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    Weird, you have the same name as me (mine is josh meyer). Anyway, heres the problem:

    [tex] \frac{a^4+9a^2}{a^2-3a+9} [/tex]

    So we set up long division as you did, with 0's in for all the non-existent powers of x. It's hard to write this out on the site, so I am doing it on paper and describing it step by step:

    1. [tex]a^2[/tex] goes in to [tex]a^4[/tex] [tex]a^2[/tex] times, so write [tex]a^2[/tex] above the [tex]9a^2[/tex]

    2. Multiply the [tex]a^2-3a+9[/tex] term by that [tex]a^2[/tex] and write all the terms obtained underneath their proper powers (cubics under cubics etc). Then change the signs on all these terms and subtract everything. (you should get [tex]0a^4+3a^3+0a^2[/tex])

    3. [tex]a^2[/tex] goes into [tex]3a^3[/tex] [tex]3a[/tex] times, so write this [tex]3a[/tex] above the 0a, multiply the [tex]a^2-3a+9[/tex] term by it, change the signs on these terms, and subtract. You ought to get
    [tex] -9a^2-27a[/tex].

    4. [tex]a^2[/tex] goes into [tex]-9a^2[/tex] -9 times, so write this above, multiply everything out, switch signs, and subtract.

    so, the answer is [tex]a^2+3a-9[/tex] with a remainder of 81.
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