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Math Help - remainder theorem

  1. #1
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    remainder theorem

    Given g(x)=ax^3-x^2+7x-2 gives the same remainder when divided by
    (x-1) and (x-2) . I found that a is -2 and the remainder is 2 .

    When g(x) is divided by mx+n , it also gives the same remainder (2) , so find m and n .

    I am not really sure about this part ..
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Given g(x)=ax^3-x^2+7x-2 gives the same remainder when divided by
    (x-1) and (x-2) . I found that a is -2 and the remainder is 2 .

    When g(x) is divided by mx+n , it also gives the same remainder (2) , so find m and n .

    I am not really sure about this part ..
    How did you get a= -2? If a= -2, so that g(x)= -2x^3- x^2+ 7x- 2, then dividing by x- 1 gives -2x^2- 3x+ 4 with remainder 2 but dividing by x-2 gives -2x^2- 5x- 3 with remainder -8, not 2.

    Do you know the "remainder theorem": If polynomial p(x) is divided by x- a, then the remainder is p(a). With a= 1, the remainder is a(1^3)- 1^2+ 7(1)- 2= a-1+ 7 -2= a+ 4 and with a= 2, the remainder is a(2^3)- 2^2+ 7(2)- 2= 8a- 4+ 14- 2= 8a+ 8. To have the same remainder when divided by x-1 and x- 2, a must satisfy a+ 4= 8a+ 8.
    Last edited by mr fantastic; September 17th 2009 at 06:01 AM. Reason: Fixed latex tags
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  3. #3
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    Quote Originally Posted by HallsofIvy View Post
    How did you get a= -2? If a= -2, so that g(x)= -2x^3- x^2+ 7x- 2, then dividing by x- 1 gives -2x^2- 3x+ 4 with remainder 2 but dividing by x-2 gives -2x^2- 5x- 3 with remainder -8, not 2.

    Do you know the "remainder theorem": If polynomial p(x) is divided by x- a, then the remainder is p(a). With a= 1, the remainder is a(1^3)- 1^2+ 7(1)- 2= a-1+ 7 -2= a+ 4 and with a= 2, the remainder is a(2^3)- 2^2+ 7(2)- 2= 8a- 4+ 14- 2= 8a+ 8. To have the same remainder when divided by x-1 and x- 2, a must satisfy a+ 4= 8a+ 8.
    sorry , i think there is a typo in the equation g(x) but i don hv the question with me now . I know about part 1 and my answers are correct .

    So can u just guide me through part 2 , how would i solve for m and n ??
    Last edited by mr fantastic; September 17th 2009 at 06:01 AM. Reason: Edited quote
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  4. #4
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    I suppose you could let g(x) = 2 (to find the solutions for x, when the remainder is 2:

    2=-2x^3-x^2+7x-2

    Factorising this would then reveal all of the possible solutions for x (including mx+n).
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