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Math Help - (Simple?) Polynomial problem

  1. #1
    Newbie Skydriver's Avatar
    Joined
    Nov 2008
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    (Simple?) Polynomial problem

    Hi, I am from Poland so please forgive my english.

    I seem to have problems solving the following assignment (translated)

    "There are polynomials:

    P(x) = (2x+1)^3
    Q(x) = 8x+a
    R(x) = 4x^2-1
    S(x) = 8x^2 + cx

    What
    a and c is required for P(x) + Q(x) - 3R(x) = S(x)?"

    After many conversions, I came up with this:

    (2x+1)^3 +
    a = 12x^2 - 3 + cx

    But after that I have no idea what next, plus I may not even be heading the right direction... help?
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
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    Hello, Skydriver!

    There must be a typo.
    As stated, the problem has no solution.


    \begin{array}{ccc}P(x) &=& (2x+1)^3 \\ Q(x) &=& 8x+a \\ R(x) &=& 4x^2-1 \\ S(x) &=& 8x^{{\color{red}3}} + cx \end{array}

    What a and c is required for: . P(x) + Q(x) - 3R(x) \:= \:S(x) ?
    The equation is:

    . . P(x) + Q(x) - 3R(x) \;=\;S(x)

    . . (2x+1)^3 + (8x+a) - 3(4x^2-1) \;=\;8x^3 + cx

    . . 8x^3 + 12x^2 + 6x + 1 + 8x + a - 12x^2 + 3 \;=\;8x^3 + cx

    . . 14x + a + 4 \;=\;cx


    We have two equal polynomials.
    Two polynomials are equal if their corresponding coefficients are equal.

    So we have: . \begin{array}{ccccccc}14 \:=\: c & \Rightarrow & \boxed{c \:=\:14} \\ a+4 \:=\:0 & \Rightarrow &\boxed{ a \:=\:\text{-}4} \end{array}


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