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Thread: Dividing equations and finding quotiant.

  1. #1
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    Dividing equations and finding quotiant.

    i)Find the quotient and the remainder when 3x^3-2x^2+x+7 is divided by x^2-2x+5#

    ii)Hence, or otherwise, determine the values of the constants a and b such that, 3x^3-2x^2+ax+b is divded by x^2-2x+5, there is no remainder.


    Been looking at this for ages and have forgotten how to dot his kind of quiestion.
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  2. #2
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    Quote Originally Posted by George321 View Post
    i)Find the quotient and the remainder when 3x^3-2x^2+x+7 is divided by x^2-2x+5#

    ii)Hence, or otherwise, determine the values of the constants a and b such that, 3x^3-2x^2+ax+b is divded by x^2-2x+5, there is no remainder.


    Been looking at this for ages and have forgotten how to dot his kind of quiestion.
    Please help!
    Thanks
    Essentially, the same as long division of numbers:
    Look at the leading term. x^2 divides into 3x^3 3x times. Now multiply the entire divisor, x^2- 2x+ 5 by that and subtract:
    \begin{array}{cccc}3x^3& - 2x^2& + x& + 7 \\ \underline{3x^3}& \underline{- 6x^2}& \underline{+ 15x} &  \\ & 4x^2 & -14x & 7\end{array}

    Now, x^2 will divide into 4x^2 4 times. Multiply x^2- 2x+ 5 by 4 and subtract:
    \begin{array}{ccc}4x^2 & -14x & 7 \\\underline{4x^2}& \underline{-8x} & \underline{20}\\ & -6x& - 13\end{array}.

    That is, x^2- 2x+ 5 divides into 3x^3- 2x^2+ x+ 7 3x+ 4 times with remainder -6x- 13.

    For the second problem, do the same thing but with "ax+ b" rather than "x+ 7". The remainder will involve a and b- set the coefficient of x and the constant term equal to 0 and solve for a and b.
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