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Thread: impossible question for me!!!!!

  1. #1
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    impossible question for me!!!!!

    when (x^4 - 6.x^3 + 16.x^2 - 25. x + 10) is divided by( x^2 - 2.x + k)
    the remainder is = x+a
    find the values of k and a
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  2. #2
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by jashansinghal View Post
    When $\displaystyle f=x^4 - 6.x^3 + 16.x^2 - 25. x + 10$ is divided by $\displaystyle g= x^2 - 2.x + k$ the remainder is $\displaystyle r = x+a$ find the values of k and a
    There are, I am sure, many subtle ways of solving this problem, but the most obvious one which springs to mind is a simple cheat:

    We know that $\displaystyle f=gh+r$ where $\displaystyle h$ is some unknown polynomial and $\displaystyle f$, $\displaystyle g$, and $\displaystyle r$ are as above, because that's what division with a remainder means. So, simply substitute unknowns into the polynomial $\displaystyle h$ and then multiply out and pair off the coefficients, and you'll get your result.
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  3. #3
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    Quote Originally Posted by jashansinghal View Post
    when (x^4 - 6.x^3 + 16.x^2 - 25. x + 10) is divided by( x^2 - 2.x + k)
    the remainder is = x+a
    find the values of k and a
    After using long division, you should find

    $\displaystyle \frac{x^4 - 6x^3 + 16x^2 - 25x + 10}{x^2 - 2x + k} = x^2 - 4x + 8 - k + \frac{(2k - 9)x + k^2 - 8k + 10}{x^2 - 2x + k}$.


    So the remainder is:

    $\displaystyle x + a = (2k - 9)x + k^2 - 8k + 10$.


    Equating like co-efficients of $\displaystyle x$ gives

    $\displaystyle 2k - 9 = 1$ and $\displaystyle k^2 - 8k + 10 = a$.


    Solving for $\displaystyle k$ in the first equation gives:

    $\displaystyle 2k = 10$

    $\displaystyle k = 5$.


    Substituting into the second equation gives:

    $\displaystyle 5^2 - 8\cdot 5 + 10 = a$

    $\displaystyle 25 - 40 + 10 = a$

    $\displaystyle a = -5$.



    So $\displaystyle a = -5$ and $\displaystyle k = 5$.
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