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Thread: Division Algorithm.....

  1. #1
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    Question Division Algorithm.....

    If the polynomial f(x) = $\displaystyle ax^3 + bx - c$ is divisible by the polynomial g(x) = $\displaystyle x^2 + bx +c$, then $\displaystyle ab = $ ?
    Please Help....
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  2. #2
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    If f(x) is divisible by g(x), the you can write

    $\displaystyle ax^3 + bx - c = (px + q)(x^2 + bx - c)$

    Expand the brackets and compare coefficients of like powers of x. Write p and q in terms of a, b and c. Then find ab.
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  3. #3
    MHF Contributor chisigma's Avatar
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    If the hypothesis id true is...

    $\displaystyle a\ x^{3} + b\ x -c = (a\ x + d)\ ( x^{2} + b\ x + c)$ (1)

    ... and that means that is...

    $\displaystyle a\ b + d=0$

    $\displaystyle c\ d = -c$ (2)

    ... so that...

    $\displaystyle d=-1$

    $\displaystyle a\ b=1$ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  4. #4
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    Question

    Quote Originally Posted by chisigma View Post
    If the hypothesis id true is...

    $\displaystyle a\ x^{3} + b\ x -c = (a\ x + d)\ ( x^{2} + b\ x + c)$ (1)

    ... and that means that is...

    $\displaystyle a\ b + d=0$

    $\displaystyle c\ d = -c$ (2)

    ... so that...

    $\displaystyle d=-1$

    $\displaystyle a\ b=1$ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    @chisigma ....
    i ve got 2 qestions regarding this....
    1)how did u ax + d to be a multiple of $\displaystyle ax^3 + bx - c$
    2) and did u get ab + d by opening the brackets?
    pls help.....
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  5. #5
    MHF Contributor chisigma's Avatar
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    Now...

    a) if $\displaystyle x^{2} + b\ x + c$ devides $\displaystyle a\ x^{3} + b\ x -c$ , then there is a first order polynomial $\displaystyle a\ x + d$ so that...

    $\displaystyle a\ x^{3} + b\ x -c = (a\ x + d)\ (x^{2} + b\ x + c) $ (1)

    b) if You develop the product (1) You obtain...

    $\displaystyle (a\ x + d)\ (x^{2} + b\ x + c) = a\ x^{3} + (a\ b + d)\ x^{2} + (a\ c + b\ d) x + d\ c = a\ x^{3} + b\ x -c $ (2)

    c) from the (2) You obtain ...

    $\displaystyle a\ b + d=0$ , $\displaystyle d= -1 \rightarrow a\ b = 1$ (3)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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