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Thread: Polynomial problem

  1. #1
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    Polynomial problem

    Hi - I have an assignment with a question that is stumping me at the first hurdle.

    Express $\displaystyle x^4 - 4x^2 + 16$ in the form
    $\displaystyle (x^2 + Ax + B)$$\displaystyle (x^2 +Cx + D)$

    where A,B,C & D are real constants.

    I just can't see how to get it in this form to complete the rest of the partial fraction question. Any ideas??
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  2. #2
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    It's telling you how to express it?

    To find A, B, C and D you can equate the coefficients
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  3. #3
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    Quote Originally Posted by dojo View Post
    Hi - I have an assignment with a question that is stumping me at the first hurdle.

    Express $\displaystyle x^4 - 4x^2 + 16$ in the form

    $\displaystyle (x^2 + Ax + B)$$\displaystyle (x^2 +Cx + D)$ just multiply this out and compare coefficients

    where A,B,C & D are real constants.

    I just can't see how to get it in this form to complete the rest of the partial fraction question. Any ideas??
    $\displaystyle x^4-4x^2+16=\left(x^2+Ax+B\right)\left(x^2+Cx+D\right)$

    $\displaystyle =x^4+Cx^3+Dx^2+Ax^3+ACx^2+ADx+Bx^2+BCx+BD$

    $\displaystyle =x^4+x^3(A+C)+x^2(B+AC+D)+x(AD+BC)+BD$

    which tells us that $\displaystyle A=-C$ and $\displaystyle AD=-BC$

    since $\displaystyle x^4-4x^2+16$ has no $\displaystyle x^3$ or $\displaystyle x$ terms.

    Also $\displaystyle B+AC+D=-4$ and $\displaystyle BD=16$

    from which the values A, B, C, D are discovered.
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  4. #4
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    We can rewrite $\displaystyle x^4- 4x^2+ 16$ as $\displaystyle x^4+ 8x^2+ 16- 12x^2= (x^2+ 4)^2- 12x^2$ and treat it as a "difference of squares". Of course, it cannot be factored with integer coefficients.
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