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Math Help - Synthetic Division with imaginary numbers!

  1. #1
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    Unhappy Synthetic Division with imaginary numbers!

    I thought I had this down, but I was wrong.

    (4x^3 -17ix^2+5x+(7-3i))/x-3i

    Help?
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  2. #2
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    Hello,

    Quote Originally Posted by mankvill View Post
    I thought I had this down, but I was wrong.

    (4x^3 -17ix^2+5x+(7-3i))/x-3i

    Help?
    This would be the same as for real numbers

    f(x)=\frac{4x^3-17ix^2+5x+(7-3i)}{x-3i}

    4x^3=4x^2*x=4x^2*x+\underbrace{4x^2*(-3i)-4x^2*(-3i)}_{=0}=4x^2(x-3i)+12ix^2
    (I made appear the factor (x-3i))

    \begin{aligned} \implies f(x) &=\frac{4x^2(x-3i)+12ix^2-17ix^2+5x+(7-3i)}{x-3i} \\ \\<br />
&=4x^2+\frac{-5ix^2+5x+(7-3i)}{x-3i} \end{aligned}

    ---------------------

    -5ix^2=-5ix(x-3i)-15 i^2 x=-5ix(x-3i)+15x

    \begin{aligned} \implies f(x) &=4x^2+\frac{-5ix(x-3i)+15x+5x+(7-3i)}{x-3i} \\ \\<br />
&=4x^2-5ix+\frac{20x+(7-3i)}{x-3i} \end{aligned}

    ---------------------

    20x=20(x-3i)+60i

    \begin{aligned} \implies f(x) &=4x^2-5ix+\frac{20(x-3i)+60i+7-3i}{x-3i} \\ \\<br />
&=4x^2-5ix+20+\frac{7+57i}{x-3i} \end{aligned}


    Is it clear enough ? :x
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  3. #3
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    ...I think so. It's still kind of confusing, but I'll work on it.

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