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June 21st, 2015, 11:05 PM   #1
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Long Division and Synthetic Division

Hi I have an homework which is I'm having hardtime to solve.

1.) Long Division

6a^4 - 43a^2b^2 + 7a^2b - 9ab^3 - 5b^4 ÷ 3a^2 + 5ab - 7b^2

2.) Synthetic Division

2x^5 + x^4 + 4x^3 + x + 1 ÷ x + 1/2
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June 22nd, 2015, 01:07 AM   #2
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Math Focus: Abstract algebra
I think there is a typo in the first question: It should be
1) $\displaystyle \frac{6a^4 - 43a^2b^2 + 7a^{\color{red}3}b - 9ab^3 - 5b^4}{3a^2 + 5ab - 7b^2}$
This would then be homogeneous in $a$ and $b$ (meaning the sum of the powers of $a$ and $b$ is constant in the numerator and denominator (separately)) so you can reduce it to an expression in just one variable: Factor out the highest power of $b$ in the numerator and the highest power of $b$ in the denominator, and let $x=\dfrac ab$.
$\displaystyle \frac{6a^4 - 43a^2b^2 + 7a^3b - 9ab^3 - 5b^4}{3a^2 + 5ab - 7b^2}$
$\displaystyle =\ \frac{b^4\left[6\left(\frac ab\right)^4-43\left(\frac ab\right)^2+7\left(\frac ab\right)^3-9\left(\frac ab\right)-5\right]}{b^2\left[3\left(\frac ab\right)^2+5\left(\frac ab\right)-7\right]}$

$\displaystyle =\ b^2\cdot\left[\frac{6x^4-43x^2+7x^3-9x-5}{3x^2+5x-7}\right]$

And now you can apply long-division techniques, substituing $x$ back to $a$ and $b$ in the end.

This is assuming there is a typo and the way I fixed it above is correct. If there is no typo, then...I don't know.
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