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June 21st, 2015, 10:05 PM  #1 
Newbie Joined: Jun 2015 From: Philippines Posts: 1 Thanks: 0  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 
June 22nd, 2015, 12:07 AM  #2 
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 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)^443\left(\frac ab\right)^2+7\left(\frac ab\right)^39\left(\frac ab\right)5\right]}{b^2\left[3\left(\frac ab\right)^2+5\left(\frac ab\right)7\right]}$ $\displaystyle =\ b^2\cdot\left[\frac{6x^443x^2+7x^39x5}{3x^2+5x7}\right]$ And now you can apply longdivision 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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division, long, synthetic 
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