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Find $a$ if the remainder is constant when $10x^3-7x^2+ax+6$ is divided by $2x^2-3x+1$.

 Feb 21, 2020
 #1
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+1

Nevermind I figured it out myself

 

We perform the polynomial division:

\[
\begin{array}{c|cc cc}
\multicolumn{2}{r}{5x} & +4 \\
\cline{2-5}
2x^2-3x+1 & 10x^3&-7x^2&+ax&+6 \\
\multicolumn{2}{r}{-10x^3} & +15x^2 & -5x \\
\cline{2-4}
\multicolumn{2}{r}{0} & 8x^2 & (a-5)x & 6 \\
\multicolumn{2}{r}{} & -8x^2 & +12x & -4 \\
\cline{3-5}
\multicolumn{2}{r}{} & 0 & (a-5+12)x & 2 \\
\end{array}
\]

 The remainder will be constant if and only if  So the answer is 7

 Feb 21, 2020

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