Final answer to the problem
$x^{5}+10+\frac{40x-90}{3x^2-4x+9}$
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Step-by-step Solution
1
Divide $3x^7-4x^6+9x^5+30x^2$ by $3x^2-4x+9$
$\begin{array}{l}\phantom{\phantom{;}3x^{2}-4x\phantom{;}+9;}{\phantom{;}x^{5}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}+10\phantom{;}\phantom{;}}\\\phantom{;}3x^{2}-4x\phantom{;}+9\overline{\smash{)}\phantom{;}3x^{7}-4x^{6}+9x^{5}\phantom{-;x^n}\phantom{-;x^n}+30x^{2}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{\phantom{;}3x^{2}-4x\phantom{;}+9;}\underline{-3x^{7}+4x^{6}-9x^{5}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}\phantom{-;x^n}}\\\phantom{-3x^{7}+4x^{6}-9x^{5};}\phantom{;}30x^{2}\phantom{-;x^n}\phantom{-;x^n}\\\phantom{\phantom{;}3x^{2}-4x\phantom{;}+9-;x^n;}\underline{-30x^{2}+40x\phantom{;}-90\phantom{;}\phantom{;}}\\\phantom{;-30x^{2}+40x\phantom{;}-90\phantom{;}\phantom{;}-;x^n;}\phantom{;}40x\phantom{;}-90\phantom{;}\phantom{;}\\\end{array}$
$x^{5}+10+\frac{40x-90}{3x^2-4x+9}$
Final answer to the problem
$x^{5}+10+\frac{40x-90}{3x^2-4x+9}$