Question:
Divide $\sqrt{3} a^{4}+2 \sqrt{3} a^{3}+3 a^{2}-6 a$ by $3 a$.
Solution:
$\frac{\sqrt{3} a^{4}+2 \sqrt{3} a^{3}+3 a^{2}-6 a}{3 a}$
$=\frac{\sqrt{3} a^{4}}{3 a}+\frac{2 \sqrt{3} a^{3}}{3 a}+\frac{3 a^{2}}{3 a}-\frac{6 a}{3 a}$
$=\frac{1}{\sqrt{3}} a^{(4-1)}+\frac{2}{\sqrt{3}} a^{(3-1)}+a^{(2-1)}-2$
$=\frac{1}{\sqrt{3}} a^{3}+\frac{2}{\sqrt{3}} a^{2}+a-2$
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