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