Question

D%20" id="TexFormula1" title=" \frac{(a-3) * (\frac{a}{3} +1)}{\frac{1}{3}} " alt=" \frac{(a-3) * (\frac{a}{3} +1)}{\frac{1}{3}} " align="absmiddle" class="latex-formula">
$\frac{a( \frac{a}{3} ) + a - 3(\frac{a}{3}) - 3 }{\frac{a}{3}}$

$\frac{\frac{3a+a}{3} + a - 3 * \frac{a}{3} - 3}{\frac{1}{3}}$

$\frac{\frac{4a}{3} + a - a - 3}{\frac{1}{3}}$

$\frac{4a}{3} / \frac{1}{3} - 3 / \frac{1}{3}$

$\frac{4a}{3} * 3 - 3 * 3$

$4a - 9$
Yet the answer I should get is $a^{2}-9$. What step am I doing wrong on this algebra problem?

Answer 1

$\bf \cfrac{(a-3)\left( \frac{a}{3}+1 \right)}{\frac{1}{3}}\implies 3\left[ (a-3)\left( \frac{a}{3}+1 \right) \right] \\\\\\ 3\left[\frac{a^2}{3}+a-a-3 \right]\implies 3\left[\frac{a^2}{3}-3 \right]\implies a^2-9$

when you have polynomials multiplication, say (x+y) (a+b+c), you can always just multiply x(a+b+c) + y(a+b+c), namely each term by all others and sum them up, like above.