Question

Simplify and maybe also state the restrictions on the variable:

Answer 1

$\frac{15a^2\left(a-1\right)}{8\left(2a+3\right)}\cdot \frac{10\left(2a+3\right)}{3a}$

$\mathrm{Apply\:the\:fraction\:rule}:$

$\frac{a}{b}\cdot \frac{c}{d}=\frac{a\:\cdot \:c}{b\:\cdot \:d}$

$=\frac{15a^2\left(a-1\right)\cdot \:10\left(2a+3\right)}{8\left(2a+3\right)\cdot \:3a}$

$\mathrm{Cancel\:the\:common\:factor:}\:2a+3$

$=\frac{15a^2\left(a-1\right)\cdot \:10}{8\cdot \:3a}$

$\mathrm{Factor\:the\:number:\:}\:15=3\cdot \:5$

$=\frac{3\cdot \:5a^2\left(a-1\right)\cdot \:10}{8\cdot \:3a}$

$\mathrm{Cancel\:the\:common\:factor:}\:3$

$=\frac{5a^2\left(a-1\right)\cdot \:10}{8a}$

$\mathrm{Factor\:the\:number:\:}\:10=2\cdot \:5$

$=\frac{5a^2\left(a-1\right)\cdot \:2\cdot \:5}{2\cdot \:4a}$

$\mathrm{Factor\:the\:number:\:}\:8=2\cdot \:4$

$=\frac{5a^2\left(a-1\right)\cdot \:2\cdot \:5}{2\cdot \:4a}$

$\mathrm{Cancel\:the\:common\:factor:}\:2$

$=\frac{5aa\left(a-1\right)\cdot \:5}{4a}$

$\mathrm{Apply\:exponent\:rule}:$

$a^{b+c}=a^b\cdot \:a^c$

$a^2=aa$

$\mathrm{Cancel\:the\:common\:factor:}\:a$

$=\frac{5a\left(a-1\right)\cdot \:5}{4}$

$\mathrm{Multiply\:the\:numbers:}\:5\cdot \:5=25$

$=\frac{25a\left(a-1\right)}{4}$

$\mathrm{I\:hope\:that\:I\:helped\:you\:!}$