# How do you simplify (x^2+12x+20)/(4x^2-9)*(6x^3-9x^2)/(x^3+10x^2)*(2x+3)?

May 5, 2017

#### Answer:

$= 3 \left(x + 2\right)$

#### Explanation:

$\frac{{x}^{2} + 12 x + 20}{4 {x}^{2} - 9} . \frac{6 {x}^{3} - 9 {x}^{2}}{{x}^{3} + 10 {x}^{2}} . \left(2 x + 3\right)$
$\text{ }$
$= \frac{{x}^{2} + 12 x + 20 + 16 - 16}{4 {x}^{2} - 9} . \frac{3 {x}^{2} \left(2 x - 3\right)}{{x}^{2} \left(x + 10\right)} . \left(2 x + 3\right)$
$\text{ }$
$= \frac{\left({x}^{2} + 12 x + 20 + 16\right) - 16}{4 {x}^{2} - 9} . \frac{3 {x}^{2} \left(2 x - 3\right)}{{x}^{2} \left(x + 10\right)} . \left(2 x + 3\right)$
$\text{ }$
$= \frac{\left({x}^{2} + 12 x + 36\right) - 16}{4 {x}^{2} - 9} . \frac{3 {x}^{2} \left(2 x - 3\right)}{{x}^{2} \left(x + 10\right)} . \left(2 x + 3\right)$
$\text{ }$
$= \frac{\left({x}^{2} + 2 \left(6\right) x + {6}^{2}\right) - 16}{4 {x}^{2} - 9} . \frac{3 {x}^{2} \left(2 x - 3\right)}{{x}^{2} \left(x + 10\right)} . \left(2 x + 3\right)$
$\text{ }$
$= \frac{{\left(x + 6\right)}^{2} - 16}{4 {x}^{2} - 9} . \frac{3 {x}^{2} \left(2 x - 3\right)}{{x}^{2} \left(x + 10\right)} . \left(2 x + 3\right)$
$\text{ }$
$= \frac{{\left(x + 6\right)}^{2} - {4}^{2}}{{\left(2 x\right)}^{2} - {3}^{2}} . \frac{3 {x}^{2} \left(2 x - 3\right)}{{x}^{2} \left(x + 10\right)} . \left(2 x + 3\right)$
$\text{ }$
Here we will apply the difference of two squares property that says:
$\text{ }$
color(blue)(a^2-b^2=(a-b)(a+b)
$\text{ }$
$= \frac{\textcolor{b l u e}{\left(x + 6 - 4\right) \left(x + 6 + 4\right)}}{\textcolor{b l u e}{\left(2 x - 3\right) \left(2 x + 3\right)}} . \frac{3 {x}^{2} \left(2 x - 3\right)}{{x}^{2} \left(x + 10\right)} . \left(2 x + 3\right)$
$\text{ }$
$= \frac{\left(x + 2\right) \textcolor{g r e e n}{\cancel{\left(x + 10\right)}}}{\textcolor{red}{\cancel{\left(2 x - 3\right)}} \textcolor{p u r p \le}{\cancel{\left(2 x + 3\right)}}} . \frac{3 \textcolor{b r o w n}{\cancel{{x}^{2}}} \textcolor{red}{\cancel{\left(2 x - 3\right)}}}{{\textcolor{b r o w n}{\cancel{x}}}^{2} \textcolor{g r e e n}{\cancel{\left(x + 10\right)}}} . \textcolor{p u r p \le}{\cancel{\left(2 x + 3\right)}}$
$\text{ }$
$= 3 \left(x + 2\right)$