# How do you simplify (7d^2 - 14d)/(8h^5) * (12h^3)/(21d^2h - 28dh^2)?

Jul 23, 2015

Here's how you could simplify this expression.

#### Explanation:

Your starting expression looks like this

$\frac{7 {d}^{2} - 14 d}{8 {h}^{5}} \cdot \frac{12 {h}^{3}}{21 {d}^{2} h - 28 {\mathrm{dh}}^{2}}$

If you take into account the fact that

$\textcolor{b l u e}{\frac{a}{b}} \cdot \textcolor{g r e e n}{\frac{c}{d}} = \frac{\textcolor{b l u e}{a} \cdot \textcolor{g r e e n}{c}}{\textcolor{b l u e}{b} \cdot \textcolor{g r e e n}{d}} = \frac{\textcolor{g r e e n}{c} \cdot \textcolor{b l u e}{a}}{\textcolor{b l u e}{b} \cdot \textcolor{g r e e n}{d}} = \frac{\textcolor{g r e e n}{c}}{\textcolor{b l u e}{b}} \cdot \frac{\textcolor{b l u e}{a}}{\textcolor{g r e e n}{d}}$

you can play around with this expression a little to get

$\frac{\stackrel{\textcolor{red}{3}}{\cancel{12}} \cancel{{h}^{3}}}{\stackrel{\textcolor{red}{2}}{\cancel{8}} {h}^{\cancel{5}}} \cdot \frac{7 {d}^{2} - 14 d}{21 {d}^{2} h - 28 {\mathrm{dh}}^{2}} = \frac{3}{2 {h}^{2}} \cdot \frac{7 {d}^{2} - 14 d}{21 {d}^{2} h - 28 {\mathrm{dh}}^{2}}$

Now take a look at the second fraction. You can write

$7 {d}^{2} - 14 d = 7 d \cdot \left(d - 2\right)$

and

$21 {d}^{2} h - 28 {\mathrm{dh}}^{2} = 7 d \cdot h \left(3 d - 4 h\right)$

Dividing these two expressions will get you

$\frac{\cancel{7 d} \cdot \left(d - 2\right)}{\cancel{7 d} \cdot h \left(3 d - 4 h\right)} = \frac{d - 2}{h \left(3 d - 4 h\right)}$

The overall expression now becomes

3/(2h^2) * (d-2)/(h(3d - 4h)) = color(green)((3(d-2))/(2h^3(3d - 4h))