1. ## Tricky Re-arrangment

Hi,

Would anyone be able to talk me through/just start me off getting from this:
$\displaystyle \frac{2}{3}(3-x)^{\frac{3}{2}} - 6 \sqrt{3-x}$

to this:

$\displaystyle -\frac{2}{3}\sqrt{3-x} (x+6)$

Thank You

2. Originally Posted by aceband
Hi,

Would anyone be able to talk me through/just start me off getting from this:
$\displaystyle \frac{2}{3}(3-x)^{\frac{3}{2}} - 6 \sqrt{3-x}$

to this:

$\displaystyle -\frac{2}{3}\sqrt{3-x} (x+6)$

Thank You
HI

$\displaystyle \frac{2}{3}(3-x)^{\frac{1}{2}}\cdot (3-x)-6(3-x)^{\frac{1}{2}}$

$\displaystyle (3-x)^{\frac{1}{2}}[\frac{2}{3}(3-x)-6]$

$\displaystyle \sqrt{3-x}(-4-\frac{2}{3}x)$

$\displaystyle \sqrt{3-x}\cdot -\frac{12+2x}{3}$

$\displaystyle -\frac{2}{3}\sqrt{3-x} (x+6)$