simplify radical

**galanm** · April 6th 2009, 05:38 AM

8x^2-8x / 2x^3?
What am I missing? I can't seem to figure out the bottom.

**Hush_Hush** · April 6th 2009, 06:17 AM

Is it

$\frac{8x^2-8x}{2x^3}$

or

$8x^2-\frac{8x}{2x^3} ?$

For the first one u have:

$\frac{8x^2-8x}{2x^3}=\frac{8x(x-1)}{2x^3}=\frac{4(x-1)}{x^2}=\frac{4x-4}{x^2}=\frac{4}{x}-\frac{4}{x^2}$

For the second u will have

$8x^2-\frac{8x}{2x^3} =8x^2-\frac{4}{x^2}$

Have a nice day ,

Hush_Hush

**earboth** · April 6th 2009, 06:19 AM

Originally Posted by galanm

8x^2-8x / 2x^3?
What am I missing? I can't seem to figure out the bottom.

$\dfrac{8x^2-8x}{2x^3} = \dfrac{2x(4x-4)}{2x\cdot x^2}=\dfrac{4(x-1)}{x^2}$

**masters** · April 6th 2009, 09:27 AM

Another look, using the greatest common factor first.

$\dfrac{8x^2-8x}{2x^3}=\frac{8x(x-1)}{2x^3}=\frac{\rlap{///}8^{\color{red}4}x(x-1)}{\rlap{//}2x^3_{\color{red}x^2}}=\frac{{\color{red}4}(x-1)}{{\color{red}x^2}}$

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