# Factoring with rational exponents

I'm not quite sure how to do this question. Every way that I tried doing it didn't yield an answer that is equivalent to the original question.

$$(2x+1)^{2/3}-4(2x+1)^{-1/3}$$

When I tried doing it, I ended up with $$(2x+1)^{2/3}\left(-\frac{8x+4}{2x+1}\right)$$ or with $$4(2x+1)^{1/3}$$

How can I factor it properly?

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Start by rewritting the equation a bit more clearly.

$$(2x+1)^{2/3} -4(2x+1)^{-1/3} = \sqrt[3]{\left ( 2x+1 \right)^2} - \frac{4}{\sqrt[3]{2x+1}}$$

Then, put everything on the same denominator.

$$\sqrt[3]{\left ( 2x+1 \right)^2} - \frac{4}{\sqrt[3]{2x+1}} = \frac{\sqrt[3]{\left ( 2x+1 \right)^2} \sqrt[3]{2x+1} - 4}{\sqrt[3]{2x+1}}$$

Simplify.

$$\frac{\sqrt[3]{\left ( 2x+1 \right)^2} \sqrt[3]{2x+1} - 4}{\sqrt[3]{2x+1}} = \frac{\sqrt[3]{\left ( 2x+1 \right)^3} - 4}{\sqrt[3]{2x+1}}$$

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More directly, factor out the smallest power of the common term, in this case $(2x+1)^{-1/3}$. – gaddy Jan 19 '14 at 4:20
I got (2x-3)/cbrt(2x+1). I think it seems right. Thanks a lot! – Alex Jan 19 '14 at 4:36
Yep, it is all right. Sorry I made a typo, $+4$ instead of $-4$. – Olivier Jan 19 '14 at 4:38

Let $\ \color{#c00}{y} = (\color{#0a0}{2x+1})^{1/\color{#c00}3}.\,$ Then $\ y^2-\dfrac{4}y\, =\, \dfrac{\color{#c00}{y^3}-4}y\, =\, \dfrac{2x-3}y\$ by $\ \color{#c00}{y^3} = \color{#0a0}{2x+1}$

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### First way

Multiply and divide both pieces by $(2 x + 1)^{1/3}$. You so obtain
$$\frac{2x+1 - 4}{(2 x + 1)^{1/3}} = \frac{2 x - 3}{(2 x + 1)^{1/3}}$$

### Second way

Factor $(2 x + 1)^{2/3}$. You so obtain
$$(2x+1)^{2/3}\cdot \frac{1-4}{2 x + 1} = (2 x + 1)^{2/3}\cdot \frac{2 x + 1 - 4}{2 x + 1} = \frac{2 x - 3}{(2 x + 1)^{1/3}}$$

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@dfeuer. Thanks for editing for me. Cheers – Claude Leibovici Jan 19 '14 at 4:45
My edit was but a slight improvement of Olivier's suggested one. – dfeuer Jan 19 '14 at 4:48
I do highly recommend using the hash-marks instead of bold for section headings so the HTML will come out with proper header elements. – dfeuer Jan 19 '14 at 4:50

As a general method you can get good results by factoring off the the binomial to the power that is leftmost on the real number line like so:

\begin{align*} (2x+1)^{\frac{2}{3}}-4(2x+1)^{-\frac{1}{3}} &= (2x+1)^{-\frac{1}{3}}\left( (2x+1)^{\frac{3}{3}} -4\right) \\ &= (2x+1)^{-\frac{1}{3}}( 2x+1 -4) \\ &= (2x+1)^{-\frac{1}{3}}( 2x-3) \\ &=\frac{2x-3}{\sqrt[3]{2x+1}}. \end{align*}

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