# Simple algebra. How can you factor out a common factor if it is $\frac{1}{\text{factor}}$ in one of the cases?

I'm sure this is simple. I want to pull out a factor as follows...

I have the expression

$$\frac{a(\sqrt{x}) - (b + c)(\frac{1}{\sqrt{x}})c}{x}.$$

It would be useful for me to pull out the $\sqrt{x}$ from the numerator and try to simplify to remove the denominator, but how can I pull out the $\sqrt{x}$ from the right-most statement $\frac{1}{\sqrt{x}}$.

Thanks for your help!

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You could "pull it out" as long as you furnish one downstairs: ${1\over\sqrt x}=\sqrt x{ 1\over\sqrt x\sqrt x}$. It would better to multiply both top and bottom by $\sqrt x$ here ("clear denominators"). – David Mitra Feb 18 '12 at 16:37

$$\frac{a(\sqrt{x}) - (b + c)({\frac{\sqrt{x}}{x}})c}{x}\;\tag{1}$$

$$=\frac{a(\color{red}{\sqrt{x}}) - (b + c)({\frac{\color{red}{\sqrt{x}}}{x}})c}{\color{red}{\sqrt x}\sqrt x}\;\tag{2}$$

$$=\frac{(\color{red}{\sqrt x})[a - (b + c)({\frac{1}{x}})c]}{\color{red}{\sqrt x}\sqrt x}\;\tag{3}$$

$$=\frac{[a - (b + c)({\frac{1}{x}})c]}{\sqrt{x}}$$

And finally:

$$=\frac{[ax - (b + c)c]}{x\sqrt{x}}$$

Hope it helps.

$(1)$: Rewriting ${1\over\sqrt x}$ as ${\sqrt x \over x}$

$(2)$: Rewriting $x$ as $\sqrt x \sqrt x$, now $\sqrt x$ is the common factor in both numerator and denominator

$(3)$: Pulling the common factor ($\sqrt x$) out in the numerator

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In general, if you have an expression $a+\frac{1}{c}b$ and you want to factor out the $\frac{1}{c}$, multiply the expression by $\frac{c}{c}$ like this:

$$a+\frac{1}{c}b \\ \frac{c}{c}\left(a+\frac{1}{c}b\right) \\ \frac{1}{c}\left(c\cdot a + c\cdot \frac{1}{c}b\right) \\ \frac{1}{c}\left( ca+b \right)$$

I find multiplying the whole expression by $\frac{c}{c}$ the clearest way to pull out these fractions, especially if the expression is part of a larger fraction, like yours. Then multiplying by $\frac{c}{c}$ is the same as just multiplying the numerator and denominator by $c$. For example,

$$\frac{a+\frac{1}{c}b}{d+e} \\ \frac{c}{c}\frac{a+\frac{1}{c}b}{d+e} \\ \frac{c\left(a+\frac{1}{c}b\right)}{c\left(d+e\right)} \\ \frac{c\cdot a+c\cdot \frac{1}{c}b}{c\cdot d+c\cdot e} \\ \frac{ca+b}{cd+ce}$$

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$$a\sqrt{x} - (b+c)\frac{1}{\sqrt{x}}c = a\sqrt{x} - (b+c)\frac{\sqrt{x}}{\sqrt{x}\sqrt{x}}c = \sqrt{x}\left( a - (b+c) \frac 1x c\right)$$

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