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by **missmorgan** on Wed Sep 07, 2011 6:36 am

I know I've done things like this before, but I don't have my book on hand and I've forgotten quite a bit over summer. I've been looking online and can't find any problems similar to this. :/ Help please?

1. Factor

2. Factor

For number two I got an answer of (x+7)^1/4[1+(x+7)^1/2]... but I'm still a little uncomfortable with the whole way I did it.

(x+7)^1/4+(x+7)^1+1/4
(x+7)^1/4+(x+7)*(x+7)^-1/4
(x+7)^1/4[1+(x+7)(x+7)^-1/4-1/4]
(x+7)^1/4[1+(x+7)(x+7)^-1/2]
(x+7)^1/4[1+(x+7)^1/2]

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by **nona.m.nona** on Wed Sep 07, 2011 12:10 pm

missmorgan wrote:1. Factor $8x^{\frac{1}{4}}\, +\, \frac{2}{x^{\frac{3}{4}}$

It might help to convert to a common denominator, combine, and then rationalize the denominator.

$\left(\frac{8x^{\frac{1}{4}}x^{\frac{3}{4}}}{x^{\frac{3}{4}}}\right)\, +\, \left(\frac{2}{x^{\frac{3}{4}}\right)$

Note also that a "2" may be factored out.

missmorgan wrote:2. Factor $\left(x\, +\, 7\right)^{\frac{1}{4}}\, +\, \left(x\, +\, 7\right)^{\frac{3}{4}}$

For number two I got an answer of (x+7)^1/4[1+(x+7)^1/2]... but I'm still a little uncomfortable with the whole way I did it.

(x+7)^1/4+(x+7)^1-1/4
(x+7)^1/4+(x+7)*(x+7)^-1/4
(x+7)^1/4[1+(x+7)(x+7)^-1/4-1/4]
(x+7)^1/4[1+(x+7)(x+7)^-1/2]
(x+7)^1/4[1+(x+7)^1/2]

It might "feel" better to convert 3/4 into 1/2 + 1/4, rather than into 1 - 1/4. But the result should be the same.

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