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by **buddy** on Wed Feb 25, 2009 10:20 pm

factor x^-1/2 + x^5/2
the -1/2 and 5/2 are powers
i dont see how anything comes out?

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by **stapel_eliz** on Wed Feb 25, 2009 11:01 pm

buddy wrote:factor x^-1/2 + x^5/2
the -1/2 and 5/2 are powers

To format this clearly, use grouping symbols:

. . . . .x^(-1/2) + x^(5/2)

You can also use the "sup" tags (from the formatting line above the message-entry box):

. . . . .x^-1/2 + x^5/2

...displays as:

. . . . .x^-1/2 + x^5/2

buddy wrote:i dont see how anything comes out?

Nothing much does, actually... :roll:

Use exponent rules to convert the negative-powered term to a fraction:

. . . . . $x^{-1/2}\, +\, x^{5/2}$

. . . . . $\frac{1}{x^{1/2}}\, +\, x^{5/2}$

You might find it easy, by the way, to convert the fractional powers to radical form, while you're doing your steps:

. . . . . $\frac{1}{\sqrt{x}}\, +\, \sqrt{x^5}$

. . . . . $\frac{1}{\sqrt{x}}\, +\, x^2 \sqrt{x}$

The common denominator is the square root of x, but of course you're not supposed to have radicals in the denominator, so let's "rationalize" that first:

. . . . . $\left(\frac{1}{\sqrt{x}}\right) \left(\frac{\sqrt{x}}{\sqrt{x}}\right)\, +\, x^2 \sqrt{x}$

. . . . . $\frac{\sqrt{x}}{x}\, +\, x^2 \sqrt{x}$

Now the common denominator would be x, so:

. . . . . $\frac{\sqrt{x}}{x}\, +\, \frac{x^3 \sqrt{x}}{x}$

. . . . . $\frac{\sqrt{x}\, +\, x^3 \sqrt{x}}{x}$

The only common factor is the radical:

. . . . . $\frac{\sqrt{x}\left(1\, +\, x^3\right)}{x}$

Factor the sum of cubes, and I think you're done!

by **buddy** on Wed Feb 25, 2009 11:35 pm

so its (square root of x)(1+x)(1+x+x^2)/x?

by **stapel_eliz** on Thu Feb 26, 2009 2:35 pm

Check your signs! There should be a "minus" somewhere in the factoring of the sum of cubes! :wink:

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