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

[buddy]

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

[stapel_eliz]

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

i dont see how anything comes out?

Nothing much does, actually...

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

. . . . .

. . . . .

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

. . . . .

. . . . .

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:

. . . . .

. . . . .

Now the common denominator would be x, so:

. . . . .

. . . . .

The only common factor is the radical:

. . . . .

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

[buddy]

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

[stapel_eliz]

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