# Factor an expression containing rational exponents

• Apr 18th 2010, 05:13 PM
desiderius1
Factor an expression containing rational exponents
Factor the expression. Express your answer so that only positive exponents occur.

$
4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)} \qquad x>= -\frac{1}{2}
$

$
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(38x+13)
$

so far I've done:
$
(24x+12)^{(1/2)}(5x+1)^{(11/7)}+(18x+9)^{(3/2)}(5x+1)^{(4/7)}
$

$
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(4x+4)(5x+1)^{(7/7)}(3x+3)
$

Is that right so far and if so, what do I do next?

Thanks.
• Apr 18th 2010, 05:35 PM
skeeter
Quote:

Originally Posted by desiderius1
Factor the expression. Express your answer so that only positive exponents occur.

$
4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)} \qquad x>= -\frac{1}{2}
$

$
(6x+3)^{(1/2)}(5x+1)^{(4/7)}(38x+13)
$

so far I've done:
$
\textcolor{red}{(24x+12)^{(1/2)}}(5x+1)^{(11/7)}+\textcolor{red}{(18x+9)^{(3/2)}}(5x+1)^{(4/7)}
$

these two factors are incorrect ... you cannot distribute the constants into factors w/ exponents not equal to 1.

$4(6x+3)^{(1/2)}(5x+1)^{(11/7)} + 3(6x+3)^{(3/2)}(5x+1)^{(4/7)}
$

common factors are $(6x+3)^{1/2}(5x+1)^{4/7}$ ...

$(6x+3)^{1/2}(5x+1)^{4/7}\left[4(5x+1) + 3(6x+3)\right]$

$(6x+3)^{1/2}(5x+1)^{4/7}\left[(20x+4) + (18x+9)\right]$

$(6x+3)^{1/2}(5x+1)^{4/7}\left[38x + 13\right]
$

...
• Apr 18th 2010, 06:12 PM
desiderius1
Another factor question:

$
8x^{(-1/2)}+\frac{4}{7}x^{(1/2)}
$

$
\frac{4(14+x)}{7x^{(1/2)}}
$

Wouldn't $8x^{(-1/2)} = \frac{1}{\sqrt{8x}}$?

I don't see how they got the $7x^{(1/2)}$ on the bottom.. when the exponent isn't negative.
• Apr 18th 2010, 06:22 PM
skeeter
Quote:

Originally Posted by desiderius1
Another factor question:

$
8x^{(-1/2)}+\frac{4}{7}x^{(1/2)}
$

$
\frac{4(14+x)}{7x^{(1/2)}}
$

Wouldn't $8x^{(-1/2)} = \frac{1}{\sqrt{8x}}$?

I don't see how they got the $7x^{(1/2)}$ on the bottom.. when the exponent isn't negative.

$
8x^{-1/2}+\frac{4}{7}x^{1/2}
$

$\frac{8}{x^{1/2}} + \frac{4x^{1/2}}{7}$

common denominator is $7x^{1/2}$ ...
• Apr 18th 2010, 06:36 PM
desiderius1
$
8 + 4x^{1/2}
$

How do you get 4(14 + x)?
• Apr 18th 2010, 06:48 PM
desiderius1
oh I see, you multiply 8 and 7 to get 56.
56/14 = 4.