# 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.

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

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

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

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

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.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Originally Posted by desiderius1
Another factor question:

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

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

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

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

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

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

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

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.