Factor an expression containing rational exponents

**desiderius1** · April 18th 2010, 05:13 PM

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} $

The answer:
$ (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.

**skeeter** · April 18th 2010, 05:35 PM

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} $

The answer:
$ (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] $

...

**desiderius1** · April 18th 2010, 06:12 PM

Another factor question:

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

Answer:
$ \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.

**skeeter** · April 18th 2010, 06:22 PM

Originally Posted by desiderius1

Another factor question:

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

Answer:
$ \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.