1. ## Dividing Rational expressions

The book says the answer is supposed to be x+2 for this question with the restrictions of x does not equal -2, $\frac12$

I couldn't get that

The question is:

$\frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9}$

2. Originally Posted by t-dot
The book says the answer is supposed to be x+2 for this question with the restrictions of x does not equal -2, $\frac12$

I couldn't get that

The question is:

$\frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9}$
$\frac{2x^2-8}{6x+3}\div\frac{6x-12}{18x+9} = \frac{2x^2-8}{6x+3} \times \frac{18x+9}{6x-12}$

$= \frac{2x^2-8}{6x+3} \times \frac{3(6x+3)}{6x-12}$

$= \frac{3(2x^2-8)}{6x - 12}$

$= \frac{6(x^2-4)}{6(x - 2)}$

$= \frac{6(x - 2)(x + 2)}{6(x - 2)}$

$= (x + 2)$

3. Hello, t-dot!

$\frac{2x^2-8}{6x+3} \div \frac{6x-12}{18x+9}\;\; {\color{blue}=\;\;\frac{2x^2-8}{6x+3} \times \frac{18x+9}{6x-12}}$

Book's answer: . $x+2$
Factor all the expressions:

. . $2x^2-8\:=\:2(x^2-4)\:=\:2(x-2)(x+2)$
. . $6x+3\:=\:3(2x+1)$
. . $6x-12 \:=\:6(x-2)$
. . $18x + 9 \:=\:9(2x+1)$

The problem becomes: . $\frac{\not{2}(x-2)(x+2)}{\not3(2x+1)} \times \frac{\not9(2x+1)}{\not6(x-2)} \;\;=\;\;x + 2$

4. thanks guys...
i just kept doing my math wrong
that was my problem
lol
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