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yeongil · #7 July 9th, 2009, 09:20 PM

Quote:

Originally Posted by spidermonkey

thank you for your explanation but what im having problems with is the actual factoring of the problem i just dont get how to do it.

Okay...

Quote:

$\frac{a^4}{2a^2 - 4a^4} \cdot \frac{8a^3 + 1}{2a + 4a^2}$

Let's look at the denominator of the first fraction:
$2a^2 - 4a^4$
I don't like how it's not in standard form (ie. descending powers), so I change it to
$-4a^4 + 2a^2$
I then factor out the greatest common monomial factor, which is $-2a^2$ :
${\color{red}-2a^2(2a^2 - 1)}$

Now, look at the numerator of the second fraction:
$8a^3 + 1$
This is a sum of cubes. There is a formula that you should know:
$x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
So the sum of cubes can be factored this way:
$8a^3 + 1 = (2a)^3 + 1^3 = (2a + 1)[(2a)^2 - 2a(1) + 1^2] = {\color{red}(2a + 1)(4a^2 - 2a + 1)}$

Finally, look at the denominator of the second fraction:
$2a + 4a^2$
Again, it's not in standard form, so I change to
$4a^2 + 2a$
Factor out the greatest common monomial factor, which is 2a:
${\color{red}2a(2a + 1)}$

So, putting it all together,
$\frac{a^4}{2a^2 - 4a^4} \cdot \frac{8a^3 + 1}{2a + 4a^2}$
$= \frac{a^4}{-2a^2(2a^2 - 1)} \cdot \frac{(2a + 1)(4a^2 - 2a + 1)}{2a(2a + 1)}$

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