# Thread: Technique for finding the correct factor

1. ## Technique for finding the correct factor

From teh answer in my textbook and the result given in wolframAlpha/Maple etc i know that this:

$\frac{16}{27}u^3 + \frac{4}{45}u^5$

Simplifies to this:

$\frac{4}{135}u^3(20 + 3u^2)$

But i don't see how they work out that $\frac{4}{135}$ is the factor to take out.

If there computer can work it out consistently im assuming theres a technique/method used to find it? Could anyone point me in the right direction?

Thank you

2. Originally Posted by aceband
From teh answer in my textbook and the result given in wolframAlpha/Maple etc i know that this:

$\frac{16}{27}u^3 + \frac{4}{45}u^5$

Simplifies to this:

$\frac{4}{135}u^3(20 + 3u^2)$

But i don't see how they work out that $\frac{4}{135}$ is the factor to take out.

If there computer can work it out consistently im assuming theres a technique/method used to find it? Could anyone point me in the right direction?

Thank you
You only have to factorize the summands and then take the graetest common factor in front of the bracket:

$\frac{4 \cdot 4}{3 \cdot 9}u^3 + \frac{4}{5 \cdot 9}u^3 \cdot u^2 = \frac{4}{3 \cdot 5 \cdot 9}u^3(20 + 3u^2)$

3. Originally Posted by earboth
You only have to factorize the summands and then take the graetest common factor in front of the bracket:

$\frac{4 \cdot 4}{3 \cdot 9}u^3 + \frac{4}{5 \cdot 9}u^3 \cdot u^2 = \frac{4}{3 \cdot 5 \cdot 9}u^3(20 + 3u^2)$
$\frac{4 \cdot 4}{3 \cdot 9}u^3 + \frac{4}{5 \cdot 9}u^3 \cdot u^2= \frac{4 \cdot 4\cdot 5}{3\cdot 5\cdot 9}u^3+ \frac{4\cdot 3}{3\cdot 5\cdot 9}u^3\cdot u^2 = \frac{4}{3 \cdot 5 \cdot 9}u^3(20 + 3u^2)$
is perhaps a little clearer. To get both "3" and "5" in both numerators I multiplied numerator and denominator of the first fraction by 5 and of the second fraction by 3.