# Factorisation

• Nov 20th 2009, 08:50 AM
wolfhound
Factorisation
Q: Factorise the numerator and denominator of each of the following and then cancel out any common factors
4x+8 4(x+2) x+2
-------- ----------- ------- = 1 ?
3x+6 3(x+2) x+2

is this correct 1(or leave it to x+2/x+2)
• Nov 20th 2009, 09:07 AM
bigwave
the (x+2) cancel out
$\displaystyle \frac{4x + 8}{3x + 6}$

$\displaystyle = \frac{4(x+2)}{3(x+2)}$

cancel out $\displaystyle (x+2)$

$\displaystyle = \frac{4}{3}$
• Nov 20th 2009, 09:09 AM
Soroban
Hello, wolfhound!

I'll guess what you're trying to say.

Quote:

Factor the numerator and denominator and cancel out any common factors.

. . $\displaystyle \frac{4x+8}{3x+6}$

. . $\displaystyle \frac{4x + 8}{3x+6} \;\;=\;\;\frac{4\,(x+2)}{3\,(x+2)} \;\;=\;\;\frac{4\,{\color{red}\rlap{//////}}(x+2)}{3\,{\color{red}\rlap{//////}}(x+2)} \;\;=\;\; \frac{4}{3}$

• Nov 20th 2009, 09:11 AM
wolfhound
Quote:

Originally Posted by Soroban
Hello, wolfhound!

. . $\displaystyle \frac{4x + 8}{3x+6} \;\;=\;\;\frac{4\,(x+2)}{3\,(x+2)} \;\;=\;\;\frac{4\,{\color{red}\rlap{//////}}(x+2)}{3\,{\color{red}\rlap{//////}}(x+2)} \;\;=\;\; \frac{4}{3}$