college algebra

[gar598]

could someone help me with this : (factoring)

ax^2-4a^2-45a^3=0

[audioaxes]

ouch- i forgot alot even though its only been 2 years since ive taken 10th grade algebra


i tried solving it but got stuck: the most i could tell you is to first pull an 'a' ::


a(x^2-4a-45a^2)=0

i know that didnt help much, if any at all

[minger]

you have two variables? i am not sure if you can just solve the equation for a without knowing what x is. if you break it down a little you can get:

-45a^3 + (x-4)a^2 = 0
if you set x = 0, you get:
-45a^3 = 0 in which case a=0
if you set x = 5, you get:
-45a^3 + a^2 = 0, which can be rewritten as:
a^2(-45a + 1) = 0
in which case a = 0, and 1/45)

i dont know, i havent taken anything multivariable yet, so maybe there is a different way to look at it.