I've been staring at the following problems for at least half an hour. Would anyone help me?
2x^2+8x=0
-9x^2+12x-4=0
4x^2-4x+2=0
3x^2+4x-1=0
x^3+2x^2-4x-8=0
For 2x^2+8x=0, both the terms have x in them so this can be factorised out:
x(2x+8)=0
I was going to do the rest of them by finding a factor, but they all appear to be decimals or surds apart from the last one. I'll show you the method so you can use it on other questions:
f(x)=x^3+2x^2-4x-8
f(2)=2^3+2^3-8-8
f(2)=8+8-8-8
f(2)=0
Since f(2)=0, f(x) has a factor x-2 ie. when x=2 f(x)=0.
This can then be factorised using this (see icemanfan's post).
A graphical method can help:
If in doubt, draw the graph and see if that helps.