First thing to do is take out any common factors.
(4x^2+4x+2) = 2(2x^2+2x+1)
If you are having no luck with facotring using integers, try completing the square.
Do you know that method?
Hello - this is two part question. First, can someone please help me factor this equation, or is this even factorable?
(4x^2+4x+2)
I've tried factoring this into either (2x-1)(2x-2) or (4x-1)(x-2), but as you can see I can't seem to find a number that will add up to 4x.
Secondly, once this is factorable, can someone show me how to solve this equation algebraically:
(4x^2+4x+2)+(3-7x)-(5-3x)
I know that I can substitute (using Princeton Review method to find answer) let's say 2 for x's and get the following, but would like to understand how this can be set up and solved for x:
(16+8+2)+(3-14)-(5-6)=
26-11+1 = 16
Many thanks
OK, I think I understand up to taking out the common factors, but I do not know the "using the squares" method. Can you please explain and going back to the entire question, once this part is solved how can (4x^2+4x+2)+(3-7x)-(5-3x) be solved algebraically?
To follow up on completing the square, it isn't difficult at all
I have linked you to a page which tells you how to do it: Completing the Square: Solving Quadratic Equations
You must know this for later math anyway, so learn it now