# factoring a basic quadratic equation

• May 9th 2011, 11:52 PM
tutormeplz
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
• May 10th 2011, 01:19 AM
pickslides
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?
• May 10th 2011, 04:35 AM
tutormeplz
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?
• May 10th 2011, 05:21 AM
pickslides
Quote:

Originally Posted by tutormeplz
how can (4x^2+4x+2)+(3-7x)-(5-3x) be solved algebraically?

There needs to be an equals sign somewhere here to 'solve' this. Or do you simply mean simplify it?

If so

$\displaystyle (4x^2+4x+2)+(3-7x)-(5-3x)$

$\displaystyle = 4x^2+4x+2+3-7x-5+3x$

$\displaystyle = 4x^2$
• May 10th 2011, 05:24 AM
jgv115
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
• May 10th 2011, 07:56 AM
ihavenonick
Quote:

Originally Posted by tutormeplz
Hello - this is two part question. First, can someone please help me factor this equation, or is this even factorable?

(4x^2+4x+2)

4x^2+4x+2
D=16-4*4*2=-16 < 0 => this equation hasnt any answers => isnt factorable.
• May 10th 2011, 02:05 PM
pickslides
Quote:

Originally Posted by ihavenonick
this equation hasnt any answers => isnt factorable.

All you have shown is that it has no real solutions.

$\displaystyle 4x^2+4x+2$

$\displaystyle 4\left[x^2+x+\frac{1}{2}\right]$

$\displaystyle 4\left[\left(x^2+x+\frac{1}{4}\right) -\frac{1}{4}+\frac{1}{2}\right]$

$\displaystyle 4\left[\left(x^2+x+\frac{1}{4}\right) +\frac{1}{4}\right]$

$\displaystyle 4\left[\left(x+\frac{1}{2}\right)^2 +\frac{1}{4}\right]$

$\displaystyle 4\left[\left(x+\frac{1}{2}\right)^2 -\frac{-1}{4}\right]$

$\displaystyle 4\left[\left(x+\frac{1}{2}\right)^2 -\left(\sqrt{\frac{-1}{4}}\right)^2\right]$

$\displaystyle 4\left[\left(x+\frac{1}{2}\right)^2 -\left(\frac{i}{2}\right)^2\right]$

$\displaystyle 4\left(x+\frac{1}{2}-\frac{i}{2}\right) \left(x+\frac{1}{2}+\frac{i}{2}\right)$
• May 10th 2011, 09:00 PM
tutormeplz
Wow - this is very helpful - I thank everyone for their time, especially pickslides and jgv115 (the link to the completing the squares is most helpful!)

Thank you once again.