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Math Help - factoring a basic quadratic equation

  1. #1
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    factoring a basic quadratic equation

    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
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  2. #2
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    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?
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  3. #3
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    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?
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  4. #4
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    Quote Originally Posted by tutormeplz View Post
    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
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  5. #5
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    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
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  6. #6
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    Quote Originally Posted by tutormeplz View Post
    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 hasn`t any answers => isn`t factorable.
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  7. #7
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    Quote Originally Posted by ihavenonick View Post
    this equation hasn`t any answers => isn`t factorable.
    Are you sure about that?

    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)
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  8. #8
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    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.
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