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Factorising quadratic

  1. Mar 21, 2008 #1
    1. The problem statement, all variables and given/known data

    factorize fully:
    4x^2 - 12x - 14

    2. Relevant equations



    3. The attempt at a solution

    Quadratic equation:
    (-(-12)+ sqrt(12^2 - (4 * 4 * -14))/ 2 * 4
    (12 + sqrt(144 + 224))/8

    in the end it became a surd
    (12+ sqrt(368))/8

    Could someone tell me how to factorize this?
    or is that as far as it will go?
     
    Last edited: Mar 22, 2008
  2. jcsd
  3. Mar 21, 2008 #2
    [tex](4x\pm a)(x\pm b)[/tex]

    or

    [tex](2x\pm a)(2x\pm b)[/tex] <--- Try this one.

    The sign of the middle term tells who the sign of the bigger value, and the last term tells you whether or not you will have 2 positive/negative values or a positive & negative value.

    If all fails, you may need the quadratic equation.
     
    Last edited: Mar 21, 2008
  4. Mar 22, 2008 #3

    Gib Z

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    The quadratic equation finds the roots of [itex]ax^2+bx+c=0, a\neq 0[/itex], and the factor theorem says that if [itex]P(a) = 0, (x-a) [/itex] is a factor of P(x). So finding the zeros will enable to factor that quadratic.
     
  5. Mar 24, 2008 #4
    Is there a typo? Because 4x^2-10x-14 factors very nicely...
    If your teachers are trying to throw you a curve ball, then you must use the quadratic equation, with zeroes (1.5+sqrt(92/16)) and (1.5- sqrt(92/16))
     
  6. Mar 24, 2008 #5
    What I did was using the solution of this quadratic when equated to zero using the quadratic formula & this is what I got;

    (x - 3/2 - 1/2*sqrt(14))*(x - 3/2 + 1/2*sqrt(14))

    I haven't checked it so maybe an error in my calculations,but I don't really think so...

    Thats the technique I use for all my quadratic factorizations & its simply the easiest.
     
  7. Mar 24, 2008 #6

    symbolipoint

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    First, do you see a factor of 2 which can be factored:

    2(2x^2 - 6x -7)

    Now, you can plan for two binomial factors to look for the other terms within each binomial:

    (2x )(x )

    Look for different ways of getting a product of -7, so use like:

    (2x -7 )(x +1 )
    OR
    (2x +7)(x -1 )
    OR
    (2x -1)(x +7)
    OR
    (you fill in the rest)(.....)
     
  8. Mar 24, 2008 #7

    symbolipoint

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    ... strange that the spaces did not get filled in on one of the lines shown above.
     
  9. Mar 24, 2008 #8

    symbolipoint

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    apparantly, none of my four suggested combinations will work. Someone else check? Typographical error in given expression?
     
  10. Mar 24, 2008 #9
    Hi,
    The question is Completely factorize [tex]4x^2 - 12x - 14[/tex]
    I used the quadratic formula and got:
    [tex](x+\frac{\sqrt23+3}{2})(x-\frac{\sqrt23+3}{2})[/tex]
    The expression does not factor nicely. It's "prime" There are no combinations that symbolipoint suggested that will work and give nice pretty integers. No factors of 14 subtract to give you 6.
    CC
     
  11. Mar 24, 2008 #10
    Don't forget to multiply with 4 again. 4x^2 - 12x - 14 has the same roots as
    x^2 - 3x - 7/2 but not the same factorization,
     
  12. Mar 24, 2008 #11

    tiny-tim

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    … the straight answer …

    Hi Learnphysics! :smile:

    The straight answer to your question is that you can't get rid of the square root, though you can simplify it.

    As happyg1 almost says, it's (3 ± √23)/2. :smile:

    (btw, you must put the ± into your (12+ sqrt(368))/8)
     
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