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Math Help - How do you factor an un-prime quadratic?

  1. #1
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    How do you factor an un-prime quadratic?

    like 6x^2+2x-5=0

    I can find solutions but they are generally wrong like (2x-5)(6x+2)=0.

    I think polynomial long-division might work, come to think of it.
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  2. #2
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    pickslides's Avatar
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    Here's a kick-off

    (ax+b)(cx+d) where ac=6 and bd=-5

    now expand (ax+b)(cx+d) out, what do you get?
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  3. #3
    Behold, the power of SARDINES!
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    Quote Originally Posted by bournouli View Post
    like 6x^2+2x-5=0

    I can find solutions but they are generally wrong like (2x-5)(6x+2)=0.

    I think polynomial long-division might work, come to think of it.
    Since the discriminant is irrational it will not factor with rational coefficients. Here is a way it can be done.

    \displaystyle  6x^2+2x-5=0 \iff 6\left[x^2+\frac{1}{3}x-\frac{5}{6}\right]=0

    Now complete the square to get

    \displaystyle 6\left[ \left(x^2+\frac{1}{3}x+\frac{1}{36}\right)-\frac{1}{36}-\frac{5}{6}\right]=6\left[ \left(x+\frac{1}{6} \right)^2-\frac{31}{36}\right]

    Now we can write this as the difference of squares

    \displaystyle 6\left[ \left(x+\frac{1}{6} \right)^2-\left(\frac{\sqrt{31}}{6}\right)^2\right]=6\left[ \left(x+\frac{1}{6} -\frac{\sqrt{31}}{6}\right)\left(x+\frac{1}{6} +\frac{\sqrt{31}}{6}\right)\right]
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  4. #4
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    It was really that easy huh? -thanks.
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