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Math Help - [SOLVED] Factoring a quadratic

  1. #1
    Forum Admin topsquark's Avatar
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    [SOLVED] Factoring a quadratic

    Can someone please tell me why this works? (It works in general apparently.)

    Let's factor
    12x^2 - 25x + 12

    Take the leading coefficient (12) and multiply it into the constant and rewrite this as
    x^2 - 25x + 144

    Now factor it:
    (x - 9)(x - 16)

    Now divide both of the constants by that same 12 as before:
    \left ( x - \frac{9}{12} \right ) \left ( x - \frac{16}{12} \right )

    = \left ( x - \frac{3}{4} \right ) \left ( x - \frac{4}{3} \right )

    Now multiply the whole thing by that same 12:
    12 \cdot \left ( x - \frac{3}{4} \right ) \left ( x - \frac{4}{3} \right )

    = 4 \cdot \left ( x - \frac{3}{4} \right )  \cdot 3 \cdot \left ( x - \frac{4}{3} \right )

    = (4x - 3)(3x - 4)
    which is the factored form of the original problem.

    I can't figure out why this works???!!

    -Dan
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    Senior Member bkarpuz's Avatar
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    Quote Originally Posted by topsquark View Post
    Can someone please tell me why this works? (It works in general apparently.)

    Let's factor
    12x^2 - 25x + 12

    Take the leading coefficient (12) and multiply it into the constant and rewrite this as
    x^2 - 25x + 144

    Now factor it:
    (x - 9)(x - 16)

    Now divide both of the constants by that same 12 as before:
    \left ( x - \frac{9}{12} \right ) \left ( x - \frac{16}{12} \right )

    = \left ( x - \frac{3}{4} \right ) \left ( x - \frac{4}{3} \right )

    Now multiply the whole thing by that same 12:
    12 \cdot \left ( x - \frac{3}{4} \right ) \left ( x - \frac{4}{3} \right )

    = 4 \cdot \left ( x - \frac{3}{4} \right )  \cdot 3 \cdot \left ( x - \frac{4}{3} \right )

    = (4x - 3)(3x - 4)
    which is the factored form of the original problem.

    I can't figure out why this works???!!

    -Dan
    Okay, let me explain this in geneneral.
    Let, a,b,c\in\mathbb{R} with a\neq0 and p(x):=ax^{2}+bx+c, then we know that
    p(x)=a(x-x_{1})(x-x_{2}),\qquad(1)
    where
    x_{1,2}:=\frac{-b\pm\sqrt{\Delta_{p}}}{2a}\text{ and }\Delta_{p}:=b^{2}-4ac.\qquad(2)
    Let q(x)=x^{2}+bx+ac, and we also know that q(x):=(x-x_{3})(x-x_{4}), where
    x_{3,4}:=\frac{-b\pm\sqrt{\Delta_{q}}}{2}\text{ and }\Delta_{q}:=b^{2}-4ac=\Delta_{p}.\qquad(3)
    So, from (2) and (3), we see that
    x_{1}=\frac{x_{3}}{a}\text{ and }x_{2}=\frac{x_{4}}{a}\qquad(4).
    Therefore, substituting the values in (4) into (1), we get the desired result.
    Last edited by bkarpuz; September 10th 2008 at 10:03 PM. Reason: Some typos are corrected.
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