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Thread: [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
    $\displaystyle 12x^2 - 25x + 12$

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

    Now factor it:
    $\displaystyle (x - 9)(x - 16)$

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

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

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

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

    $\displaystyle = (4x - 3)(3x - 4)$
    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
    $\displaystyle 12x^2 - 25x + 12$

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

    Now factor it:
    $\displaystyle (x - 9)(x - 16)$

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

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

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

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

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