# [SOLVED] Factoring a quadratic

• Sep 10th 2008, 07:16 PM
topsquark
[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
• Sep 10th 2008, 09:36 PM
bkarpuz
Quote:

Originally Posted by topsquark
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. (Wink)