# Why (ax^2 + bx + c) = a(x-α)(x-β)?

1. Jun 14, 2013

### Atran

Say, 3x2 - 2x - 1 = 0
Then: x1 = -1/3, x2 = 1
Therefore: k * (x + 1/3) (x - 1) = 3x2 - 2x - 1
Why is k = 3?

Why, $Ax^{2} + Bx + C = A (x-\alpha) (x-\beta)$?

Thanks for help.

Last edited by a moderator: Jun 14, 2013
2. Jun 14, 2013

### pwsnafu

Well the polynomial $3x^2 - 2x -1 = (3x+1)(x-1) = 3 (x+1/3)(x-1)$.

Also solving the equation $3x^2-2x-1 = 0$ is equivalent to $x^2 - \frac{2}{3}x - \frac{1}{3} = 0$.

And even further notice that when you take $A(x-\alpha)(x-\beta)$ and expand the coefficient of $x^2$ is equal to A.

3. Jun 14, 2013

### Dods

Well, you have in general
$$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

and

$$x_1=\frac{-b+ \sqrt{b^2-4ac}}{2a}$$

$$x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}$$

We also have

$$a(x_1 + x_2)=a(\frac{-b+ \sqrt{b^2-4ac}}{2a} + \frac{-b- \sqrt{b^2-4ac}}{2a}) = a(\frac{-2b}{2a}) = a(\frac{-b}{a}) = -b$$

and if $a(x_1+x_2) = -b$,

$$a(-x_1-x_2) = b$$

We also have

$$(x_1)\cdot(x_2)=(\frac{-b+ \sqrt{b^2-4ac}}{2a})\cdot(\frac{-b- \sqrt{b^2-4ac}}{2a})$$

which after a little work gets you $(x_1) \cdot (x_2)= \frac{c}{a}$. Equivalently we can have $$a(x_1)\cdot(x_2) = c$$

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If we substitute our b and c into $ax^2+bx+c$ we get

$$ax^2+bx+c = ax^2+ a(-x_1-x_2)x + a(x_1)\cdot (x_2)$$
$$= a(x^2 -x\cdot x_1 -x\cdot x_2 + (x_1)\cdot(x_2))$$
$$= a(x-x_2)(x-x_1)$$

In the above step you factorize, it should be clear how you get there if you try to expand$(x-x_2)(x-x_1)$.

So that's why $ax^2+bx+c = a(x-x_2)(x-x_1)$

Last edited: Jun 14, 2013
4. Jun 14, 2013