Why (ax^2 + bx + c) = a(x-α)(x-β)?

[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.

 

[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.

 

[Dods]

To answer the question:

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

Thanks for help.

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

-----------------

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)

 

[Atran]

Thank you for the answers.