# Quod Erat Demonstrandum

## 2010/10/21

### 因式分解二次多項式

Filed under: NSS — johnmayhk @ 5:37 下午
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$ax^2 + bx + c \equiv a(x - \alpha)(x - \beta)$ ………. (*)

$ax^2 + bx + c \equiv ax^2 - a(\alpha + \beta)x + a\alpha\beta$

$\alpha + \beta = \frac{-b}{a}$
$\alpha\beta = \frac{c}{a}$

$ax^2 + bx + c = 0$ 的根是 $\alpha$$\beta$
$a(x - \alpha)(x - \beta)$ 的根也是 $\alpha$$\beta$

$ax^2 + bx + c = 0$ 中，$x^2$ 的係數是 $a$
$a(x - \alpha)(x - \beta)$ 中，$x^2$ 的係數也是 $a$

$ax^2 + bx + c = 0$$a(x - \alpha)(x - \beta)$ 是恆等的。」

$\alpha, \beta$ 為二次方程 $ax^2 + bx + c = 0$ 的根，證明

$ax^2 + bx + c \equiv a(x - \alpha)(x - \beta)$

｛解釋一｝

$\alpha$$ax^2 + bx + c = 0$ 的其中一個根，

$a\alpha^2 + b\alpha + c = 0$

$ax^2 + bx + c \equiv ax^2 + bx + c - 0$
$ax^2 + bx + c \equiv ax^2 + bx + c - (a\alpha^2 + b\alpha + c)$
$ax^2 + bx + c \equiv a(x^2 - \alpha^2) + b(x - \alpha)$
$ax^2 + bx + c \equiv a(x - \alpha)(x + \alpha) + b(x - \alpha)$
$ax^2 + bx + c \equiv (x - \alpha)(a(x + \alpha) + b)$ ………. (*)

$a\beta^2 + b\beta + c = 0$

$a\beta^2 + b\beta + c = (\beta - \alpha)(a(\beta + \alpha) + b)$
$0 = (\beta - \alpha)(a(\beta + \alpha) + b)$

$a(\beta + \alpha) + b = 0$　［看，這裡順便得到 $\alpha + \beta = -\frac{b}{a}$

$ax^2 + bx + c \equiv (x - \alpha)(a(x + \alpha) + b - 0)$
$ax^2 + bx + c \equiv (x - \alpha)(a(x + \alpha) + b - (a(\beta + \alpha) + b))$
$ax^2 + bx + c \equiv (x - \alpha)(a(x - \beta))$
$ax^2 + bx + c \equiv a(x - \alpha)(x - \beta)$

$(x - \alpha)(a(x + \alpha) + b) = 0$ 有重根 $\alpha$

$(x - \alpha)(a(x + \alpha) + b) = 0$

$x - \alpha = 0$$a(x + \alpha) + b = 0$

$a(\alpha + \alpha) + b = 0$

$ax^2 + bx + c \equiv (x - \alpha)(a(x + \alpha) + b - 0)$
$ax^2 + bx + c \equiv (x - \alpha)(a(x + \alpha) + b - (a(\alpha + \alpha) + b))$
$ax^2 + bx + c \equiv (x - \alpha)(a(x - \alpha))$
$ax^2 + bx + c \equiv a(x - \alpha)(x - \alpha)$
$ax^2 + bx + c \equiv a(x - \alpha)^2$

｛解釋二｝