Factor Problem 1

Using Prof Po-Shen Loh's different way to solve quadratic equation (video) .

Let the roots of $x^2 - \red 6x - 112 = 0$ be $\dfrac {\red 6}2-u$ and $\dfrac {\red 6}2+u$ . Then by Vieta's formula, the sum of the two roots $r+s = 3-u+3+u = 6$ and the product of the two roots:

$\begin{aligned} (3-u)(3+u) & = -112 \\ (u-3)(u+3) & = 112 \\ u^2 - 9 & = 112 \\ u^2 & = 121 \\ \implies u & = \pm 11 \end{aligned}$

Implying the two roots are $-8$ and $14$ and the factors are $\boxed{(x-14)(x+8)}$ .

$x^2-6x-112=x^2-14x+8x-112=x(x-14)+8(x-14)=\boxed {(x-14)(x+8)}$ .