Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-1280$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-8$$ $$ab=1\left(-1280\right)=-1280$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product $-1280$.
Rewrite $x^{2}-8x-1280$ as $\left(x^{2}-40x\right)+\left(32x-1280\right)$.
$$\left(x^{2}-40x\right)+\left(32x-1280\right)$$
Factor out $x$ in the first and $32$ in the second group.
$$x\left(x-40\right)+32\left(x-40\right)$$
Factor out common term $x-40$ by using distributive property.
$$\left(x-40\right)\left(x+32\right)$$
Steps Using the Quadratic Formula
Quadratic polynomial can be factored using the transformation $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$, where $x_{1}$ and $x_{2}$ are the solutions of the quadratic equation $ax^{2}+bx+c=0$.
$$x^{2}-8x-1280=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
