Factor the expression by grouping. First, the expression needs to be rewritten as $-5x^{2}+ax+bx-8$. To find $a$ and $b$, set up a system to be solved.
$$a+b=14$$ $$ab=-5\left(-8\right)=40$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $40$.
$$1,40$$ $$2,20$$ $$4,10$$ $$5,8$$
Calculate the sum for each pair.
$$1+40=41$$ $$2+20=22$$ $$4+10=14$$ $$5+8=13$$
The solution is the pair that gives sum $14$.
$$a=10$$ $$b=4$$
Rewrite $-5x^{2}+14x-8$ as $\left(-5x^{2}+10x\right)+\left(4x-8\right)$.
$$\left(-5x^{2}+10x\right)+\left(4x-8\right)$$
Factor out $5x$ in the first and $-4$ in the second group.
$$5x\left(-x+2\right)-4\left(-x+2\right)$$
Factor out common term $-x+2$ by using distributive property.
$$\left(-x+2\right)\left(5x-4\right)$$
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 $ax^{2}+bx+c=0$.
$$-5x^{2}+14x-8=0$$
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.
Now solve the equation $x=\frac{-14±6}{-10}$ when $±$ is plus. Add $-14$ to $6$.
$$x=-\frac{8}{-10}$$
Reduce the fraction $\frac{-8}{-10}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{4}{5}$$
Now solve the equation $x=\frac{-14±6}{-10}$ when $±$ is minus. Subtract $6$ from $-14$.
$$x=-\frac{20}{-10}$$
Divide $-20$ by $-10$.
$$x=2$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{4}{5}$ for $x_{1}$ and $2$ for $x_{2}$.
