Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-8$. To find $a$ and $b$, set up a system to be solved.
$$a+b=15$$ $$ab=2\left(-8\right)=-16$$
Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-16$.
$$-1,16$$ $$-2,8$$ $$-4,4$$
Calculate the sum for each pair.
$$-1+16=15$$ $$-2+8=6$$ $$-4+4=0$$
The solution is the pair that gives sum $15$.
$$a=-1$$ $$b=16$$
Rewrite $2x^{2}+15x-8$ as $\left(2x^{2}-x\right)+\left(16x-8\right)$.
$$\left(2x^{2}-x\right)+\left(16x-8\right)$$
Factor out $x$ in the first and $8$ in the second group.
$$x\left(2x-1\right)+8\left(2x-1\right)$$
Factor out common term $2x-1$ by using distributive property.
$$\left(2x-1\right)\left(x+8\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}+15x-8=0$$
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 $x=\frac{-15±17}{4}$ when $±$ is plus. Add $-15$ to $17$.
$$x=\frac{2}{4}$$
Reduce the fraction $\frac{2}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{1}{2}$$
Now solve the equation $x=\frac{-15±17}{4}$ when $±$ is minus. Subtract $17$ from $-15$.
$$x=-\frac{32}{4}$$
Divide $-32$ by $4$.
$$x=-8$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{1}{2}$ for $x_{1}$ and $-8$ for $x_{2}$.
