Factor the expression by grouping. First, the expression needs to be rewritten as $2x^{2}+ax+bx-5$. To find $a$ and $b$, set up a system to be solved.
$$a+b=-3$$ $$ab=2\left(-5\right)=-10$$
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 $-10$.
$$1,-10$$ $$2,-5$$
Calculate the sum for each pair.
$$1-10=-9$$ $$2-5=-3$$
The solution is the pair that gives sum $-3$.
$$a=-5$$ $$b=2$$
Rewrite $2x^{2}-3x-5$ as $\left(2x^{2}-5x\right)+\left(2x-5\right)$.
$$\left(2x^{2}-5x\right)+\left(2x-5\right)$$
Factor out $x$ in $2x^{2}-5x$.
$$x\left(2x-5\right)+2x-5$$
Factor out common term $2x-5$ by using distributive property.
$$\left(2x-5\right)\left(x+1\right)$$
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 $ax^{2}+bx+c=0$.
$$2x^{2}-3x-5=0$$
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 $x=\frac{3±7}{4}$ when $±$ is plus. Add $3$ to $7$.
$$x=\frac{10}{4}$$
Reduce the fraction $\frac{10}{4}$ to lowest terms by extracting and canceling out $2$.
$$x=\frac{5}{2}$$
Now solve the equation $x=\frac{3±7}{4}$ when $±$ is minus. Subtract $7$ from $3$.
$$x=-\frac{4}{4}$$
Divide $-4$ by $4$.
$$x=-1$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $\frac{5}{2}$ for $x_{1}$ and $-1$ for $x_{2}$.
