Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-15$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=1\left(-15\right)=-15$$
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 $-15$.
$$-1,15$$ $$-3,5$$
Calculate the sum for each pair.
$$-1+15=14$$ $$-3+5=2$$
The solution is the pair that gives sum $2$.
$$a=-3$$ $$b=5$$
Rewrite $x^{2}+2x-15$ as $\left(x^{2}-3x\right)+\left(5x-15\right)$.
$$\left(x^{2}-3x\right)+\left(5x-15\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(x-3\right)+5\left(x-3\right)$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+2x-15=0$$
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.
$$x=\frac{-2±\sqrt{2^{2}-4\left(-15\right)}}{2}$$
Square $2$.
$$x=\frac{-2±\sqrt{4-4\left(-15\right)}}{2}$$
Multiply $-4$ times $-15$.
$$x=\frac{-2±\sqrt{4+60}}{2}$$
Add $4$ to $60$.
$$x=\frac{-2±\sqrt{64}}{2}$$
Take the square root of $64$.
$$x=\frac{-2±8}{2}$$
Now solve the equation $x=\frac{-2±8}{2}$ when $±$ is plus. Add $-2$ to $8$.
$$x=\frac{6}{2}$$
Divide $6$ by $2$.
$$x=3$$
Now solve the equation $x=\frac{-2±8}{2}$ when $±$ is minus. Subtract $8$ from $-2$.
$$x=-\frac{10}{2}$$
Divide $-10$ by $2$.
$$x=-5$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $3$ for $x_{1}$ and $-5$ for $x_{2}$.
