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 negative, the negative number has greater absolute value than the positive. 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 $-2$.
$$a=-5$$ $$b=3$$
Rewrite $x^{2}-2x-15$ as $\left(x^{2}-5x\right)+\left(3x-15\right)$.
$$\left(x^{2}-5x\right)+\left(3x-15\right)$$
Factor out $x$ in the first and $3$ in the second group.
$$x\left(x-5\right)+3\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(x+3\right)$$
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.
