Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx-35$. To find $a$ and $b$, set up a system to be solved.
$$a+b=2$$ $$ab=1\left(-35\right)=-35$$
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 $-35$.
$$-1,35$$ $$-5,7$$
Calculate the sum for each pair.
$$-1+35=34$$ $$-5+7=2$$
The solution is the pair that gives sum $2$.
$$a=-5$$ $$b=7$$
Rewrite $x^{2}+2x-35$ as $\left(x^{2}-5x\right)+\left(7x-35\right)$.
$$\left(x^{2}-5x\right)+\left(7x-35\right)$$
Factor out $x$ in the first and $7$ in the second group.
$$x\left(x-5\right)+7\left(x-5\right)$$
Factor out common term $x-5$ by using distributive property.
$$\left(x-5\right)\left(x+7\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+2x-35=0$$
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.
$$x=\frac{-2±\sqrt{2^{2}-4\left(-35\right)}}{2}$$
Square $2$.
$$x=\frac{-2±\sqrt{4-4\left(-35\right)}}{2}$$
Multiply $-4$ times $-35$.
$$x=\frac{-2±\sqrt{4+140}}{2}$$
Add $4$ to $140$.
$$x=\frac{-2±\sqrt{144}}{2}$$
Take the square root of $144$.
$$x=\frac{-2±12}{2}$$
Now solve the equation $x=\frac{-2±12}{2}$ when $±$ is plus. Add $-2$ to $12$.
$$x=\frac{10}{2}$$
Divide $10$ by $2$.
$$x=5$$
Now solve the equation $x=\frac{-2±12}{2}$ when $±$ is minus. Subtract $12$ from $-2$.
$$x=-\frac{14}{2}$$
Divide $-14$ by $2$.
$$x=-7$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $5$ for $x_{1}$ and $-7$ for $x_{2}$.
