Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+10$. To find $a$ and $b$, set up a system to be solved.
$$a+b=7$$ $$ab=1\times 10=10$$
Since $ab$ is positive, $a$ and $b$ have the same sign. Since $a+b$ is positive, $a$ and $b$ are both positive. List all such integer pairs that give product $10$.
$$1,10$$ $$2,5$$
Calculate the sum for each pair.
$$1+10=11$$ $$2+5=7$$
The solution is the pair that gives sum $7$.
$$a=2$$ $$b=5$$
Rewrite $x^{2}+7x+10$ as $\left(x^{2}+2x\right)+\left(5x+10\right)$.
$$\left(x^{2}+2x\right)+\left(5x+10\right)$$
Factor out $x$ in the first and $5$ in the second group.
$$x\left(x+2\right)+5\left(x+2\right)$$
Factor out common term $x+2$ by using distributive property.
$$\left(x+2\right)\left(x+5\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+7x+10=0$$
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.
$$x=\frac{-7±\sqrt{7^{2}-4\times 10}}{2}$$
Square $7$.
$$x=\frac{-7±\sqrt{49-4\times 10}}{2}$$
Multiply $-4$ times $10$.
$$x=\frac{-7±\sqrt{49-40}}{2}$$
Add $49$ to $-40$.
$$x=\frac{-7±\sqrt{9}}{2}$$
Take the square root of $9$.
$$x=\frac{-7±3}{2}$$
Now solve the equation $x=\frac{-7±3}{2}$ when $±$ is plus. Add $-7$ to $3$.
$$x=-\frac{4}{2}$$
Divide $-4$ by $2$.
$$x=-2$$
Now solve the equation $x=\frac{-7±3}{2}$ when $±$ is minus. Subtract $3$ from $-7$.
$$x=-\frac{10}{2}$$
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 $-2$ for $x_{1}$ and $-5$ for $x_{2}$.
