Factor the expression by grouping. First, the expression needs to be rewritten as $x^{2}+ax+bx+24$. To find $a$ and $b$, set up a system to be solved.
$$a+b=10$$ $$ab=1\times 24=24$$
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 $24$.
$$1,24$$ $$2,12$$ $$3,8$$ $$4,6$$
Calculate the sum for each pair.
$$1+24=25$$ $$2+12=14$$ $$3+8=11$$ $$4+6=10$$
The solution is the pair that gives sum $10$.
$$a=4$$ $$b=6$$
Rewrite $x^{2}+10x+24$ as $\left(x^{2}+4x\right)+\left(6x+24\right)$.
$$\left(x^{2}+4x\right)+\left(6x+24\right)$$
Factor out $x$ in the first and $6$ in the second group.
$$x\left(x+4\right)+6\left(x+4\right)$$
Factor out common term $x+4$ by using distributive property.
$$\left(x+4\right)\left(x+6\right)$$
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 $ax^{2}+bx+c=0$.
$$x^{2}+10x+24=0$$
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.
$$x=\frac{-10±\sqrt{10^{2}-4\times 24}}{2}$$
Square $10$.
$$x=\frac{-10±\sqrt{100-4\times 24}}{2}$$
Multiply $-4$ times $24$.
$$x=\frac{-10±\sqrt{100-96}}{2}$$
Add $100$ to $-96$.
$$x=\frac{-10±\sqrt{4}}{2}$$
Take the square root of $4$.
$$x=\frac{-10±2}{2}$$
Now solve the equation $x=\frac{-10±2}{2}$ when $±$ is plus. Add $-10$ to $2$.
$$x=-\frac{8}{2}$$
Divide $-8$ by $2$.
$$x=-4$$
Now solve the equation $x=\frac{-10±2}{2}$ when $±$ is minus. Subtract $2$ from $-10$.
$$x=-\frac{12}{2}$$
Divide $-12$ by $2$.
$$x=-6$$
Factor the original expression using $ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right)$. Substitute $-4$ for $x_{1}$ and $-6$ for $x_{2}$.
