Let's start by factoring out the greatest common factor. Then, we will factor the resulting trinomial.
To factor a quadratic expression with a leading coefficient of $1,$ we first need to identify the values of $b$ and $c.$ $\begin{aligned} \textbf{General Expression:}&\ x^2+\textcolor{#ff00ff}{b}x+\textcolor{#ff8c00}{c} \\ \textbf{Our Expression:}&\ x^2+\textcolor{#ff00ff}{10}x+\textcolor{#ff8c00}{16} \end{aligned}$ Next, we have to find a factor pair of $\textcolor{#ff8c00}{c}$ $=$ $\textcolor{#ff8c00}{16}$ whose sum is $\textcolor{#ff00ff}{b}$ $=$ $\textcolor{#ff00ff}{10}.$ Note that $16$ is a positive number, so for the product of the factors to be positive, they must have the same sign — both positive or both negative.
Factor Pair | Product of Factors | Sum of Factors |
---|---|---|
$1$ and $16$ | $\textcolor{#ff8c00}{16}$ | $17$ |
$\text{-}1$ and $\text{-}16$ | $\textcolor{#ff8c00}{16}$ | $\text{-} 17$ |
$2$ and $8$ | $\textcolor{#ff8c00}{16}$ | $\textcolor{#ff00ff}{10}$ |
$\text{-}2$ and $\text{-}8$ | $\textcolor{#ff8c00}{16}$ | $\text{-} 10$ |
$4$ and $4$ | $\textcolor{#ff8c00}{16}$ | $8$ |
$\text{-}4$ and $\text{-}4$ | $\textcolor{#ff8c00}{16}$ | $\text{-} 8$ |
The factors whose product is $\textcolor{#ff8c00}{16}$ and whose sum is $\textcolor{#ff00ff}{10}$ are ${\color{#0000FF}{2}}$ and $\textcolor{deepskyblue}{8}.$ With this information, we can now factor the trinomial. $\begin{gathered} x^2+\textcolor{#ff00ff}{10}x+\textcolor{#ff8c00}{16}\quad\Leftrightarrow\quad (x+{\color{#0000FF}{2}})(x+\textcolor{deepskyblue}{8}) \end{gathered}$ Before we finish, remember that we factored out the greatest common factor from the original expression. Therefore, we need to include it again. $\begin{gathered} x^3+10x^2+16x\quad \Leftrightarrow \quad x(x+2)(x+8) \end{gathered}$