# How do you find the product of  (2x +1)(x + 3)?

May 23, 2016

$\left(2 x + 1\right) \left(x + 3\right) = \textcolor{b l u e}{2 {x}^{2} + 7 x + 3}$

#### Explanation:

Method 1: Distribution
$\left(\textcolor{red}{2 x + 1}\right) \left(\textcolor{g r e e n}{x + 3}\right)$
$\textcolor{w h i t e}{\text{XXX}} = \textcolor{red}{2 x} \left(\textcolor{g r e e n}{x + 3}\right) \textcolor{red}{+ 1} \left(\textcolor{g r e e n}{x + 3}\right)$

$\textcolor{w h i t e}{\text{XXX}} = \left(\textcolor{t e a l}{2 x \cdot x + 2 x \cdot 3}\right) + \left(\textcolor{b r o w n}{1 \cdot x + 1 \cdot 3}\right)$

$\textcolor{w h i t e}{\text{XXX}} = \textcolor{t e a l}{2 {x}^{2} + 6 x} + \textcolor{b r o w n}{x + 3}$

$\textcolor{w h i t e}{\text{XXX}} = \textcolor{b l u e}{2 {x}^{2} + 7 x + 3}$

Method 2: FOIL
FOIL: First-Outside-Inside-Last is often taught as a method for multiplying two binomials.
Given
$\textcolor{w h i t e}{\text{XXX}} \left(2 x + 1\right) \left(x + 3\right)$
the First terms are $2 x$ and $x$
$\textcolor{w h i t e}{\text{XXX}}$Multiplying the First terms gives: $2 {x}^{2}$
the Outside terms are $2 x$ and $3$
$\textcolor{w h i t e}{\text{XXX}}$Multiplying the Outside terms gives: $6 x$
the Inside terms are $1$ and $x$
$\textcolor{w h i t e}{\text{XXX}}$Multiplying the Inside terms gives: $1 x$
the Last terms are $1$ and $3$
$\textcolor{w h i t e}{\text{XXX}}$Multiplying the Last Terms gives: $3$

$\textcolor{w h i t e}{\text{XXX}} 2 {x}^{2} + 6 x + 1 x + 3$
$\textcolor{w h i t e}{\text{XXXXXXXXX}} = 2 {x}^{2} + 7 x + 3$