# How do you foil 2 (x+2) (x+5)?

Aug 10, 2015

FOIL can only be applied to the 2 binomials; the result of that multiplication can then be multiplied by $2$ to get:
$\textcolor{w h i t e}{\text{XXXX}}$$2 {x}^{2} + 14 x + 20$

#### Explanation:

Evaluating the product of two binomials using FOIL:

Sum
$\textcolor{w h i t e}{\text{XXXX}}$product of First terms
$\textcolor{w h i t e}{\text{XXXX}}$product of Outside terms
$\textcolor{w h i t e}{\text{XXXX}}$product of Inside terms
$\textcolor{w h i t e}{\text{XXXX}}$product of Last terms

For the given example: $2 \left(x + 2\right) \left(x + 5\right)$:

We can apply FOIL to $\left(x + 2\right) \left(x + 5\right)$ as
$\textcolor{w h i t e}{\text{XXXX}}$First terms: $x , x$$\textcolor{w h i t e}{\text{XXXX}}$Product $= {x}^{2}$
$\textcolor{w h i t e}{\text{XXXX}}$Outside terms: $x , + 5$$\textcolor{w h i t e}{\text{XXXX}}$Product $= + 5 x$
$\textcolor{w h i t e}{\text{XXXX}}$Inside terms: $+ 2 , x$$\textcolor{w h i t e}{\text{XXXX}}$Product $= + 2 x$
$\textcolor{w h i t e}{\text{XXXX}}$Last terms: $+ 2 , + 5$$\textcolor{w h i t e}{\text{XXXX}}$Product $= + 10$

Sum
$\textcolor{w h i t e}{\text{XXXX}}$$= {x}^{2} + 5 x + 2 x + 10$

$\textcolor{w h i t e}{\text{XXXX}}$$= {x}^{2} + 7 x + 10$

So
$\textcolor{w h i t e}{\text{XXXX}}$$\left(x + 2\right) \left(x + 5\right) = {x}^{2} + 7 x + 10$

and
$\textcolor{w h i t e}{\text{XXXX}}$$2 \left(x + 2\right) \left(x + 7\right) = 2 {x}^{2} + 14 x + 20$