How do you find the product of (8x + 7)(x + 2) ?

Sep 15, 2017

$8 {x}^{2} + 23 x + 14$

Explanation:

$\left(8 x + 7\right) \left(x + 2\right)$
First, we will distribute parentheses/brackets
from $\left(8 x + 7\right) \left(x + 2\right)$ to
$8 x \cdot x + 8 x \cdot 2 + 7 \cdot x + 7 \cdot 2$
Example:
$\left(w + x\right) \left(y + z\right) = w y + w z + x y + x z$
In this case, $w$ would be $8 x$, $x$ would be $7$, $y$ would be $x$ and $z$ would be $2$.

$8 x \cdot x + 8 x \cdot 2 + 7 \cdot x + 7 \cdot 2$ is the same as
$8 x x + 8 x \cdot 2 + 7 x + 7 \cdot 2$
Now we will do:
$8 {x}^{2} + 8 x \cdot 2 + 7 x + 7 \cdot 2$
Why did $8 x x$ change to $8 {x}^{2}$?
It's because of this:
$a a = {a}^{1 + 1} = {a}^{2}$

So, now you have $8 {x}^{2} + 8 x \cdot 2 + 7 x + 7 \cdot 2$
You want to solve bit by bit. First we'll start with $8 x \cdot 2$. $8 \cdot 2 = 16$ So we can replace $8 x \cdot 2$ with $16 x$.
$8 {x}^{2} + 16 x + 7 x + 7 \cdot 2$
$16 x + 7 x = 23 x$
$8 {x}^{2} + 23 x + 7 \cdot 2$
$7 \cdot 2 = 14$

$8 {x}^{2} + 23 x + 14$