# What is the standard form of  y=(x + 6)(x + 2) ?

Mar 12, 2018

The polynomial in standard form is $y = {x}^{2} + 8 x + 12$.

#### Explanation:

Use the distributive property to expand the multiplication (I color-coded each part so it's easier to follow):

$\textcolor{w h i t e}{=} \left(\textcolor{red}{x} + \textcolor{b l u e}{6}\right) \left(\textcolor{g r e e n}{x} + \textcolor{p u r p \le}{2}\right)$

$= \textcolor{red}{x} \cdot \textcolor{g r e e n}{x} + \textcolor{red}{x} \cdot \textcolor{p u r p \le}{2} + \textcolor{b l u e}{6} \cdot \textcolor{g r e e n}{x} + \textcolor{b l u e}{6} \cdot \textcolor{p u r p \le}{2}$

$= {x}^{2} + \textcolor{red}{x} \cdot \textcolor{p u r p \le}{2} + \textcolor{b l u e}{6} \cdot \textcolor{g r e e n}{x} + \textcolor{b l u e}{6} \cdot \textcolor{p u r p \le}{2}$

$= {x}^{2} + 2 x + \textcolor{b l u e}{6} \cdot \textcolor{g r e e n}{x} + \textcolor{b l u e}{6} \cdot \textcolor{p u r p \le}{2}$

$= {x}^{2} + 2 x + 6 x + \textcolor{b l u e}{6} \cdot \textcolor{p u r p \le}{2}$

$= {x}^{2} + 8 x + \textcolor{b l u e}{6} \cdot \textcolor{p u r p \le}{2}$

$= {x}^{2} + 8 x + 12$
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