# How do you multiply (3x- 1)^2?

Jul 24, 2017

See a solution process below:

#### Explanation:

Use this rule to multiply this expression:

${\left(\textcolor{red}{a} - \textcolor{b l u e}{b}\right)}^{2} = {\textcolor{red}{a}}^{2} - 2 \textcolor{red}{a} \textcolor{b l u e}{b} + {\textcolor{b l u e}{b}}^{2}$

Substituting $3 x$ for $a$ and $1$ for $b$ gives:

${\left(\textcolor{red}{\left(3 x\right)} - \textcolor{b l u e}{1}\right)}^{2} \implies {\textcolor{red}{\left(3 x\right)}}^{2} - \left(2 \cdot \textcolor{red}{3 x} \cdot \textcolor{b l u e}{1}\right) + {\textcolor{b l u e}{1}}^{2} \implies$

$9 {x}^{2} - 6 x + 1$

Another method is to first rewrite the expression as:

$\left(3 x - 1\right) \left(3 x - 1\right)$

To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{3 x} - \textcolor{red}{1}\right) \left(\textcolor{b l u e}{3 x} - \textcolor{b l u e}{1}\right)$ becomes:

$\left(\textcolor{red}{3 x} \times \textcolor{b l u e}{3 x}\right) - \left(\textcolor{red}{3 x} \times \textcolor{b l u e}{1}\right) - \left(\textcolor{red}{1} \times \textcolor{b l u e}{3 x}\right) + \left(\textcolor{red}{1} \times \textcolor{b l u e}{1}\right)$

$9 {x}^{2} - 3 x - 3 x + 1$

We can now combine like terms:

$9 {x}^{2} + \left(- 3 - 3\right) x + 1$

$9 {x}^{2} + \left(- 6\right) x + 1$

$9 {x}^{2} - 6 x + 1$