# How do you find the product (b+7)(b-7)?

Jul 29, 2017

See a solution process below:

#### Explanation:

Process 1
This problem has a special rule you can follow:

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

Substituting gives:

$\left(\textcolor{red}{b} + \textcolor{b l u e}{7}\right) \left(\textcolor{red}{b} - \textcolor{b l u e}{7}\right) = {\textcolor{red}{b}}^{2} - {\textcolor{b l u e}{7}}^{2} = {b}^{2} - 49$

Process 2
We can also use this process to hopefully get to the same answer.

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

$\left(\textcolor{red}{b} + \textcolor{red}{7}\right) \left(\textcolor{b l u e}{b} - \textcolor{b l u e}{7}\right)$ becomes:

$\left(\textcolor{red}{b} \times \textcolor{b l u e}{b}\right) - \left(\textcolor{red}{b} \times \textcolor{b l u e}{7}\right) + \left(\textcolor{red}{7} \times \textcolor{b l u e}{b}\right) - \left(\textcolor{red}{7} \times \textcolor{b l u e}{7}\right)$

${b}^{2} - 7 b + 7 b - 49$

We can now combine like terms:

${b}^{2} + \left(- 7 + 7\right) b - 49$

${b}^{2} + 0 b - 49$

${b}^{2} - 49$