# How do you multiply b(15 + -3b)(15 + 5b)?

Apr 21, 2017

See the entire solution process below:

#### Explanation:

First, rewrite this expression as:

$b \left(15 - 3 b\right) \left(15 + 5 b\right)$

There are a couple of ways to go next. I am first going to multiply the $b$ term on the left of the expression by the first term in parenthesis by the left. We need to multiply the term outside the parenthesis by each term within the parenthesis:

$\textcolor{red}{b} \left(15 - 3 b\right) \left(15 + 5 b\right)$

$\left(\left(15 \times \textcolor{red}{b}\right) - \left(3 b \times \textcolor{red}{b}\right)\right) \left(15 + 5 b\right)$

$\left(15 b - 3 {b}^{2}\right) \left(15 + 5 b\right)$

Next, we need to multiply both terms in parenthesis. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$\left(\textcolor{red}{15 b} - \textcolor{red}{3 {b}^{2}}\right) \left(\textcolor{b l u e}{15} + \textcolor{b l u e}{5 b}\right)$ becomes:

$\left(\textcolor{red}{15 b} \times \textcolor{b l u e}{15}\right) + \left(\textcolor{red}{15 b} \times \textcolor{b l u e}{5 b}\right) - \left(\textcolor{red}{3 {b}^{2}} \times \textcolor{b l u e}{15}\right) - \left(\textcolor{red}{3 {b}^{2}} \times \textcolor{b l u e}{5 b}\right)$

$225 b + 75 {b}^{2} - 45 {b}^{2} - 15 {b}^{3}$

We can now group and combine like terms:

$- 15 {b}^{3} + 75 {b}^{2} - 45 {b}^{2} + 225 b$

$- 15 {b}^{3} + \left(75 - 45\right) {b}^{2} + 225 b$

$- 15 {b}^{3} + 30 {b}^{2} + 225 b$