# How do you simplify (3+ 2c ) ( 7- 4c )?

Jul 29, 2017

See a solution process below:

#### Explanation:

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

$\left(\textcolor{red}{3} + \textcolor{red}{2 c}\right) \left(\textcolor{b l u e}{7} - \textcolor{b l u e}{4 c}\right)$ becomes:

$\left(\textcolor{red}{3} \times \textcolor{b l u e}{7}\right) - \left(\textcolor{red}{3} \times \textcolor{b l u e}{4 c}\right) + \left(\textcolor{red}{2 c} \times \textcolor{b l u e}{7}\right) - \left(\textcolor{red}{2 c} \times \textcolor{b l u e}{4 c}\right)$

$21 - 12 c + 14 c - 8 {c}^{2}$

We can now combine like terms:

$21 + \left(- 12 + 14\right) c - 8 {c}^{2}$

$21 + 2 c - 8 {c}^{2}$

Or in standard form:

$- 8 {c}^{2} + 2 c + 21$

Jul 29, 2017

$21 + 2 c - 8 {c}^{2}$

#### Explanation:

$\text{simplify means to multiply the factors together}$

$\text{each term in the second factor is multiplied by each term}$
$\text{in the first factor}$

$\Rightarrow \left(\textcolor{red}{3 + 2 c}\right) \left(7 - 4 c\right)$

$= \textcolor{red}{3} \left(7 - 4 c\right) \textcolor{red}{+ 2 c} \left(7 - 4 c\right)$

$\text{distributing the brackets gives}$

$= \left(\textcolor{red}{3} \times 7\right) + \left(\textcolor{red}{3} \times - 4 c\right) + \left(\textcolor{red}{2 c} \times 7\right) + \left(\textcolor{red}{2 c} \times - 4 c\right)$

$= 21 + \left(- 12 c\right) + 14 c + \left(- 8 {c}^{2}\right)$

$= 21 + 2 c - 8 {c}^{2}$