# How do you simplify the product (6x - 5)(3x + 1) and write it in standard form?

Jun 3, 2018

$18 {x}^{2} - 9 x - 5$

#### Explanation:

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

$= \left(\textcolor{red}{6 x - 5}\right) \left(3 x + 1\right)$

$= \textcolor{red}{6 x} \left(3 x + 1\right) \textcolor{red}{- 5} \left(3 x + 1\right) \leftarrow \textcolor{b l u e}{\text{distribute}}$

$= \left(\textcolor{red}{6 x} \times 3 x\right) + \left(\textcolor{red}{6 x} \times 1\right) + \left(\textcolor{red}{- 5} \times 3 x\right) + \left(\textcolor{red}{- 5} \times 1\right)$

$= 18 {x}^{2} \textcolor{b l u e}{+ 6 x} \textcolor{b l u e}{- 15 x} - 5 \leftarrow \textcolor{b l u e}{\text{collect like terms}}$

$= 18 {x}^{2} - 9 x - 5 \leftarrow \textcolor{m a \ge n t a}{\text{in standard form}}$

Try using the FOIL method (First, Outside, Inside, Last)
$18 {x}^{2} - 9 x - 5$

#### Explanation:

You have to multiply each term in the left bracket by each term in the right bracket. The easiest way to make sure you don't miss anything is to use a systematic method such as FOIL.
The FOIL method means selecting those pairs of numbers to multiply together
Eg. $\left(a + b\right) \left(c + d\right) = a c + a d + b c + b d$

Question $\left(6 x - 5\right) \left(3 x + 1\right)$
First: $6 x \cdot 3 x = 18 {x}^{2}$
Outside: $6 x \cdot 1 = 6 x$
Inside:$- 5 \cdot 3 x = - 15 x$
Last: $- 5 \cdot 1 = - 5$
$\left(6 x - 5\right) \left(3 x + 1\right) = 18 {x}^{2} + 6 x + \left(- 15 x\right) + \left(- 5\right)$
$6 x$ and $- 15 x$ are like terms so they simplify
$\left(6 x - 5\right) \left(3 x + 1\right) = 18 {x}^{2} - 9 x - 5$