# What is the standard form of y=(x+4)(2x-3) -3x^2+6x?

Apr 29, 2017

See the entire solution process below:

#### Explanation:

First, multiply the two 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.

$y = \left(\textcolor{red}{x} + \textcolor{red}{4}\right) \left(\textcolor{b l u e}{2 x} - \textcolor{b l u e}{3}\right) - 3 {x}^{2} + 6 x$ becomes:

$y = \left(\textcolor{red}{x} \times \textcolor{b l u e}{2 x}\right) - \left(\textcolor{red}{x} \times \textcolor{b l u e}{3}\right) + \left(\textcolor{red}{4} \times \textcolor{b l u e}{2 x}\right) - \left(\textcolor{red}{4} \times \textcolor{b l u e}{3}\right) - 3 {x}^{2} + 6 x$

$y = 2 {x}^{2} - 3 x + 8 x - 12 - 3 {x}^{2} + 6 x$

We can now group and combine like terms:

$y = 2 {x}^{2} - 3 {x}^{2} - 3 x + 8 x + 6 x - 12$

$y = \left(2 - 3\right) {x}^{2} + \left(- 3 + 8 + 6\right) x - 12$

$y = - 1 {x}^{2} + 11 x - 12$

$y = - {x}^{2} + 11 x - 12$