# How do you write  y=-3(x-3)(x+2) in standard form?

Jul 25, 2017

See a solution process below:

#### Explanation:

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

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

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

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

We can next combine like terms within the parenthesis:

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

$y = - 3 \left({x}^{2} + \left[- 1\right] x - 6\right)$

$y = - 3 \left({x}^{2} - 1 x - 6\right)$

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

Now, we can multiply each term within the parenthesis by the term outside the parenthesis:

$y = \textcolor{red}{- 3} \left({x}^{2} - x - 6\right)$

$y = \left(\textcolor{red}{- 3} \cdot {x}^{2}\right) - \left(\textcolor{red}{- 3} \cdot x\right) - \left(\textcolor{red}{- 3} \cdot 6\right)$

$y = - 3 {x}^{2} - \left(- 3 x\right) - \left(- 18\right)$

$y = - 3 {x}^{2} + 3 x + 18$