# How do you write y = −3(x−7)(x+4) in standard form?

Apr 23, 2017

See the entire solution process below:

#### Explanation:

To write this equation in standard form we must first multiply the two terms in parenthesis with each other. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

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

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

$y = - 3 \left({x}^{2} + 4 x - 7 x - 28\right)$

We can then combine like terms within the parenthesis:

$y = - 3 \left({x}^{2} + \left(4 - 7\right) x - 28\right)$

$y = - 3 \left({x}^{2} + \left(- 3\right) x - 28\right)$

$y = - 3 \left({x}^{2} - 3 x - 28\right)$

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

$y = \left(- 3 \cdot {x}^{2}\right) + \left(- 3 \cdot - 3 x\right) + \left(- 3 \cdot - 28\right)$

$y = - 3 {x}^{2} + 9 x + 84$