# How do you write y=5/2(x-6)(x-2) in standard form?

Jul 26, 2017

See a solution process below:

#### Explanation:

To begin writing this in standard form, first, multiply the two terms within parenthesis. To multiply these two terms you multiply each individual term in the left parenthesis by each individual term in the right parenthesis.

$y = \frac{5}{2} \left(\textcolor{red}{x} - \textcolor{red}{6}\right) \left(\textcolor{b l u e}{x} - \textcolor{b l u e}{2}\right)$ becomes:

$y = \frac{5}{2} \left[\left(\textcolor{red}{x} \times \textcolor{b l u e}{x}\right) - \left(\textcolor{red}{x} \times \textcolor{b l u e}{2}\right) - \left(\textcolor{red}{6} \times \textcolor{b l u e}{x}\right) + \left(\textcolor{red}{6} \times \textcolor{b l u e}{2}\right)\right]$

$y = \frac{5}{2} \left({x}^{2} - 2 x - 6 x + 12\right)$

We can next combine like terms:

$y = \frac{5}{2} \left({x}^{2} + \left(- 2 - 6\right) x + 12\right)$

$y = \frac{5}{2} \left({x}^{2} + \left(- 8\right) x + 12\right)$

$y = \frac{5}{2} \left({x}^{2} - 8 x + 12\right)$

Now, multiply each term within the parenthesis by the term outside of the parenthesis:

$y = \textcolor{red}{\frac{5}{2}} \left({x}^{2} - 8 x + 12\right)$

$y = \left(\textcolor{red}{\frac{5}{2}} \times {x}^{2}\right) - \left(\textcolor{red}{\frac{5}{2}} \times 8 x\right) + \left(\textcolor{red}{\frac{5}{2}} \times 12\right)$

$y = \frac{5}{2} {x}^{2} - \left(5 \times 4 x\right) + \left(5 \times 6\right)$

$y = \frac{5}{2} {x}^{2} - 20 x + 30$