# What is the standard form of y= (3x – 4) (x^2+16) ?

Jun 25, 2018

See a solution process below:

#### Explanation:

To put this equation in standard form we must multiply the two terms on the right side of the equation by multiplying each individual term in the left parenthesis by each individual term in the right parenthesis.

$y = \left(\textcolor{red}{3 x} - \textcolor{red}{4}\right) \left(\textcolor{b l u e}{{x}^{2}} + \textcolor{b l u e}{16}\right)$ becomes:

$y = \left(\textcolor{red}{3 x} \times \textcolor{b l u e}{{x}^{2}}\right) + \left(\textcolor{red}{3 x} \times \textcolor{b l u e}{16}\right) - \left(\textcolor{red}{4} \times \textcolor{b l u e}{{x}^{2}}\right) - \left(\textcolor{red}{4} \times \textcolor{b l u e}{16}\right)$

$y 3 {x}^{3} + 48 x - 4 {x}^{2} - 64$

Now, put the terms in order by largest exponent on the left to the lowest exponent or constant on the right:

$y 3 {x}^{3} - 4 {x}^{2} + 48 x - 64$#