# What is the standard form of y= (2/5x^2-1/12)(1/3x+5/8) ?

Feb 16, 2016

$y = 2 {x}^{3} / 15 + {x}^{2} / 4 - \frac{x}{36} - \frac{5}{96}$

#### Explanation:

use the distribution property of multiplication over addition

$y = \frac{2}{5} {x}^{2} \cdot \left(\frac{1}{3} x + \frac{5}{8}\right) - \frac{1}{12} \cdot \left(\frac{1}{3} x + \frac{5}{8}\right)$
$y = 2 {x}^{3} / 15 + 10 {x}^{2} / 40 - \frac{x}{36} - \frac{5}{96}$

simplify some of the fractions to get

$y = 2 {x}^{3} / 15 + {x}^{2} / 4 - \frac{x}{36} - \frac{5}{96}$

hope it helps.. feel free to ask questions if you have any

Feb 16, 2016

$\left(\frac{2}{15}\right) {x}^{3} + \left(\frac{1}{4}\right) {x}^{2} - \left(\frac{1}{36}\right) x - \frac{5}{96}$

#### Explanation:

As y=(2/5x^2−1/12)(1/3x+5/8) is multiplication of one quadratic expression and one linear expression and hence of the form $a {x}^{3} + b {x}^{2} + c x + d$.

So, multiplying y=(2/5x^2−1/12)(1/3x+5/8) i.e.

$\left(\frac{2}{5} \cdot \frac{1}{3}\right) {x}^{3} + \left(\frac{2}{5} \cdot \frac{5}{8}\right) {x}^{2} - \left(\frac{1}{12} \cdot \frac{1}{3}\right) x - \left(\frac{1}{12} \cdot \frac{5}{8}\right)$

= $\left(\frac{2}{15}\right) {x}^{3} + \left(\frac{1}{4}\right) {x}^{2} - \left(\frac{1}{36}\right) x - \frac{5}{96}$