# What is the standard form of y= ( x^4-1) (6-x)?

Feb 16, 2017

In standard form $y = - {x}^{5} + 6 {x}^{4} + x - 6$

#### Explanation:

$y = \left({x}^{4} - 1\right) \left(6 - x\right)$

= ${x}^{4} \left(6 - x\right) - 1 \left(6 - x\right)$

= $6 {x}^{4} - {x}^{5} - 6 + x$

and hence in standard form $y = - {x}^{5} + 6 {x}^{4} + x - 6$