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

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Answer 1 · 2017-02-16T16:05:59

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

$y=(x^4-1)(6-x)$

= $x^4(6-x)-1(6-x)$

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

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

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