# What is the standard form of y= x^2(x-9) (6-x)?

Jun 30, 2016

$y = - {x}^{4} + 15 {x}^{3} - 54 {x}^{2}$

#### Explanation:

In $y = {x}^{2} \left(x - 9\right) \left(6 - x\right)$, the RHS is a polynomial of degree $4$ in $x$, as $x$ gets multiplied four times.

The standard form of a polynomial in degree $4$ is $a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + f$, for which we should expand ${x}^{2} \left(x - 9\right) \left(6 - x\right)$ by multiplying.

${x}^{2} \left(x - 9\right) \left(6 - x\right)$

= ${x}^{2} \left(x \left(6 - x\right) - 9 \left(6 - x\right)\right)$

= ${x}^{2} \left(6 x - {x}^{2} - 54 + 9 x\right)$

= ${x}^{2} \left(- {x}^{2} + 15 x - 54\right)$

= $- {x}^{4} + 15 {x}^{3} - 54 {x}^{2}$

Note that here coefficient of $x$ and constant terms are both zero in this case.