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

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Answer 1 · 2016-06-30T12:18:36

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

In $y=x^2(x-9)(6-x)$ , 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 $ax^4+bx^3+cx^2+dx+f$ , for which we should expand $x^2(x-9)(6-x)$ by multiplying.

$x^2(x-9)(6-x)$

= $x^2(x(6-x)-9(6-x))$

= $x^2(6x-x^2-54+9x)$

= $x^2(-x^2+15x-54)$

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

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

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