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.