Show that if the polynomial #f(x)=ax^3+3bx^2+3cx+d# is divided exactly by #g(x)=ax^2+2bx+c#, then #f(x)# is a perfect cube, while #g(x)# is a perfect square?

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Answer 1 · 2018-02-02T10:03:54

See below.

Given #f(x)# and #g(x)# as

#f(x)=ax^3+3bx^2+3cx+d#
#g(x)=ax^2+2bx+c#

and such that #g(x)# divides #f(x)# then

#f(x) = (x+e)g(x)#
Now grouping coeficients

#{(d-c e=0),(c-b e = 0),(b-a e=0):}#

solving for #a,b,c# we obtain the condition

#{(a=d/e^3),(b=d/e^2),(c=d/e):}#

and substituting into #f(x)# and #g(x)#

#f(x) = (d(x+e)^3)/e^3 = (root(3)(d)(x+e)/e)^3#

#g(x) = (d(x+e)^2)/e^3 = (sqrt(d/e)(x+e)/e)^2#