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?

1 Answer
Feb 2, 2018

See below.

Explanation:

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