What is the reminder when f(x)=x^5+x^4+1 is divided by g(x)=(x-1)^3 ?

Since the divisor is of 3rd order, the reminder has to be
of 2nd order.

r=ax^2 + bx +
c

We 'll write the reminder
theorem:

f=g*q + r, where q is the quotient of the
division.

We notice that x=1 is a multiple root of the
polynomial g.

f(1)=g(1)*q(1) +
r(1)

By substituting the value x=1 into all polynomials,
we'll obtain:

f(1)= 1^5 + 1^4 +
1

f(1)=1+1+1

f(1)=3

g(1)=
(1-1)^3=0

r(1)=a+b+c

So, we'll
have:

3=0*q(1) +
a+b+c

3=a+b+c

Because of the
fact that x=1 is a multiple root, it will cancel the first derivative of the expression:
f=g*c + r

f' =(g*c)' + r'

5x^4
+4 x^3=3*(x-1)^2*q+(x-1)^3*q' +2*ax+ b

By substituting the
value x=1 into all polynomials, we'll
obtain:

5+4=2*a+b

9=2*a+b

We'll
calculate the first derivative of the expression

5x^4 +4
x^3=3*(x-1)^2*q+(x-1)^3*q' +2*ax+ b

(5x^4 +4
x^3)'=(3*(x-1)^2*q+(x-1)^3*q' +2*ax+ b)'

5*4* x^3 + 4*3*
x^2=6*(x-1)*q+3*(x-1)^2*q'+3*(x-1)^2*q'+(x-1)^3*q"+2a

By
substituting the value x=1 into all polynomials, we'll
obtain:

20+12=2*a

We'll divide
by 2:

a = 10 + 6

a =
16

2a+b=9

2*16 + b =
9

b = 9 - 32

b =
-23

a+b+c =
3

16-23+c=3

-7+c=3

c=3+7

c=10

The
reminder is: 16x^2 - 23x + 10