Determine the reminder when f(x)= (x-1)^2 + x^5 + x^4 - x^2 + 2x is divided by g(x)=x^3+3x^2+3x+1 ?

We'll expand the square from
f(x):

f(x) = x^2 - 2x + 1 + x^5 + x^4 - x^2 +
2x

We'll eliminate like
terms:

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

We'll recognize in expression of g(x) the binomial raised
to cube:

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)=1

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

r(-1)=a-b+c

So,
we'll have:

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

1=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

1=-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 =
-4

-2a+b=1

-2*(-4) + b =
1

b = 1 - 8

b =
-7

a-b+c =
1

-4+7+c=1

3+c=1

c=1-3

c=-2

The
reminder is: r(x) = -4x^2 - 7x - 2.