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

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We have f(x) = x^5+x^4+1 and g(x) = (x- 1)^3 = x^3 - 3x^2 + 3x - 1.

We can find the remainder by dividing them.

x^3 - 3x^2 + 3x - 1 | x^5+x^4+1 |x^2 + 4x + 9

................................x^5 - 3x^4 + 3x^3 - x^2

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.................................0 + 4x^4 - 3x^3 + x^2 +1

.......................................4x^2 - 12x^3 + 12x^2 - 4x

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......................................0 + 9x^3 - 11x^2 + 4x + 1

............................................9x^3 - 27x^2 + 27x - 9

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.............................................0 + 16x^2 - 23x + 10

**Therefore the remainder is 16x^2 - 23x +10.**

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**