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
-----------------------------------------------------------
.................................0 + 4x^4 - 3x^3 + x^2 +1
.......................................4x^2 - 12x^3 + 12x^2 - 4x
----------------------------------------------------------------
......................................0 + 9x^3 - 11x^2 + 4x + 1
............................................9x^3 - 27x^2 + 27x - 9
---------------------------------------------------------------
.............................................0 + 16x^2 - 23x + 10
Therefore the remainder is 16x^2 - 23x +10.
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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
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