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#1 2016-12-02 21:43:51

bobbym
bumpkin
From: Bumpkinland
Registered: 2009-04-12
Posts: 109,606

Because she asked.

Hi;

This problem appears in another thread and someone asked to see my solution.

1) Find a fifth degree polynomial f(x) with the conditions (x-1)^3 |  f(x) -1  and x^3 | f(x).

The outline of how it is done.

We get the remainders:

Those remainders on the RHS must be set to 0 and the coeficients equated to 0.

From that we get 6 simultaneous equations that are easy to solve:

So the polynomial we seek is

Of course it is much easier using Mathematica:

Way 1:

p[x_, a_, b_, c_, d_, e_, f_] := a*x^5 + b*x^4 + c*x^3 + d*x^2 + e*x + f

rul = FindInstance[
   ForAll[x, 
    PolynomialRemainder[p[x, a, b, c, d, e, f] - 1, (x - 1)^3, x] == 
      0 && PolynomialRemainder[p[x, a, b, c, d, e, f], x^3, x] == 
      0], {a, b, c, d, e, f}] // First;

p[x, a, b, c, d, e, f] /. rul // TraditionalForm

yields:

6 x^5-15 x^4+10 x^3

Way 2:

p[x_, a_, b_, c_, d_, e_, f_] := 
  a*x^5 + b*x^4 + c*x^3 + d*x^2 + e*x + f;

PolynomialRemainder[p[x, a, b, c, d, e, f] - 1, (x - 1)^3, x];
PolynomialRemainder[p[x, a, b, c, d, e, f], (x)^3, x];

rul = Solve[{-1 + 6 a + 3 b + c + f == 0, (-15 a - 8 b - 3 c + e) == 
      0, (10 a + 6 b + 3 c + d) == 0, f == 0, e == 0, d == 0}, {a, b, 
     c, d, e, f}][[1]];
p[x, a, b, c, d, e, f] /. rul

10 x^3 - 15 x^4 + 6 x^5


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

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