Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,156

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.**

**Online**

Pages: **1**