Polynomial

**ellenbaggao** · July 17th 2008, 05:58 AM

how's this?
thanks.

y^3-gy+5
y^3-3y-2

and

x^5+y^5
x+y

If you can show the solution, better.

THANKS!

**Isomorphism** · July 17th 2008, 06:12 AM

Originally Posted by ellenbaggao

x^5+y^5
x+y

If you can show the solution, better.

THANKS!

You could have simply tried long division

But I am going to try a different approach...

$\frac{x^5 + y^5}{x+y} = y^4\frac{\left(\frac{x}{y}\right)^5 + 1}{\left(\frac{x}{y}\right)+1} = y^4\frac{-a^5 + 1}{-a+1}$

Where $-a = \frac{x}{y}$ .

You know the geometric progression formula, $\frac{a^n - 1}{a - 1} = 1 + a + a^2 + a^3+ \cdots + a^{n-1}$

Thus, now we can substitute $a = -\frac{x}{y}$ :

$y^4\frac{-a^5 + 1}{-a+1} = y^4(1 + a + a^2 + a^3 + a^4) =$ $y^4(1 - \frac{x}{y} + \left(-\frac{x}{y}\right)^2 - \left(\frac{x}{y}\right)^3 + \left(-\frac{x}{y}\right)^4) = y^4 - xy^3+ x^2y^2 - x^3y + x^4$

So:

$\frac{x^5 + y^5}{x+y} = y^4 - xy^3+ x^2y^2 - x^3y + x^4$