1. ## Polynomial

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!

2. 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...

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

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

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

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

$\displaystyle y^4\frac{-a^5 + 1}{-a+1} = y^4(1 + a + a^2 + a^3 + a^4) =$$\displaystyle 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:

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

3. Hello, ellenbaggao!

$\displaystyle \frac{y^3-{\color{red}g}y+5}{y^3-3y-2}$ . . . . What is that $\displaystyle {\color{blue}g}$ ?

$\displaystyle \frac{x^5+y^5}{x+y}$
Factor: .$\displaystyle \frac{(x+y)(x^4-x^3y + x^2y^2 - xy^3 + y^4)}{x+y}$

Reduce: .$\displaystyle x^4 - x^3y + x^2y^2 - xy^3 + y^4$

4. ## hello.

y^3-6y+5
y^3-3y-2

thanks.

5. Originally Posted by ellenbaggao
y^3-6y+5
y^3-3y-2

thanks.
note $\displaystyle y^3-6y+5=y^3-3y-3y-2+7=(y^3-3y-2)-3y+7$

Hence $\displaystyle \frac{y^3-6y+5}{y^3-3y-2}=\frac{(y^3-3y-2)-3y+7}{y^3-3y-2}=1+\frac{-3y+7}{y^3-3y-2}$