Finding all real zeros of the polynomial $$x^5 - 5x^4 +6x^3 -30x^2 +8x - 40 = 0$$
So far I have...
$$r/s: +- 1, +- 40, +- 2, +- 20, +- 4, +- 10, +-5, +- 8$$
Only $+ 5$ works.
Then I have $$(x + 5)(            ) = x^5 - 5x^4 +6x^3 -30x^2 +8x - 40$$
Then you have to use long devision between $x + 5$ and $ x^5 - 5x^4 +6x^3 -30x^2 +8x - 40$
That's where I get lost. Help?!

here is how to do the long division
x^5 - 5x^4 + 6x^3 - 30x^2 + 8x - 40
x^5 - 5x^4                          subtract x^4(x-5)
-----------------------------------
             6x^3 - 30x^2 + 8x - 40
             6x^3 - 30x^2           subtract 6x^2(x-5)
-----------------------------------
                            8x - 40
                            8x - 40 subtract 8(x-5)
-----------------------------------
                                  0

and since we got 0 remainder we can say that $$x^5 - 5x^4 + 6x^3 - 30x^2 + 8x - 40 = (x^4+6x^2+8)(x-5).$$
(Now you can find the roots of the quadratic $z^2+6z+8$ and square root them to find the real roots of the quartic)
