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?!

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If $x=+5$ works, you should divide by $x-5$. You should get a result that only has terms in $x^4, x^2, 1$. – Ross Millikan Nov 14 '12 at 18:45

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)

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Hint:

$$x^5-5x^4+6x^3-30x^2+8x-40=x^4(x-5)+6x^2(x-5)+8(x-5)$$

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