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$x^5 - 5x^4 +6x^3 -30x^2 +8x - 40 = 0$

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 $x^4, x^2, 1$.2012-11-14

2 Answers 2

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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:

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