Question 533623
Can you please state the positive real zeros, negative real zeros, and imaginary zero for this functions : r(x)=x^4-x^3-5x^2+6x+1
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Using the Rational Roots Theorem to find the zeros:

....0....|....1........-1........-5.........6.........1
....1....|....1..........0........-5.........1.........2
....2....|....1......... 1........-3.........0.........1
....3....|....1........  2........ 1.........6.........19 (3 is upper bound)

....0....|....1........-1........-5.........6.........1
..-1....|....1........-2........-3.........3........-2 (real root between 0 and -1)
..-2....|....1........-3........1.........4.........-7
..-3....|....1........-4........7......-15.........46 (-3 is lower bound)  (real root between-2 and -3)

Procedure above shows there are two real roots between 0 and -1 and between -2 and -3,
but they are not rational. I don't know how to find these two irrational real roots algebraically. The graphics program on my computer show these roots to be -0.2929 and -2.2740. The other two roots are non-real or imaginary. See graph below as a visual check.
{{{ graph( 300, 300, -5, 5, -5, 5,x^4-x^3-5x^2+6x+1) }}}