SOLUTION: solve using the rational zero theorem and descarte rule:x^4-5x^3-54x^2-80x-32=0

Question 255299: solve using the rational zero theorem and descarte rule:x^4-5x^3-54x^2-80x-32=0
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
x^4 - 5x^3 - 54x^2 - 80x - 32 = 0
step 1 - set up P, N, i
P are the number of sign changes you see when (x) is placed into the equation. We get P = 1.
N are the number of sign changes you see when (-x) is placed into the equation. We get N = 3.
Now 1 + 3 = 4.
P . . . 1 . . . . 1
N . . . 3 . . . . 1
i . . . .0 . . . . .2
Notice that i = 0 or 2. Imaginary numbers always travel in conjugate pairs.
Now on to rational zeros. or P/Q
P = +-(1, 2, 4, 8, 16, 32)
Q = +-(1)
We can find that x = -4 and x = -1 give us 2 zeros.
By synthetic division, using x = -4, we get
(x^3-9x^2-18x-8)
and this divided by x = -1 gets us
(x^2-10x-8)
setting this = 0 and solving gets us

or simplified as

SO our four answers are:
x = -4, x = -1, x = 5 +-sqrt(33)
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