Question 250698: What are the zeros of p(x)=x^4+7x^3+9x^2-17x-20 solved algebraically?
Answer by drk(1908)   (Show Source):
You can put this solution on YOUR website! We are given this polynomial:
p(x) = x^4 + 7x^3 + 9x^2 - 17x - 20.
Using PNI, we get
P . . . N . . . I
1 . . . 3 . . . 0
1 . . . 1 . . . 2.
So, I know there MUST be 1 positive root.
Now I look at P/Q; all the factors of 20 over all the factors of 1.
+-20 +-10 +-5 +-4 +-2 +-1.
I use synthetic division until I find a hit: (X + 4) and (X + 1) are factors of P(x).
So, (X + 4)(X + 1)[X^2 + 2X - 5] = P(x).
The zero's become
X = -4, -1, -1 + sqrt(6), -1 - sqrt(6).
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