SOLUTION: Find all real and complex zeros of P(x)=x^3-x^2-4x-6. Show all work.

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find all real and complex zeros of P(x)=x^3-x^2-4x-6. Show all work.      Log On


   

Question 695069: Find all real and complex zeros of P(x)=x^3-x^2-4x-6. Show all work.
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
Rational zeros of P(x) must be factors of 6 (with + or - signs.
Choices are: -6, -3, -2, -1, 1, 2, 3, and 6.
P(3)= 3%5E3-3%5E2-4%2A3-6=27-9-12-6=0 so x=3 is a zero.
That means that %28x-3%29 is a factor of P(x).
Dividing, we find
%28x%5E3-x%5E2-4x-6%29%2F%28x-3%29=x%5E2%2B2x%2B2 <--> x%5E3-x%5E2-4x-6=%28x-3%29%28x%5E2%2B2x%2B2%29
p(x)= %28x-3%29%28x%5E2%2B2x%2B2%29
Besides, highlight%28x=3%29, the zeros of P(x) are the solutions to x%5E2%2B2x%2B2=0
That is easy to solve by "completing the square", but feel free to use the quadratic formula, if you like it better.
x%5E2%2B2x%2B2=0 --> x%5E2%2B2x%2B1=-1 --> %28x%2B1%29%5E2=-1 --> highlight%28x=-1+%2B-+i%29