SOLUTION: Find all the zeros of the polynomial function. f(x) = {{{ x^3-4x^2-11x+2}}}

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Question 825385: Find all the zeros of the polynomial function.
f(x) = +x%5E3-4x%5E2-11x%2B2
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
As per the rational zero theorem,
the possible rational zeros of f%28x%29=+x%5E3-4x%5E2-11x%2B2 are integers factors of 2 , meaning
-2, -1, 1, and 2.

We can try them all, starting by the easiest ones.
Trying 1 and -1 by substitution is easy enough:
f%281%291-4-11%2B2=-3-11%2B2=-14%2B2=-12
f%28-1%29=-1-4%2B11%2B2=-5%2B11%2B2=-6%2B2=8 .
That means that x=1 is not a zero of f%28x%29 , and neither is x=-1 .
If x=2 is a zero of f%28x%29 , x-2 must be a factor of f%28x%29 . If x=-2 is a zero of f%28x%29 , x-%28-2%29=x%2B2 must be a factor of f%28x%29 .
We could try -2 and 2 by substitution, but since in the end we will need to divide, we may choose just to try dividing by x-2 and by x%2B2 .
We would find that dividing f%28x%29 by x-2 leaves a remainder,
but that f%28x%29 divides evenly by x%2B2 , and
f%28x%29=%28x%2B2%29%28x%5E2-6x%2B1%29 .
That means that highlight%28x=-2%29 is a zero of f%28x%29 ,
and the remaining zeros are the zeros of x%5E2-6x%2B1 .

We can use find the zeros of x%5E2-6x%2B1 by solving the quadratic equation
x%5E2-6x%2B1=0 either by using the quadratic formula or by "completing the square":
x%5E2-6x%2B1=0
x%5E2-6x=-1
x%5E2-6x%2B9=-1%2B9
%28x-3%29%5E2=8
So either x-3=sqrt%288%29=2sqrt%282%29 --> highlight%28x=3%2B2sqrt%282%29%29 ,
or x-3=-sqrt%288%29=-2sqrt%282%29 --> highlight%28x=3-2sqrt%282%29%29 .

In sum, the zeros of f%28x%29=+x%5E3-4x%5E2-11x%2B2 are
highlight%28x=-2%29 , highlight%28x=3%2B2sqrt%282%29%29 , and highlight%28x=3-2sqrt%282%29%29 .