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

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Find all the zeros of the polynomial function. f(x) = {{{ x^3-4x^2-11x+2}}}      Log On


   

Question 825414: Find all the zeros of the polynomial function.
f(x) = +x%5E3-4x%5E2-11x%2B2
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
The possible rational zeros are +1/1 and +2/1 (which simplify to +1 and +2). Trying these we find that only -2 is actually a zero:
-2  |   1   -4   -11   2
-----       -2    12  -2
       ------------------
        1   -6     1   0
The zero in the lower right corner is the remainder. (It is also f(-2)!) Since it is zero then what we divided by (x-(-2)) [or (x+2)] divides evenly. So (x+2) is a factor of f(x). Not only that but the rest of the bottom row tells us what the other factor is. The "1 -6 1" translates into x%5E2-6x%2B1. So now
f%28x%29+=+x%5E3-4x%5E2-11x%2B2+=+%28x%2B2%29%28x%5E2-6x%2B1%29
The other zeros of f(x) will come from the second factor. Since it is a quadratic that won't factor, we can use the quadratic formula:
x+=+%28-%28-6%29%2B-sqrt%28%28-6%29%5E2-4%281%29%281%29%29%29%2F2%281%29
Simplifying...
x+=+%286%2B-sqrt%2836-4%281%29%281%29%29%29%2F2
x+=+%286%2B-sqrt%2836-4%29%29%2F2
x+=+%286%2B-sqrt%2832%29%29%2F2
x+=+%286%2B-sqrt%2816%2A2%29%29%2F2
x+=+%286%2B-sqrt%2816%29%2Asqrt%282%29%29%2F2
x+=+%286%2B-+4%2Asqrt%282%29%29%2F2
x+=+%282%283%2B-+2%2Asqrt%282%29%29%29%2F2
x+=+%28cross%282%29%283%2B-+2%2Asqrt%282%29%29%29%2Fcross%282%29
x+=+3%2B-+2%2Asqrt%282%29
which is short for x+=+3%2B2%2Asqrt%282%29 or x+=+3-2%2Asqrt%282%29

So the three zeros for f(x) are: -2, 3%2B2%2Asqrt%282%29 and 3-2%2Asqrt%282%29