SOLUTION: F(x)= x^4-6x^3+7x^2+6x-8 .....find all the zeros of the polynomial function

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Question 547646: F(x)= x^4-6x^3+7x^2+6x-8 .....find all the zeros of the polynomial function
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
If the zeros are rational, they would be fractions whose numerator is a factor of 8 (the constant term) and whose denominator is a factor of 1 (the leading coeficient. The choices are 1, -1, 2, -2, 4, -4, 8, and -8.
We can see that
F%281%29=+1%5E4-6%2A1%5E3%2B7%2A1%5E2%2B6%2A1-8=1-6%2B7%2B6-8=0 and

So, we can divide the polynomial by x-1 and x=1, or by their product
%28x-1%29%28x%2B1%29=x%5E2-1
to get a second degree polynomial, whose roots (real or not) we know we can find.
Alternately, we can try our luck with the other possible integer roots.
Dividing, by whatever method we choose, we get
%28x%5E4-6x%5E3%2B7x%5E2%2B6x-8%29%2F%28%28x-1%29%28x%2B1%29%29=x%5E2-6x%2B8
That second degree polynomial is factoring friendly, and we easily see that
x%5E2-6x%2B8=%28x-2%29%28x-4%29
So we've found all four roots: -1, 1, 2, and 4,
and they were all integers.
A meaner problem would have you ending up with a second degree polynomials with irrational roots, or with no real roots.