Question 547646
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(1)= 1^4-6*1^3+7*1^2+6*1-8=1-6+7+6-8=0}}} and
{{{F(-1)= (-1)^4-6*(-1)^3+7*(-1)1^2+6*(-1)-8=1+6+7-6-8=0}}}
So, we can divide the polynomial by {{{x-1}}} and {{{x=1}}}, or by their product
{{{(x-1)(x+1)=x^2-1}}}
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
{{{(x^4-6x^3+7x^2+6x-8)/((x-1)(x+1))=x^2-6x+8}}}
That second degree polynomial is factoring friendly, and we easily see that
{{{x^2-6x+8=(x-2)(x-4)}}}
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.