(HINT: A graph of the function h(x) is a good guide to the solution.)
Look for rational zeroes first, that is, factors of ±8, which are 1, 2, 4, 8. Start with x=1:
66-60+137-130-101+80+8=0, so 1 is a zero, so we can divide by it using synthetic division:
1 | 66 -60 137 -130 -101 80 8
66 66 6 143 13 -88 | -8
66 6 143 13 -88 -8 | 0 = 66x5+6x4+143x3+13x2-88x-8.
Note that the magnitude of the constant is still 8. We also have to consider the leading coefficient 66. This has factors: 1, 2, 3, 6, 11, 22, 33, 66. And we still have the remaining factors of 8: 2, 4, 8 (1 has been eliminated already and will not work again, neither will -1). Note that the factors are all even while the factors of 66 contain a mixture of odds and evens. Since the whole expression is a mixture of odds and evens in the coefficients, we can't mix evens in the same factor.
Observe that 66, 88 and 143 are all divisible by 11, so perhaps 1/11 is a zero. Also, 6×11=66 and 13×11=143 and 8×11=88. 6, 13 and 8 are the other coefficients. Let's see if -1/11 is a zero:
-66/115+6/114-143/113+13/112+88/11-8=
-6/14641+6/14641-13/121+13/121+8-8=0. Bingo! 11x+1 is a zero.
Divide by 11x+1 using algebraic division:
6x4 13x2 -8
11x+1 ) 66x5+6x4+143x3+13x2 -88x-8
66x5+6x4 143x3+13x2 -88x-8
0
We are left with 6x4+13x2-8=(3x2+8)(2x2-1).
So now we have the means to find the remaining zeroes.
2x2=1 has two zeroes: ±1/√2, that is, -√2/2 and √2/2.
The remaining zeroes are imaginary: ±i√(8/3), or -2i√6/3 and 2i√6/3.
The zeroes are: 1, -1/11, -√2/2, √2/2, -2i√6/3, 2i√6/3.