Find rational zeroes first:
Take factors of coefficients of the highest and lowest degree:
Factors of 2 are 1 and 2; factors of 56 are 1, 2, 4, 7, 8, 14, 28, 56.
Rational zeroes are:
1, 2, 4, 7, 8, 14, 28, 56, 7, 7/2 (positive and negative forms).
Plug in these one by one:
x=2: 32-120+128-60+56=36;
x=4: 512-960+512-120+56=0. So x=4 is a zero.
Use synthetic division to reduce the degree of the polynomial:
4 | 2 -15 32 -30 56
2 8 -28 16 | -56
2 -7 4 -14 | 0
So now we have the cubic: 2x3-7x2+4x-14.
The rational zeroes are:
1, 2, 7, 14, ½, 7/2. We can see that ±1 is not a zero and we have already tried 2, and -2 would make the polynomial negative. 7 and -7 are not zeroes. 14 is not a zero, because 143 is too large, just as 7 was.
x=½: ¼-7/4+2-14≠0;
x=7/2: 343/4-343/4+14-14=0; so 7/2 is another zero. Divide by this zero:
7/2 | 2 -7 4 -14
2 7 0 | 14
2 0 4 | 0
The quadratic is 2x2+4=2(x2+2). The distributed 2 converts the zero from x-7/2 to 2x-7 as a factor.
Therefore we have factorised the polynomial=(x-4)(2x-7)(x2+2). The two complex solutions are x=i√2 and -i√2.