2x4+3x3-14x2+8x=x(2x3+3x2-14x+8). x=0 is one of the zeroes.
The other zeroes are the zeroes of 2x3+3x2-14x+8, and they may be irrational.
One way of finding them is to use Newton's iterative Method which includes calculus.
If f(x)=2x3+3x2-14x+8, then f'(x)=6x2+6x-14.
xn+1=xn-f(xn)/f'(xn), where x0 is close to a zero of f(x). A sketch of the graph of the cubic shows that there are zeroes near x=-4, x=1 and x=2. f(-4)=-128+48+56+8=-16; f(1)=-1; f(2)=16+12-28+8=8.
x1=x0-f(x0)/f'(x0).
x0=-4:
x1=-3.7241..., x2=-3.6909..., x3=-3.69047..., ..., x=-3.690471 after a few more iterations.
x0=1:
x=0.755134 after a few iterations.
x0=2:
x=1.435337 after a few iterations.
The zeroes (all real): -3.690471, 0, 0.755134, 1.435337 to 6 decimal place accuracy.