I can tell you how I would solve it if that helps. First draw a graph so that you can see roughly where the curve crosses the x axis. This will show you how many real zeroes there are, if any. When you do this you will see that x=-3 and x=1 are close to the zeroes. The graph doesn’t appear to cross the x-axis anywhere else, but now we need to apply calculus. Find the derivative: 4x³+18x²+54x+32. We can now use the approximate zeroes we have found so far and apply Newton’s Method to get accurate measures of these zeroes. You will need a calculator.
We use a simple iterative process for homing in on the zeroes. The formula to use is:
x₁=x₀-(x₀⁴+6x₀³+27x₀²+32x₀-60)/(4x₀³+18x₀²+54x₀+32), where x₀ is -3 or 1 and x₁ is the first iteration result. When we find x₁, we substitute it in place of x₀ and we get x₂, the second iteration result. And so on for subsequent iterations. When the result of two consecutive iterations seems to be the same, we have reached the maximum accuracy of the calculating device and we stop, so the final iteration is the best, most accurate zero for x near the point where we started. In other words, we converge to one specific value for each zero.
I got the results after only a few iterations: x=-2.918749059 and 0.9429231745.
To proceed to find the remaining two zeroes expected to be complex we can divide the original expression by the quadratic we can now calculate from knowing the two zeroes. This is the bumpy part of the ride!
Let z₁ and z₂ represent the zeroes we just found, then the quadratic will be (x-z₁)(x-z₂)=x²-(z₁+z₂)x+z₁z₂.
Let s=z₁+z₂ (sum of zeroes) and p=z₁z₂ (product of zeroes).
Using algebraic long division we can work out the other quadratic. I worked this out to be:
x²+(6+s)x+27-p+6s+s²=0 and we use the formula to find the remaining zeroes:
x=½(-(6+s)±√(-72-12s-3s²+4p)). The discrimant is expected to be negative, so we will have an imaginary number. When we substitute true values for s and p we get the complex zeroes:
-2.012087058±4.213382776i.