Test for rational zeroes first. Factors of the constant 26 are ±1, ±2, ±13, ±26.
Setting x to these in turn does not give us a zero expression, but, because the polynomial is a cubic, there must be at least one real zero.
If f(x)=-x3+2x2+x-26, then f(-3)=16 and f(-2)=-12, so there is a zero when -3<x<-2. Newton's Method provides an iterative way to find the zero, using xn+1=xn-f(xn)/f'(xn).
f'(x)=-3x2+4x+1. Using the intermediate Value Theorem we can start the process by x0=(-3-2)/2=-2.5.
x1=-2.5135..., x2=-2.51345..., ..., x=-2.5134514815737 approx.
Represent this zero by X, and use synthetic division to find the quadratic factor:
X | -3 2 1 -26
-3 -3X 2X-3X2 | 26
-3 2-3X 1+2X-3X2 | 0 = -3x2+(2-3X)x+(1+2X-3X2).
Let a=-3, b=2-3X, c=1+2X-3X2, then:
x=(-b±√(b2-4ac))/2a.
b=9.540354 approx, c=-22.979218 approx.
4ac=275.750616 approx, b2=91.018363 approx, so b2-4ac<0.
x=(-b±i√(4ac-b2))/2a=(-9.540354±i√184.732253)/-6.
x=1.5901±2.2653i approx are the complex zeroes.