First, we need to find which number when substituted into the equation will give the answer zero. \(f(1) = {(1)^3} + 4{(1)^2} + (1) - 6 = 0\) Therefore \((x - 1)\)is a factor. Factorise the quadratic ...

Locating scattering resonances is a standard task in certain areas of physics and engineering. This often can be reduced to finding zeros of complex analytic functions. In this talk, I will discuss a ...