Evaluating $2x^3+2x^2-7x+72$ where $x= \frac{3-5i}{2} $

Question

So I have to find the value of : $$2x^3+2x^2-7x+72$$ at $$x= \frac{3-5i}{2} $$ Where $i$ stands for $\sqrt{-1}$. I know the obvious approach would be to substitute the value of $x$ in the equation, but the equation gets extremely messy in that process. Is there any easier or quicker approach for the sum.

P.s. The answer is $4$.

Answer 1

You have $\frac{3-5i}{2}$ is the solution of $2x^2-6x+17 = 0$.

Then, $$2x^3 + 2x^2-7x+72 = x(2x^2-6x+17) + 4(2x^2-6x+17) + 4 = 4.$$

Answer 2

Hint

$$x= \frac{3-5i}{2}$$ $$x^2=x \times x=-4-\frac{15 i}{2}$$ $$x^3=x \times x^2=-\frac{99}{4}-\frac{5 i}{4}$$

Answer 3

By the polynomial remainder theorem $f(\frac{3-5i}{2})$ is simply the remainder we get when we divide $f(x)$ by $x-\frac{3-5i}{2}$. So perform synthetic division, which I think is easier.