I encountered the following question while learning about complex numbers. I did reach a solution, yet I feel like I took a long route. I am sure that there are better ways to solve this question. I will share my solution down . If there are any new suggestions with explanation, I would be more than thankful.
find the roots for the following equation
$$x^4-2x^3-x^2+2x+10$$
Complex Numbers – How to Find the Roots for This Equation?
complex numbersroots
Best Answer
When I have to solve a quartic equation $x^4+\alpha x^3+\beta x^2+\gamma x+\delta=0$, usually the first thing that I do is to apply the substitution $x=y-\frac\alpha4$, because then I get a quartic equation without a cubic term.
In your case, that means to make the substitution $x=y+\frac12$. Note that$$p\left(y+\frac12\right)=y^4-\frac{5y^2}2+\frac{169}{16}.$$Well, not only there is no cubic term, as there is no term of degree $1$. And the roots of this quartic equation (which is a biquadratic equation) are$$\frac32-i,\frac32+i,-\frac32-i,\text{ and }-\frac32+i.$$So, you have\begin{align}p(x)=0&\iff p\left(y+\frac12\right)=0\\&\iff y^4-\frac{5y^2}2+\frac{169}{16}=0\\&\iff y=\frac32-i\vee y=\frac32+i\vee{}\\&\phantom{\iff}\vee y=-\frac32-i\vee y=-\frac32+i\\&\iff x-\frac12=\frac32-i\vee x-\frac12=\frac32+i\vee{}\\&\phantom{\iff}\vee x-\frac12=-\frac32-i\vee x-\frac12=-\frac32+i\\&\iff x=2-i\vee x=2+i\vee x=-1-i\vee x=-1+i.\end{align}