how to find the roots for this equation?

Question

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 $$x^4-2x^3-x^2+2x+10$$

Answer 1

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}

Answer 2

I tried solutions such as x=$$\pm1,\pm10,\pm5,\pm2 $$ found no real roots could not write (x+a)*( ) so went with a different approach reached 2 equations $$(x^2+ax\pm 2)(x^2+bx\pm 5)$$ $$(x^2+ax\pm 1)(x^2+bx\pm 10)$$ the coefficient of $$x^3= -2 a+b=-2 b=-2-a$$ checked each one $$(x^2+ax\pm 1)(x^2+bx\pm 10)$$ I know that the multiplication should result in coefficient 2 for x $$2=\pm10a\pm(a+2)$$ results in a=0 as a solution ,since the teacher demanded that a and b are members of z

coefficient of $$x^3 =1 $$ $$\pm1=\pm10\pm(1-a)(a+2)$$ no solution

$$(x^2+ax\pm 2)(x^2+bx\pm 5)$$ $$2=\pm5a\pm2(a+2)$$ $$\pm(3a-4)$$ $$-1=\pm5\pm2-a(a+2)$$ -1=7-2*4 there is a solution $$(x^2+2x+ 2)(x^2-4x+ 5)$$ $$((x+1)^2+ 1)((x-2)^2+1)$$ $$(−−2)(−+1)(+−2)(++1)=0 $$ this took a lot of time and a couple of hints to solve any new approaches are welcome