سوال ۲۶ کنکور ارشد علوم کامپیوتر ۱۴۰۵

The quartic is palindromic (its coefficients $1,2,2,2,1$ read the same forwards and backwards), so it factors neatly.

Factor the polynomial

Seek a factorization into two monic quadratics with constant term $1$: $z^4+2z^3+2z^2+2z+1=(z^2+az+1)(z^2+bz+1).$ Expanding and matching coefficients gives $a+b=2$ (the $z^3$ and $z$ terms) and $2+ab=2\Rightarrow ab=0$ (the $z^2$ term). Hence $\{a,b\}=\{0,2\}$, so $z^4+2z^3+2z^2+2z+1=(z^2+1)(z+1)^2.$

Read off the roots

• $z^2+1=0 \Rightarrow z=i,\ z=-i$
• $(z+1)^2=0 \Rightarrow z=-1$ (a double root)

The distinct roots are $\{\,i,\,-i,\,-1\,\}$, so their sum is $i+(-i)+(-1)=-1.$

Pitfall: Summing the roots with multiplicity gives $i-i-1-1=-2$ (equivalently Vieta's $-\tfrac{2}{1}$), which is the trap option (1). The word distinct (متمایز) means the repeated root $-1$ is counted only once, giving $-1$.