Here's the question you clicked on:
ketz
Solve: z4 + 2z3 + 5z2 + 8z + 4 = 0 It's urgent...
www.wolframalpha.com
Using rational root theorem u can realize that z=-1 is a solution...
so u can factor out z+1\[ z^4 + 2z^3 + 5z^2 + 8z + 4=z^4+z^3+z^3+z^2+4z^2+4z+4z+4\]\[=(z+1)(z^3+z^2+4z+4)\]
now for the cubic part ..easily u can see that z=-1 is a solution for that so u can factor out z+1 from that too\[(z+1)(z^3+z^2+4z+4)=(z+1)(z^2(z+1)+4(z+1))=(z+1)^2(z^2+4)\]we're done
\[z^4+2z^3+5z^2+8z+4=(z+1)^2(z^2+4)\]
\[z=-1, \pm2i\]i think u want real root so \[z=-1\]