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shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
www.wolframalpha.com
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
Using rational root theorem u can realize that z=1 is a solution...
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
so u can factor out z+1\[ z^4 + 2z^3 + 5z^2 + 8z + 4=z^4+z^3+z^3+z^2+4z^2+4z+4z+4\]\[=(z+1)(z^3+z^2+4z+4)\]
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

deena123 Group TitleBest ResponseYou've already chosen the best response.0
\[z^4+2z^3+5z^2+8z+4=(z+1)^2(z^2+4)\]
 one year ago

deena123 Group TitleBest ResponseYou've already chosen the best response.0
\[z=1, \pm2i\]i think u want real root so \[z=1\]
 one year ago
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