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## ketz Solve: z4 + 2z3 + 5z2 + 8z + 4 = 0 It's urgent... one year ago one year ago

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1. shubhamsrg

www.wolframalpha.com

2. mukushla

Using rational root theorem u can realize that z=-1 is a solution...

3. mukushla

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)$

4. mukushla

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

5. deena123

$z^4+2z^3+5z^2+8z+4=(z+1)^2(z^2+4)$

6. deena123

$z=-1, \pm2i$i think u want real root so $z=-1$