Hello.
Find the other two zeroes of P(z) if -2+i is a zero or p(z)=2z^3+9z^2+14z+5.
My attempt:
-2+i is a zero, so -2-i is also a zero.
P(z)=2z^3+9z^2+14z+5
.....=(z-[-2-i])(z-[-2-i])(z-z3)
.....=((z+2)^2+1)(z-z3)
.....=(z^2+4z+5)(z-z3)
5*-z3= -5z3 (I do NOT mean -15z here)
......5=-5z3
.....-1=z3
Two other Zeroes are -2-i , -1
However, the correct answer given in the book is
-2-i and -0.5
Thank you
Hello masters.
I will show a worked solution from the textbook which i tried to follow.
And thank you , I did not know about the rational roots theorem , but googled it after you mentioned it and it helped very much. But if possible could you also show me how to do the the way the the books does it.
At this point you have dropped the "2" multiplying z^3.
You should have "= 2((z+2)^2+ 1)(z- z3)"......=((z+2)^2+1)(z-z3)
2*5*-z3= -10z3= 5 so z3= -1/2......=(z^2+4z+5)(z-z3)
5*-z3= -5z3 (I do NOT mean -15z here)
......5=-5z3
.....-1=z3
Two other Zeroes are -2-i , -1
However, the correct answer given in the book is
-2-i and -0.5
Thank you