Question

z4-z3+3z2-2z+2;z2+2. by long division method​

Answer 1

Step-by-step explanation:

As Thomas has said, if 2+i is a root then 2−i is also root. Hence you have two factors: z−(2−i) and z−(2+i) whose product is z2−4z+5. So we can can assume that there is another quadratic factor z2+bz+c where b and c are constants to be determined. Obviously their product should yield the given polynomial. Thus,

(z2−4z+5)(z2+bz+c)=z4−2z3−z2+2z+10.

Equating the constant terms you obtain 5c=10 which gives c=2. Now equating the coefficients of z, we obtain the equation;

−4c+5b=2

from which we obtain b=2. Replacing b and c we now solve for z in the equation

z2+2z+2=0

as follows:

(z+1)2+1=0.

Thus, z+1=i and z+1=−i, which gives z=−1+i,z=−1−i. Hence we have obtained all the factors.

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