Question

Let 1 + i be one of the root of z^3 + Az^2 + Bz + 26 = 0, then find the real roots of the equation and determine the value of a and b.

Answer 1

Answer:

The real root is -13

The value of A is 11 and B is -24

Step-by-step explanation:

The given equation is $z^3+Az^2+Bz+26=0$.

If $1+i$ is a root then then the complex conjugate $(1-i)$ is also a root.

Let us now derive the quadratic equation that gave rise to this complex roots.

By the factor theorem, we get;

$(z+(-1-i))(z+(-1+i)=z^2-2z+2$

Now the  

Now the fully factored cubic polynomial will be of the form:

$(z+c)(x^2-2x+2)=0$

By comparing constant terms; 2c=26. This means c=13

Hence  the required polynomial is $(z+13)(x^2-2x+2)=0$

Therefore the real root is z=-13

We now expand fully to get:

$z^3+11z^2-24z+26=0$

By comparing coefficients; A=11, and B=-24