Answer to Question #251904 in Algebra for moe

Question

You are given that $1+i$  is a root of the equation $z^{3}+2z^{2}+az+b=0$ , where a and b are real numbers. Which of the following are true?

1. $a=−6$ and $b=8$
2. $z=4$  is also a root of the equation.
3. $z=−1−i$  is also a root of the equation.
4. $a=8$ and $b=-6$
5. $z=−4$  is also a root of the equation.

Answer 1

$z^3+2z^2+az+b=0\\(1+i)^3+2(1+i)^2+a(1+i)+b=0\\1+i^3+3i(1+i)+2(1+i^2+zi)+ai+a+b=0$

$1-i+3i-3+2-2+4i+ai+a+b=0\\-2+a+b+6i+ai=0$

$-2+a+b+i(a+6)=0$ (1)

therefore a +6=0

a=-6

now put this in equation 1

b=8

$z^3+2z^2+az+b=0$

now putting value of a and b

$z^3+2z^2-6z+8=0$

now lets check if -4 is a root of the equation put z=-4

$(-4)^3+2(-4)^2-6(-4)+8=0\\-64+32+24+8=0\\0=0$

1. hence $z=−4$  is also a root of the equation.

(and 2,3 ,4 are incorrect)