Solved eq. x^3-x^2+8x+10=0 and had solutions 3i+1, 1-3i is correct?

We recall the property of complex roots: If a complex
number is the root of an equation, then it's conjugate is also the root of the
equation.

Fot the beginning, you'v get as roots 2 complex
numbers: z and z'(z' is the conjugate of z).

z = a +
bi

z' = a - bi

Now,l we'll
verify if the complex numbers are the roots of the equation by substituting them into
the original equation.

(1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 =
0

We'll expand the cube using the
formula:

(a+b)^3 = a^3 + b^3 +
3ab(a+b)

a = 1 and b =
3i

(1+3i)^3 = 1^3 + (3i)^3 +
3*1*3i*(1+3i)

(1+3i)^3 = 1 - 27i +
9i(1+3i)

We'll remove the
brackets:

(1+3i)^3 = 1 - 27i + 9i -
27

We'll combine real parts and imaginary
parts:

(1+3i)^3 = -26 +
i(9-27)

(1+3i)^3 = -26 -
18i

We'll expand the square using the
formula:

(a+b)^2 = a^2 + 2ab +
b^2

(1+3i)^2 = 1^2 + 2*1*3i +
(3i)^2

(1+3i)^2 = 1 + 6i -
9

(1+3i)^2 = -8 + 6i

We'll
substitute the results into the expression (1+3i)^3 - (1+3i)^2 + 8(1+3i) + 10 =
0.

 -26 - 18i - (-8 + 6i) + 8 + 24i + 10 =
0

We'll combine like
terms:

(-26 + 8 + 8 + 10) + i(-18 - 6 + 24) =
0

0 + 0*i = 0

It
is obvious that 1 + 3i is the root of the
equation.

According to the rule, the
conjugate of 1 + 3i, namely 1 - 3i, is also the root of the equation. So it is not
necessary to verify if 1 - 3i is the root, since we've demonstrated that 1 + 3i is the
root of the equation.

Conclusion: 1 + 3i and
1 - 3i are the roots of the
equation

 x^3-x^2+8x+10=0

Note:
In calculus, we've substituted i^2 by it's value,
-1.