f(x) = x^3-x^2+8x+10 = 0
Given that one root is x1 = 3i+1.
To find the other roots.
We know that 3i+4 is a complex root. The complex roots always occur in pairs. If a+bi is a root, then its complex conjugate a-bi is also a root of an equation. Therefore x2 =1-3i is also a root.
Therefore (x-x1)(x-x2) = (x-1-3i) (x-4+3i) =(x-1)^2-(3i)^2 = (x^2-2x+10) is a factor of x^3-x^2+8x+10.
Therefore (x^3-x^2+8x+10) (x^2-2x+10)(x+k).... (1)
Put x = 0 in (1) : 10 = 10k. So k = 1.
Also put x= -1 in x^3-x^2+8x+10. Then -1-1-8+10 = 0.
Therefore x= -1 is a root.
Thus the roots (or solutions) of x^3-x^2+8x+10 = 0 are 3i+1, -3i+1 and -1.
Since the equation has a complex root, we'll 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.
So, we have as root the number z = a + bi => z' = a - bi is also the root of the equation.
Now, we'll verify if the complex numbers is the root of the equation by substituting it into the original equation.
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.