Determine the expression of the equation 2x^3-x^2+ax+b=0 if a solution is 1+i .

In other words, we'll have to determine a and
b.

The equation has 3 solutions and 2 of them are complex,
the 3rd solution being a real one.

If the equation has has
a complex solution, that means that the conjugate of the complex solution is also a
solution for equation.

So, the given equation has as
solutions:

x1 = 1 + i and x2 = 1 -
i

The polynomial 2x^3-x^2+ax+b is divided by (x - 1 - i)(x
-1 + i).

P(1 + i) = 2(1+i)^3 - (1+i)^2 + a(1+i) + b =
0

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

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

P(1 + i) = -4 + 4i - 2i + a + ai + b =
0

-4 + a + b + i(a + 2) = 0

-4
+ a + b = 0

a + b = 4

a + 2 =
0

a = -2

b = 4 -
a

b = 4 + 2

b =
6

The expression of the equation is: 2x^3 -
x^2 - 2x + 6 = 0.