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

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### 2 Answers

We have the equation: 2x^3 - x^2 + ax + b = 0 with one of its solutions given as 1 + i.

Substituting x with 1 + i,

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

=> 2*( 1 + 3i^2 + 3i + i^3) - 1 - i^2 - 2i + a + ai + b = 0

=> 2*( 1 - 3 + 3i - i) - 1 + 1 - 2i + a + ai + b = 0

=> 2*( - 2 + 2i ) - 2i + a + ai + b = 0

=> -4 + 4i - 2i + a + ai + b = 0

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

Equating the real and complex coefficients, we have

-4 + a + b = 0

2 + a = 0

=> a = -2

substituting a = 6 in -4 + a + b = 0

=> -4 - 2 + b = 0

=> b = 6

**The required equation is 2x^3 - x^2 - 2x + 6 = 0**

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.**