To create a polynomial equation of degree 3 with roots 2, ‘a + bi’, and its conjugate ‘a – bi’, we need to follow these steps:
1. **Identify the roots**: We have one real root, which is 2, and two imaginary roots: let’s say they are ‘a + bi’ and ‘a – bi’ (the conjugate pair).
2. **Construct the polynomial using the roots**: The polynomial can be formed using the factors derived from the roots:
- For the root 2, the factor is (x – 2).
- The factors for the complex roots ‘a + bi’ and ‘a – bi’ are (x – (a + bi)) and (x – (a – bi)).
This leads us to the polynomial:
Polynomial: (x – 2) * (x – (a + bi)) * (x – (a – bi))
3. **Multiply out the factors**: To get the polynomial in standard form, we need to expand this expression.
The product of the complex factors can be simplified using the formula for multiplying conjugates:
(x – (a + bi))(x – (a – bi)) = (x – a)^2 + b^2
So, we have:
Polynomial: (x – 2) * ((x – a)^2 + b^2)
4. **Final Form**: Now let’s choose specific values for ‘a’ and ‘b’. If we take ‘a = 1’ and ‘b = 1’, the roots would then be 2, 1 + i, and 1 – i. The polynomial becomes:
((x – 2) * ((x – 1)^2 + 1))
5. **Expand this**: When you expand this polynomial, you get:
(x – 2) * (x^2 – 2x + 2) = x^3 – 4x^2 + 6x – 4
So, the polynomial equation of degree 3 where the roots are 2 and an imaginary number (along with its conjugate) is:
x^3 – 4x^2 + 6x – 4 = 0