find a third degree polynomial equation with coefficients that has roots -4 and 6+i

possible answers
x^3-8x^2-11x+148=0
x^3-8x^2-12x+37=0
x^3-12x^2+37x=0
x^3-8x^2-11x=0

The answer is

A. X^3-8x^2-11x+148=0

PLease explain

To find a third-degree polynomial equation with the given roots, we can use the fact that complex roots come in conjugate pairs. Given that one of the roots is 6 + i, we know that the conjugate root is 6 - i.

To find the polynomial equation, we can use the following steps:

1. Start by setting up the factors using the roots:
(x - (-4)) * (x - (6 + i)) * (x - (6 - i)) = 0

2. Simplify:
(x + 4) * (x - 6 - i) * (x - 6 + i) = 0

3. Expand the expression:
(x + 4) * ((x - 6) - i) * ((x - 6) + i) = 0

4. Use the difference of squares formula to simplify:
(x + 4) * ((x - 6)^2 - i^2) = 0
(x + 4) * ((x - 6)^2 + 1) = 0

5. Expand and simplify further:
(x + 4) * (x^2 - 12x + 36 + 1) = 0
(x + 4) * (x^2 - 12x + 37) = 0

So, the third-degree polynomial equation with roots -4 and 6+i is:
x^3 - 12x^2 + 37x + 148 = 0

Therefore, the correct answer from the given options is:
x^3 - 12x^2 + 37x = 0