find a third degree polynomial equation with rational coefficants that has roots -4 and 6+i
To find a third degree polynomial equation with rational coefficients that has roots -4 and 6+i, we know that complex roots always come in conjugate pairs. Therefore, the conjugate of 6+i is 6-i.
Using the fact that the sum and product of the roots of a polynomial equation can be found using Vieta's formulas, let's construct the equation:
The roots are -4, 6+i, and 6-i.
The sum of the roots is: -4 + 6 + i + 6 - i = 8.
The product of the roots is: (-4) * (6 + i) * (6 - i) = (-4)(36 - 1) = -140.
A polynomial equation can be expressed as: (x - root1)(x - root2)(x - root3) = 0.
Expanding this equation, we have:
(x - (-4))(x - (6 + i))(x - (6 - i)) = 0
(x + 4)(x - 6 - i)(x - 6 + i) = 0
Multiplying this out gives us:
(x + 4)(x^2 - 12x + 37) = 0
Now, we can expand further:
x^3 - 8x^2 - 5x + 148 = 0
So, a third degree polynomial equation with rational coefficients that has roots -4 and 6+i is: x^3 - 8x^2 - 5x + 148 = 0.