find a third degree polynomial equation with rational coefficients that has roots -4 and 6+i
A cubic equation always has three roots.
These three roots are:
1) three real roots
2) 1 real root and two complex roots
If one of the two complex roots is a + ib, then the other complex root is a - ib.
We are given two of the roots as -4 and 6+i.
Since one of the roots is complex and equals 6+i, the the other complex root is 6-i.
Our three roots then are: -4, 6+i, 6-i.
Our three solutions to the cubic equation are: x = -4, x = 6+i, x = 6-i.
Which can be rewritten as: x + 4 = 0, x - (6+i) = 0, x - (6-i) = 0
Multiplying these together gives us the original cubic equation.
(x + 4)(x - (6+i))(x - (6-i)) = 0
Multiplying this out,
(x + 4)(x^2 - (6-i)x - (6+i)x + (6-i)(6+i)) = 0
(x + 4)(x^2 - 12x + 6^2 - i^2) = 0
(x + 4)(x^2 - 12x + 37) = 0
x^3 - 8x^2 - 11x +148 = 0
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