Answer:
x ∈ {-1, 3, 3i, -3i}
Explanation:
Polynomials with real coefficients have complex roots in conjugate pairs. That means -3i is also a solution to this quartic. Then two factors are ...
(x -3i)(x +3i) = (x^2 +9)
Dividing the given polynomial by this using your favorite method, you find the other quadratic factor to be ...
(x^2 -2x -3) = (x +1)(x -3)
The solutions are the values of x that make these factors zero, so are x = -1, and x = 3.
The full set of solutions to this quartic is ...
x ∈ {-1, 3, -3i, 3i}
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Comment on dividing by x^2+9
Since there is no x term in this divisor, the first two terms of the quotient will be (x^4 -2x^3)/x^2 = x^2 -2x. The last term of the quotient will be -27/9 = -3. Thus, we can do the long division with hardly any effort to find ...
x^4 -2x^3 +6x^2 -18x -27 = (x^2 -2x -3)(x^2 +9)
= (x +1)(x -3)(x -3i)(x +3i)
The above makes the assumption that there is no remainder, a reasonable assumption for a problem of this nature. In any event, checking the factorization by multiplying it out shows it to be correct.