Given -6i is a root, determine all other roots of f(x)=x³

Question

Given -6i is a root, determine all other roots of f(x)=x³ - 2x² + 36x - 72.

Quizplus · Accepted Answer

Since -6i is a root, its conjugate 6i must also be a root (since complex roots always come in conjugate pairs). From there, we can use polynomial division to find the quadratic factor that corresponds to these two roots: (x + 6i)(x - 6i) = x^2 + 36 So we can rewrite the original polynomial as: f(x) = (x^2 + 36)³ - 2(x^2 + 36)² + 36x - 72 = (x^2 + 36)(³ - 2²) + 36(x - 2) Since the quadratic factor has no real roots, the only possible rational roots of the polynomial are the divisors of 36. Testing these values, we find that x = 2 is a root, which means that (x - 2) is a linear factor of the polynomial. To find the remaining quadratic factor, we can use polynomial division again: (x^2 + 36)(³ - 2²) + 36(x - 2) = (x - 2)(x^2 + 6x + 18)(³ - 2²) + 0 So the roots of f(x) are -6i, 6i, and -2 - 3i, which corresponds to the choice A).

Given -6i Is a Root, Determine All Other Roots of F(x)=x3