Question

Find the cubic equation with integral coefficients, some of whose roots are 3 and 2i.

a) x^3 - 9x^2 + 27x - 18
b) x^3 + 9x^2 - 27x - 18
c) x^3 - 9x^2 - 27x + 18
d) x^3 + 9x^2 + 27x + 18

Answer 1

To find the cubic equation with integral coefficients that has roots 3 and 2i, we can use Vieta's formulas to solve for the coefficients. The resulting equation is x^3 - 9x^2 + 27x - 18.

Step-by-step explanation:

The cubic equation with integral coefficients, some of whose roots are 3 and 2i can be found using the fact that complex roots occur in conjugate pairs. Therefore, the other root must be the conjugate of 2i, which is -2i.

To find the cubic equation, we can set up the equation using Vieta's formulas, which relate the coefficients of a polynomial to its roots. The sum of the roots is given by -b/a, the product of the roots is given by c/a, and the sum of the products of pairs of roots is given by -d/a. We can substitute the known values into these formulas and solve for the coefficients.

Using the given roots of 3, -2i, and 2i, we find that the cubic equation with integral coefficients is: x^3 - 9x^2 + 27x - 18, which corresponds to option (a).