Question

Write a polynomial equation with roots 5 and -9i

x^3-___x^2+___x-___=0

Answer 1

I get
(x - 5)(x + 9i)(x - 9i) = 0
(x-5) (x^2 + 81) = 0
x^3+ 81 x -5 x^2 - 405 = 0
or
x^3 -5 x^2 + 81 x - 405 = 0

Answer 2

To form a polynomial equation with the given roots, we know that the roots must be in the form (x - a) where a is each individual root.

Since we are given two roots: 5 and -9i, we can set up the equation as follows:

(x - 5)(x - (-9i))(x - (9i))

Now, simplify each factor:

(x - 5)(x + 9i)(x - 9i)

Expand the equation using the FOIL (First, Outer, Inner, Last) method:

(x * x) + (x * -9i) + (-5 * x) + (-5 * -9i) + (9i * x) + (9i * -9i)

This simplifies to:

x^2 - 9ix - 5x + 45i - 9ix + 81i^2

Since i^2 is equal to -1, this can be further simplified to:

x^2 - 9ix - 5x + 45i - 9ix + 81(-1)

x^2 - 9x^2 - 5x + 45i - 9ix - 81

Finally, rearranging the terms:

x^3 - 14x^2 - 14ix - 81x + 45i - 81 = 0

Therefore, the polynomial equation is:

x^3 - 14x^2 - (81 + 14i)x + (45i - 81) = 0

Answer 3

You are correct. The polynomial equation with roots 5 and -9i is:

x^3 - 5x^2 + 81x - 405 = 0