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
Write a polynomial equation with roots 5 and -9i
x^3-___x^2+___x-___=0
3 answers
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
(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
You are correct. The polynomial equation with roots 5 and -9i is:
x^3 - 5x^2 + 81x - 405 = 0
x^3 - 5x^2 + 81x - 405 = 0
1 answer