Answer to Question #100005 in Algebra for Diamond Sturdevant

Question

Use the given zero to find the remaining zeros of the function.

f(x)=x^3-2x^2+36x-72, zero: 6i

Answer 1

f (x) = x³ - 2x² + 36x - 72 = 0, x₁ = 6i

Since there is no i in the equation, suppose that the root is -6i

f (x) = x³ - 2x² + 36x - 72 = 216i + 72 - 216i -72 = 0

The assumption turned out to be true: x₂ = -6i

(x - 6i)*(x + 6i) = x² - 6ix + 6ix + 36 = x² + 36

Divide the function f(x) by the product x²+36 of the roots

x³ -2x² + 36x - 72 | x²+36

x³ + 36x | x - 2

________

-2x² -72

-2x² -72

__

0

x³ - 2x² + 36x - 72 = (x - 6i)*(x + 6i)*(x - 2) = 0

x-2= 0

x₃ = 2

Answer:

x₁ = 6i; x₂ = -6i; x₃ = 2.