Question

If f(x)=2x^3-16x^2+28x-24 and f(6)=0, then find all of the zeros of f(x) algebraically.

Answer 1

6, 1 - i, and 1 + i

Step-by-step explanation:

The given function is f(x) = 2x³ - 16x² + 28x - 24. To find all the zeros, we first need to factorize the function. If f(6) = 0, them (x - 6) is a factor, so we can find the other factor dividing 2x³ - 16x² + 28x - 24 by (x - 6) as follows

Therefore,

f(x) = 2x³ - 16x² + 28x - 24

f(x) = (x - 6)(2x² - 4x + 4)

So, the zeros are the number that satisfied

f(x) = (x - 6)(2x² - 4x + 4) = 0

Then,

x - 6 = 0

x = 6

or

2x² - 4x + 4 = 0

To solve 2x² - 4x + 4 = 0, we will use the quadratic equation, so the solutions are

`\begin{gathered} x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(2)(4)}}{2(2)} \\ x=\frac{4\pm\sqrt[]{-16}}{4} \\ x=(4\pm4i)/(4)=1_{}\pm i \\ \text{Then} \\ x=1+i \\ or \\ x=1-i \end{gathered}`

Therefore, the zeros of f(x) are x = 6, x = 1 - i and x = 1 + i