Answer:
6, 1 - i, and 1 + i
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
[tex]\begin{gathered} x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(2)(4)}}{2(2)} \\ x=\frac{4\pm\sqrt[]{-16}}{4} \\ x=\frac{4\pm4i}{4}=1_{}\pm i \\ \text{Then} \\ x=1+i \\ or \\ x=1-i \end{gathered}[/tex]Therefore, the zeros of f(x) are x = 6, x = 1 - i and x = 1 + i
