Find all of the zeros in the function. Use a comma to separate answers. Use I to express complex numbers.
y=x^3-3x^2+16x-48
To find the zeros of the function y = x^3 - 3x^2 + 16x - 48, we set y equal to zero and solve for x:
0 = x^3 - 3x^2 + 16x - 48
This equation can be factored by grouping:
0 = x^2(x - 3) + 16(x - 3)
Now, we can factor out the common binomial (x - 3):
0 = (x - 3)(x^2 + 16)
Setting each factor equal to zero:
x - 3 = 0 or x^2 + 16 = 0
From the first equation, x = 3.
To solve the second equation, we can rearrange it as follows:
x^2 = -16
Taking the square root of both sides, we get:
x = ±√(-16)
Since the square root of a negative number is not a real number, we express the solutions as complex numbers with the imaginary unit i:
x = ±4i
Therefore, the zeros of the function y = x^3 - 3x^2 + 16x - 48 are x = 3, x = 4i, and x = -4i.