• First way to solve:
We'll manipulate the expression of the equation:
[tex]
y=x^3-3x^2+16x-48\\\\
y=x^2(x-3)+16(x-3)\\\\
y=(x-3)(x^2+16)[/tex]
If we have y=0:
[tex] x-3=0~~~or~~~x^2+16=0\\\\
x=3~~~or~~~x^2=-16\\\\
x=3~~~or~~~x=\pm\sqrt{-16}\\\\
x=3~~~or~~~x=\pm4i
[/tex]
Then, the function has one real zero (x=3) and two imaginary zeros (4i and -4i).
Answer: B
• Second way to solve:
The degree of the function is 3. So, the function has 3 complex zeros.
Since the coefficients of the function are reals, the imaginary roots are in a even number (a imaginary number and its conjugated)
The function "has only one non-repeated x-intercept", then there is only one real zero.
The number of zeros is 3 and there is 1 real zero. So, there are 2 imaginary zeros.
Answer: B.
