Question

The polynomial function y = x3 -3x2 + 16x - 48 has only one non-repeated x-intercept. What do you know about the complex zeros of the function?
A)
The function has one imaginary zero.


B)
The function has one real zero and two imaginary zeros.


C)
The function has one real zero and one imaginary zero.


D)
The function has one imaginary zero and two real zeros.

Answer 1

• First way to solve:
We'll manipulate the expression of the equation:

$y=x^3-3x^2+16x-48\\\\ y=x^2(x-3)+16(x-3)\\\\ y=(x-3)(x^2+16)$

If we have y=0:

$x-3=0~~~or~~~x^2+16=0\\\\ x=3~~~or~~~x^2=-16\\\\ x=3~~~or~~~x=\pm\sqrt{-16}\\\\ x=3~~~or~~~x=\pm4i$

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.