A point in Quadrant I has all negative values
1 answer:
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• First way to solve:
We'll manipulate the expression of the equation:
If we have y=0:
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.
We'll manipulate the expression of the equation:
If we have y=0:
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.