zeros mean in the numerator you have...
(x-8)*(x+6+i)
that is degree two.
Now real numbers mean the complex cannot be allowed, so one adds the conjugate of the complex, so
y(x)=(x-8)*(x+6+i)*(x+6-i)
Please help!! I do not understand any of this!!
Information is given about a polynomial f(x) whose coefficients are real numbers. Find the remaining zeros of f.
Degree 3; zeros: 8, -6-i
2 answers
i need help i have a polynomial f(x) whose coefficeint are real numbers. Find the remaining zeros of f. Degree 3, zeros -4,-2,-i