i need help on this i am stuck

Question

using the imaginary zero -6i The question is. "What is the complex conjugate for your number? Explain how you know.

and using the real zeros 4,-1,-6. It says "Assign each of your real zeros a different multiplicity list all your zeros below with their multiplicities."

Answer 1

conjugate zero for -6i is +6i

factors are (x-6i)(x+6i) = x^2+36

f(x) = x^2+36

set it = 0 and solve for x to find the zeros

x^2+36 = 0

x^2 =-36

x = + or - sqr(-36) = + or - 6i = +6i or -6i

If you graphed the equation, it would never intersect the x-axis, thus there are no real zeros, just imaginary ones.

IF you have zeros 4, -1,-6, then factors are (x-4)(x+1)(x+6) which = 0 when x=4, -1, -6

multiplicity just puts an exponent over any of the factors, such as

(x-4)^2(x+1)^3(x+6)^4 or any exponents you prefer

Multiplicity means that graphically the curve does not intersect the x axis, but instead just touches it, then changes direction