Write an equation in standard form that has the following zeros. x = –3, 2, –3i

Question

ive tried a bunch to solve it but i just dont know how

Answer 1

-3, 2 -3i

you have to add a 4th zero, the conjugate of -3i = +3i (just change the sign)

then the factors are just sticking an x in front of each zero with their sign changed.

(x+3)(x-2)(x^2+9) multiply them out to get

(x^2 -12x -6)(x^2+9) = x^4 -12x^3 +3x^2 -108x -54

or keep it in factored form if you want where you can read off the zeros

f(x) = y = (x+3)(x-2)(x^2+9) = (x+3)(x-2)(x-3i)(x+3i)

equation is

f(x) = y = x^4 -12x^3 + 3x^2 -108x -54