Use the given zero to find the remaining zeros. h(x)=3x^4+10x^3+19x^2+90x-73 zero:-3i
There’s an error in your quartic. It should be: h(x)=3x^4+10x^3+19x^2+90x-72
You are given that one of the zeros is -3i.
i.e. one of the factors is (x + 3i)
If a polynomial has a complex factor, or root, then it must have a 2nd root which is a complex conjugate of the 1st root.
i.e. a 2nd factor is (x – 3i).
Since both (x + 3i) and (x – 3i) are factors of h(x) then so also is their product,
Which is (x + 3i)(x – 3i) = x^2 + 9
Which means that (x^2 + 9) divides into (3x^4+10x^3+19x^2+90x-72).
The result of this division is: 3x^2 + 10x – 8
I.e. (x^2 + 9)(3x^2 + 10x – 8) = 3x^4+10x^3+19x^2+90x-72
The factor (3x^2 + 10x – 8) factorises as (3x – 2)(x + 4)
Giving roots: x = 2/3 and x = -4.
The four roots are: x = 3i, x = -3i, x = 2/3, x = -4