Answer:
[tex]f(x) = -5x^{4} -5x^{3} +25x^{2} -5x+30[/tex]
Step-by-step explanation:
A fourth degree polynomial is of the form
f(x) = K(x-a)(x-b)(x-c)(x-d)
Where K is a constant and a,b,c, and d are roots of the equation.
We have three roots already: -3, 2, i
Because we are told that the coefficients are real and we have an imaginary zero, i, we need to obtain its conjugate which is = -i
So, the four roots are: a = -3, b = 2, c = i, d = -i
f(x) = K(x-(-3))(x-2)(x-i)(x-(-i))
f(x) = K(x+3)(x-2)(x-i)(x+i)
[tex]f(x)=K(x^{2} -2x+3x-6)(x^{2} -ix+ix-i^{2})[/tex]
But [tex]i^{2} = -1[/tex]
[tex]f(x) = K(x^{2}+x-6)(x^{2} +1)\\\\f(x) = K(x^{4}+x^{3} -6x^{2} + x^{2} +x-6)\\\\f(x) = K(x^{4} +x^{3} - 5x^{2} +x-6)[/tex]
The polynomial passes through (-2, 100). That is, at x = -2, f(x) = 100
[tex]100 = K[(-2)^{4}+(-2)^{3} -5(-2)^{2}+(-2)-6]\\100 = K(16-8-20-2-6)\\100 = -20K\\\\K = -5[/tex]
[tex]f(x) = -5(x^{4} +x^{3} - 5x^{2} +x-6)\\f(x) = -5x^{4} -5x^{3} +25x^{2} -5x+30[/tex]
