This will be a 4th degree polynomial. Our root of x = 7 in factorization form is (x-7). Our root of x = -11 in factorization form is (x+11) and the last one is a complex number. According to the conjugate root theorem, if we have 2+8i, we also HAVE to have 2-8i. In factorization form that first one is (x-(2+8i)) which simplifies to (x-2-8i). Its conjugate in factorization form is (x-2+8i). Now we will FOIL all that out. Let's start with the (x-2-8i)(x-2+8i). That multiplies out to [tex]x^2-2x+8ix-2x+4-16i-8ix+16i-64i^2[/tex]. We have to combine like terms here to shorten that a bit. [tex]x^2-4x+4-64i^2[/tex]. i^2 is equal to -1, and -1(64) = -64. Now we have [tex]x^2-4x+4-(-64)[/tex]. That is [tex]x^2-4x+68[/tex]. Now let's FOIL in another factorization. [tex](x^2-4x+68)(x-7)[/tex]. That comes out to [tex]x^3-11x^2+96x-476[/tex]. One more term to go! [tex]x^3-11x^2+96x-476)(x+11)[/tex]. That, finally, is [tex]x^4-25x^2+580x-5236[/tex].
