Use the given zero to completely factor P(x) into linear factors. Zero: 5i; p(x)=x^4-x^3+23x^2-25x-50

Note that if [tex] z [/tex] is a complex solution of a real-coefficient polynomial, then also the conjugate [tex]\overline{z}[/tex] is a solution of the polynomial.So, both [tex] \pm5i [/tex] are solutions of this polynomial, which in turn means that the polynomial can be divided by[tex](x-5i)(x+5i)=x^2+25 [/tex]So, if we long divide[tex]\dfrac{x^4-x^3+23x^2-25x-50}{x^2+25} = x^2-x-2[/tex]So far, we can factor our polynomial as[tex] x^4-x^3+23x^2-25x-50 = (x^2+25)(x^2-x-2) [/tex]But we already know that [tex](x-5i)(x+5i)=x^2+25 [/tex], so our factorization actually looks like this:[tex] x^4-x^3+23x^2-25x-50 = (x-5i)(x+5i)(x^2-x-2) [/tex]We only need to factor the second parenthesis and we're done: the solutions to [tex] x^2-x-2=0 [/tex] are [tex]x=-1[/tex] and [tex]x=2[/tex]. So, the complete factorization is[tex] x^4-x^3+23x^2-25x-50 = (x-5i)(x+5i)(x+1)(x-2) [/tex]Note that we can't go any further, because all factors have degree one, and are factorization "atoms". Or, if you prefer, the fundamental theorem of algebra states that, if we use complex number, a polynomial of degree [tex]n[/tex] has [tex]n[/tex] solutions, and we found them all, and thus there can't be more.