Question

Find the nth degree polynomial function with real coefficients satisfying the given conditions.

n=4
-1,3 and 2+4i are zeros
f(1)=-68

Can someone show steps to solve this?
this is what i have so far
(x+1)(x-3)(x-2-4i)(x-2+4i)

Answer 1

`\bf (x+1)(x-3)(x-2-4i)(x-2+4i) \\\\ \textit{now, let's take a peek at }(x-2-4i)(x-2+4i)\\ \textit{and recall our }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -----------------------------\\\\ (x-2-4i)(x-2+4i)\implies [(x-2)-(4i)][(x-2)+(4i)] \\\\\ [(x-2)^2-(4i)^2]\implies [(x^2-2x+4)-(4^2\boxed{i^2})] \\\\\ [(x^2-2x+4)-(16\cdot \boxed{-1})]\implies [(x^2-2x+4)+16] \\\\ (x^2-2x+20)\\\\ -----------------------------`


`\bf thus \\\\ \begin{array}{llll} (x+1)(x-3)\\(x-2-4i)(x-2+4i) \end{array}\implies (x+1)(x-3)(x^2-2x+20) \\\\\\ (x^2-2x-3)(x^2-2x+20)\implies \begin{array}{llll} x^4-2x^3+20x^2-2x^3\\+4x^2-40x+3x^2-6x+60 \end{array} \\\\\\ x^4-4x^3+27x^2-46x+60`