Mathematics, 08.12.2019 09:31 sam1625
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)
Answers: 3
![\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)\\\\ -----------------------------](/tpl/images/0408/8130/63139.png)
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Find the nth degree polynomial function with real coefficients satisfying the given conditions.
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