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Maksim231197 [3]
2 years ago
10

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

Mathematics
1 answer:
RUDIKE [14]2 years ago
4 0
\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
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Answer:

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Zeros of a polynomial is the values at which the polynomial becomes zero. They are also called the roots of the polynomial.

When (x - a)(x - b) = 0, we can say that either (x - a) = 0 or (x - b) = 0. At least one zero renders the whole equation to be zero.

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