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

$\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$

4 0

3 years ago

Where does a Sine Graph start if it is Postive?<br> -Midline<br> -Min<br> -Max

Luden [163]

The sin graph starts at zero,

8 0

3 years ago