If $P(x) = x^4 - 2x^3 - 2x^2 + 8x - 8$, then
a.) Find all zeros of $P$, and state their multiplicities.
b.) Sketch the graph of $P$.
a.) To find the zeros of $P$, we apply synthetic division with the possible rational zeros of the factor of $8$ which are $\pm 1, \pm 2, \pm 4$ and $\pm 8$. Then,
By trial and error,
Again, by applying Synthetic Division
Thus,
$
\begin{equation}
\begin{aligned}
P(x) =& x^4 - 2x^3 - 2x^2 + 8x - 8
\\
\\
=& (x - 2)(x^3 - 2x + 4)
\\
\\
=& (x -2)(x + 2)(x^2 - 2x + 2)
\end{aligned}
\end{equation}
$
Therefore, rational zeros of $P$ are $2$ and $-2$. Also, all the zeros have multiplicity of $1$.
b.) To sketch the graph of $P$, we must know first the intercepts of the function. The values of the $x$ intercepts are the zeros of the function, that is $ 2$ and $-2$. Next, to determine the $y$ intercept, we set $x = 0$ so, $P(0) = ( 0 - 2)(0 + 2)(0^2 - 2(0) + 2) = (-2)(2)(2) = -8$.
Since the function has an even degree and a positive leading coefficient, then its end behavior is $y \to \infty$ as $x \to -\infty$ and $y \to \infty$ as $x \to \infty$. Then, the graph is
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