(a) Factor $P(x) = x^3 - 2x - 4$ into linear and irreducible quadratic factors with real coefficients. (b) Factor $P$ completely into linear factors with complex coefficients.
a.) The possible rational zeros of $P$ are the factors of 4 which are $\pm 1, \pm 2$ and $\pm 4$. Then by using synthetic division and trial and error,
Thus,
$P(x) = (x-2)(x^2 +2x + 2)$
The factor $x^2 + 2x + 2$ is irreducible, since it has no real zeros.
b.) To get the complete factorization, we use quadratic formula to get the complex zeros of the quadratic equation,
$\displaystyle x= \frac{-2\pm \sqrt{2^2 - 4(1)(2)}}{2(1)} = - 1 \pm i$
Thus,
$
\begin{equation}
\begin{aligned}
P (x) &= (x-2) [ x- ( -1 + i) ] [ x - (-1 - i) ]\\
\\
&= (x - 2) [ x + (1 - i) ] [ x + (1 + i)]
\end{aligned}
\end{equation}
$
Thursday, February 7, 2013
College Algebra, Chapter 4, 4.5, Section 4.5, Problem 64
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