Find all rational, irrational and complex zeros (and state their multiplicities) of the polynomial function $P(x) = 18x^3 + 3x^2 - 4x - 1$. Use Descartes' Rule of signs, the Upper and Lower Bounds Theorem, the Quadratic Formula or other factoring techniques.
The possible rational zeros of $P$ are the factors of $1$ divided by the factors of $18$ which are $\displaystyle \pm 1, \pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{1}{6}, \pm \frac{1}{9}$ and $\displaystyle \pm \frac{1}{18}$. Then, by applying Synthetic Division and trial and error,
Again, by applying synthetic division
Thus,
$
\begin{equation}
\begin{aligned}
P(x) =& 18x^3 + 3x^2 - 4x - 1
\\
\\
=& \left(x + \frac{1}{3} \right) (18x^2 - 3x - 3)
\\
\\
=& \left(x + \frac{1}{3} \right) \left(x + \frac{1}{3} \right) (18x - 9)
\\
\\
=& \left(x + \frac{1}{3} \right)^2 (18x - 9)
\end{aligned}
\end{equation}
$
Thus, the zeros of $P$ are $\displaystyle \frac{-1}{3}$ and $\displaystyle \frac{1}{2}$. The zeros have multiplicity of $2$ and $1$ respectively.
No comments:
Post a Comment