Answer
$P(x)=(x-2)(x-5)(x+3)^{2}$
Zeros: $ \ -3,\ 2, \ 5. \ $
Work Step by Step
$P(x)=x^{4}-x^{3}-23x^{2}-3x+90$
$P(-x)=x^{4}+x^{3}-23x^{2}+3x+90$
Descart's rule of signs:
P(x) has 2 sign changes $\Rightarrow$ 2 or 0 positive zeros.
P(-x) has 2 sign changes $\Rightarrow$ 2 or 0 negative zeros.
Rational Zeros Theorem:
candidates for p:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 45,\pm 90$
candidates for q:$\quad \pm 1,$
Possible rational zeros $\displaystyle \frac{p}{q}$:$\quad \pm 1,\pm 2,\pm 3,\pm 5,\pm 6,\pm 9,\pm 10,\pm 15,\pm 30,\pm 45,\pm 90$
Testing with synthetic division,
$\left.\begin{array}{l}
2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1 & -1 & -23 & -3 & 90 \\\hline
& 2 & 2 & -42 & -90\\\hline
1& 1 & -21 & -45&|\ \ 0\end{array}$
$P(x)=(x-2)(x^{3}+x^{2}-21x-45)$
Again, testing with synthetic division,
$\left.\begin{array}{l}
5 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrr}
1& 1 & -21 & -45\\\hline
& 5 & 30& 45\\\hline
1& 6 & 9 &|\ \ 0\end{array}$
$P(x)=(x-2)(x-5)(x^{2}+6x+9)$
... recognize a perfect square
$P(x)=(x-2)(x-5)(x+3)^{2}$
Zeros: $ \ -3,\ 2, \ 5. \ $
