<s>[INST] Solve the equation $12x^5+16x^4-17x^3-19x^2+5x+3=0$ Solve the equation $$12x^5+16x^4-17x^3-19x^2+5x+3=0$$ The divisors of $3$ are $\pm1;\pm3$ and the divisors of $12$ are $\pm1;\pm2;\pm3;\pm4;\pm6;\pm12$, so the possible rational roots are $$\pm1;\pm\dfrac12;\pm\dfrac13;\pm\dfrac14;\pm\dfrac16;\pm\dfrac{1}{12};\pm3;\pm\dfrac32;\pm\dfrac34$$ which makes $22$ possible roots. We can use Horner, but how can I reduce the number of possible roots?
 [/INST] We first try the simplest $x= \pm 1$, then $12+16-17-19+5+3=0 \Rightarrow x-1 $is a factor and $-12+16+17-19-5+3=0 \Rightarrow x+1 $ is also a factor.
By synthetic division, we have

$$
\begin{aligned}12 x^{5}+16 x^{4}-17 x^{3}-19 x^{2}+5 x+3 =(x+1)(x-1)\left(12 x^{3}+16 x^{2}-5 x-3\right)
\end{aligned}
$$
Now we can try $x=\dfrac{1}{2},
12\left(\frac{1}{2}\right)^{3}+16\left(\frac{1}{2}\right)^{2}-5\left(\frac{1}{2}\right)-3=0 \Rightarrow x-\frac{1}{2}$ is also a factor.
By synthetic division again,
$$
\begin{aligned}
& 12 x^{5}+16 x^{4}-17 x^{3}-19 x^{2}+5 x+3\\
=&(x+1)(x-1)(2 x-1)\left(6 x^{2}+11 x+3\right) \\
=&(x+1)(x-1)(2 x-1)(3 x+1)(2 x+3)
\end{aligned}
$$
Therefore the roots of the equation are $\pm 1,\dfrac{1}{2} ,-\dfrac{1}{3} \text{ and }-\dfrac{3}{2}.$
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