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?

Step 1:
Try $\pm1$, they work so we can reduce the problem to a cubic by dividing by $(x-1)(x+1)=x^2-1$:
$$
P(x)=(12x^5+16x^4-17x^3-19x^2+5x+3)/(x^2-1)\\=12 x^3 + 16 x^2 - 5 x - 3
$$
Step 2: To find a 3rd solution we find the extrema of $P(x)$ we solve
$$
P'(x)=36x^2-32x-5=0
$$
Finding that it has 2 solutions at $x=-1.02..$ and $x=0.135..$. The first one is a maximum and the $P$ is positive there so there must be a zero of $P(x)$ to the left of $-1.02...$ and it must be $-3/2$.
Step 3: Now we can divide by $x+3/2$, to get a quadratic polynomial:
$$
P(x)/(x+3/2)=12x^2-12x-2
$$
whose zeros are the remaining 2 zeros: $-1/3$ and $1/2$.
