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?
 A: 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 $1/2$.
A: Foreword:
I don't have what I would consider a "complete solution." Rather, some discussion on how to narrow down the candidates you get from the rational root theorem. The end result is still more tedious arithmetic, but it'll be a little better than checking $22$ evaluations of a quintic polynomial by (presumably) hand.
In brief, my "solution" goes as so:

*

*A quick note that $\pm 1$ are roots simply on inspection. (Nice for polynomials in integer coefficients.)

*Descartes' rule of signs. (Basically tells you how many roots are of a certain sign.)

*Sturm's Theorem. (Uses calculus but lets you easily narrow down the intervals on which roots lie, just by checking the sign alternations in a sequence.)

Maybe there's a more elegant method you can use, but that's something I'll leave to others, because none come to mind for me.
A warning should also be given in that the approach enlisted generally only works if you have only rational roots. Finding irrational roots would usually prove a lot more troublesome with these ideas alone. Luckily the polynomial in question only has rational roots, but we shouldn't assume such without a basis that isn't WolframAlpha.

A brief initial attempt - Inspection:
While not the most enlightening answer, whenever $\pm 1$ is a possibility, it is worth trying it at least since it's easy to do the arithmetic for with polynomials like this. You can then reduce this one to a cubic of the form
$$12x^3 + 16x^2 - 5x - 3$$
From here the rational root theorem gives that the roots are possibly... well, sadly, still of the same forms, and clearly $\pm 1$ were eliminated as possibilities.
So this won't give the most pleasing answer on its own, but the arithmetic will be easier. Or you could just go straight to using the cubic formula, but that would be less than pleasant.
Depending on your personal thought, we could stop here. But let's go further.

Descartes' Rule of Signs:
One aide would be Descartes' rule of signs (Wikipedia). Consider the polynomial
$$p(x) = \sum_{n=0}^N a_n x^n = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0$$
and define
$$q(x) = \sum_{n=0}^N (-1)^n a_n x^n$$
Consider the number of changes of sign between consecutive terms of $p$ when written in the form above (descending powers of $x$). Call it $s(p)$. Then:

*

*The number of positive roots is equal to $s(p)$, $s(p) - 2$, $s(p) - 4$, $\cdots$, etc., and as few as zero or one.

*The number of negative roots is equal to $s(q)$, $s(q) - 2$, $\cdots$, etc.

For notational purposes, let $n_+(p)$ be the number of positive roots of $p$, and $n_-(p)$ the number of negative roots of $p$.
Now, with this in mind, consider our original quintic, $f$, and the derived cubic, $g$:
$$\begin{align*}
f(x) &= 12x^5+16x^4-17x^3-19x^2+5x+3 \\
g(x) &= 12x^3 + 16x^2 - 5x - 3
\end{align*}$$
Then we have the following:
$$
n_+(f) \in \{0,2\} \qquad
n_-(f) \in \{1,3\} \qquad
n_+(g) = 1 \qquad
n_-(g) \in \{0,2\}$$
As with inspection, this isn't a ton on its own, but it can make testing values slightly more efficient depending on the order you do them in.

Sturm's Theorem:
$\newcommand{\set}[1]{\left\{#1\right\}}$
$\newcommand{\SS}{\mathcal{S}}$
For this, we reference Sturm's theorem. It is a method that requires calculus, but we'll just have to make do. Combined with the rational root theorem, and a reduction to the cubic equation, with Descartes' rule slightly making things more efficient, this will definitely be a bit of a kicker.
Define a sequence of polynomials $\SS(p) := \set{P_i}_{i=0}^{\deg(p)-1}$ to be the Sturm sequence for $p(x)$ by the following rules:
$$P_0 = p \qquad P_1 = p' \qquad P_i = -\text{rem}(P_{i-1},P_{i-2}) \text{ for } i \ge 2$$
where for $P_i$ we are taking the remainder on Euclidean division of $P_{i-1}/P_{i-2}$.
We then define a function $V_{\SS(p)}(\xi)$ which encodes the number of sign variations in the Sturm sequence $\SS$ when each member is evaluated at $\xi$.
Sturm's theorem then says this: if $p$ is a square-free polynomial, then $p$ has $V_{\SS(p)}(b) - V_{\SS(p)}(a)$ distinct real roots in the interval $(a,b]$.
So, we can easily get the Sturm sequence $\SS(g)$ for our reduced cubic $g$ as so:
$$\begin{align*}
P_0(x) &= 12x^3 + 16x^2 - 5x - 3 = g(x) \\
P_1(x) &= 36x^2 + 32x - 5 = \frac{dg}{dx}\\
P_2(x) &= \frac{218}{27} x + \frac{61}{27} \\
P_3(x) &= \frac{132,200}{11,881} > 0
\end{align*}$$
For clarity, if we visualize the positive candidate roots we get from the rational root theorem (ignoring $\pm 1$ since those are easily tested on inspection), we get

(The negative roots have a symmetrical situation. We can't ignore them, but they make the visual a bit cramped.)
From here, it is a matter of finding the change in $V_{\SS(g)}$ on appropriate intervals, and cross-checking this against the roots in said interval and the number of roots we get per Descartes' rule of signs.
Thus this is reduced to largely a matter of arithmetic. Tedious arithmetic, but you can quickly eliminate candidates fairly quickly like this.
After all, bear in mind you only have $1$ positive real root for $g$, and $0$ or $2$ for negative roots, so you can test relatively large intervals and knock out many candidates at once, especially if you are clever with dealing with overlapping intervals or determining when the $P_i$ have certain signs and so on.

Again, I don't feel this constitutes a "complete" or "elegant" solution, but I feel in the end it makes things a little more efficient and less tedious for those working by hand.
A: 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 $\pm 1,\dfrac{1}{2} ,-\dfrac{1}{3} \text{ and }-\dfrac{3}{2}.$
A: Polynomials like this, at most: Trial and error is an optimum option to find its roots. Note, not the best.
Denote $f(x) = 12x^5 + 16x^4 − 17x^3 − 19x^2 + 5x + 3$
Best way (I do it) is to start of is by a pair of a positive, for example $2$, and its negative, $-2$ and substitute it into the function.
Also, before we continue, try recognising the polynomial. There are more positive terms to negative terms. Also, the 2 highest degree terms are positive. This means big $x$-values will likely not result in a $f(x) = 0$. Hence, the roots are more likely near the origin. Secondly, you have more roots for $x < 0$ than $x > 0$. Note, your highest degree term is an odd, $5$. Any $x < 0$ will result in a negative value for $12x^5$. This means your likely to balance out $f(x) = 0$, for $x < 0$, rather than $x > 0$. For, $x < 0$, your polynomial will resemble this: $-,+,+,-,-,+3$
With all that assumption done: Let us choose our first trial. We choose $x = \pm 2$. This is close to the origin.
$f(2) = 441, f(-2) = -75$. $f(2) > 0$ and $f(-2) < 0$, hence there is root(s) for $x \in [-2,2]$.
Lets go closer to the origin: Let us try for $x = \pm 1$
$f(-1) = 0, f(1) = 0$. Woo hoo, we already scored $2$ roots, $(x - 1), (x + 1)$. Lucky!
Now, from here we do not need to do trial and error. With the $2$ roots, we can reduce the polynomial to a cubic, which is easy and hopefully can be factored. This can be done through polynomial long division:
$$\frac{f(x)}{(x - 1)(x + 1)} = 12x^3 + 16x^2 - 5x - 3$$
So, unfortunately we cannot factor it. But because we have reduced our polynomial to a smaller degree, we can try using the Rational Root Theorem.
The rational root theorem allows us to test and find roots of polynomial, generally by comparing the factors for its first and last terms.
A cubic has at least $1$ root, which can be shown in the form $(ax - b)(cx^2 + dx + e)$. Note that when multiplied to get the original cubic, $be$ must equal the constant (last term) and $ac$ must equal the first term (highest degree term). If the cubic has $2$ or $3$ roots, then the quadratic should be further reducible into linear factors.
Hence, we can find the factors and see which ones match, using synthetic division. I am not going go into detail on the rational root theorem nor am I going through the steps of synthetic division - because its already done in the one of the answers here (it was posted while I was writing this one).
The other factors should be $(2x - 1)(2x + 3)(3x + 1)$, which gives $x = \frac{1}{2}$, $-\frac{3}{2}$, $-\frac{1}{3}$ along with the roots $x = \pm 1$ found above.
See, Rational Root Theorem on Wiki.
Reflection On Method:
Straight forward not the best method (trial and error), but it is a possible one, mostly when considering you cannot factor your polynomial or your polynomial does not have all roots as integers. With that being said, the method, along with its name which clearly describes its lack in efficiency, a possible advantage could be to use it to find the integer roots, which sometimes reduces the polynomial enough for one to factor to find other non-integer roots.
