Solving 5th degree polynomial Solve the equation, $$6x^5 + 5x^4 − 51x^3 + 51x^2 − 5x − 6 = 0$$(hints: the pattern of the coefficients)
How I attempted this problem is to use the rational root theorem to obtain the factors. The possible roots are $$±(1, 1/2, 1/3, 1/6, 2, 2/3, 3, 3/2, 6)$$
By substituting the values inside the equation, only the following roots satisfy the equation:
$$x=1,\:x=\frac{3}{2},\:x=\frac{2}{3}$$
Henceforth, I came up with the following based on the above roots:
$$\left(x-1\right)\left(2x-3\right)\left(3x-2\right)$$
Clearly expanding them would give me a 3rd degree polynomial as follows:
$$6x^3 - 19x^2 + 19x - 6$$
Using polynomial division where I divided the original 5th degree equation with the above equation,
I obtained the following equation:
$$x^2+4x+1$$
Now, solving the above equation using quadratic formula, I am able to get the roots. Hence, I have obtained all the roots of the solution.
However, looking at the hint that asks me to observe the pattern of the coefficients, I think that the method used to arrive at my solution might have been long-winded (it was indeed tedious plugging in the rational .roots one-by-one into the equation to see which returns value of 0). Is there something that I'm missing in the way I approached the question? 

The "pattern" is that w hen you reverse the order of the coefficients you get the negative of the original polynomial:
$6x^5+5x^4-51x^3+51x^2-5x-6\to-6x^5-5x^4+51x^3-51x^2+5x+6$
When the polynomial has this pattern you can render it this way:
$6x^5+5x^4-51x^3+51x^2-5x-6=6\color{blue}{(x^5-1)}+5\color{blue}{(x^4-x)}-51\color{blue}{(x^3-x^2)}$
where the blue factors are all multiples of $x-1$ forcing $x=1$ to be a root.  When you divide out this factor (using the above rearrangement to help things along) you get
$6x^5+5x^4-51x^3+51x^2-5x-6=(x-1)\color{blue}{(6x^4+11x^3-40x^2+11x+6)}$
where the blue quartic factor now reads exactly the same when the coefficients are reversed.  This arrangement, in an even degree polynomial, forces a "symmetric" factorization into quadratics:
$6x^4+11x^3-40x^2+11x+6=6(x^2+ax+1)(x^2+bx+1)$
Multiplying the right side out and matching terms with either odd power of $x$ gives $a+b=11/6$, and matching the terms with $x^2$ gives $ab=-26/3$.  From the known sum and product we infer that $a$ and $b$ solve
$w^2-(11/6)w-(26/3)=0$
$6w^2-11w-52=0$
Thereby the roots for $a$ and $b$ are $4$ and $-13/6$, from which we now have
$6x^5+5x^4-51x^3+51x^2-5x-6=6(x-1)(x^2+4x+1)(x^2-(13/6)x+1)=(x-1)(x^2+4x+1)(6x^2-13x+6)$
and the remaining quadratic factors are solved by conventional methods.
It looks more complicated, but note that no trial and error is needed.
