4
$\begingroup$

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?

$\endgroup$
8
$\begingroup$

If you ask me, you did the good thing ignoring the hint and solving it your way. However, there is an "algorithm" for solving these type of problems that the hint is probably referring to. Notice that sum of coefficients is $0$, so $x = 1$ is root. Dividing by $x - 1$ leaves $$6x^4 + 11x^3 - 40x^2 +11x +6 = 0$$ and noticing that $0$ isn't a root and dividing by $x^2$ you get $$6(x^2 + \frac{1}{x^2}) + 11(x + \frac{1}{x}) - 40 = 0$$ Now substitute $y = x + \frac{1}{x}$, and notice $x^2 + \frac{1}{x^2} = y^2 - 2$, you are left with two quadratic equations to solve.

$\endgroup$
  • $\begingroup$ Interesting! Do you happen to know if there is a name for this "algorithm"? $\endgroup$ – Alex Jul 11 at 4:14
  • $\begingroup$ I don't know. Seems that Wikipedia is referring to those as palindromic and antipalindromic polynomials if that helps. Also, the answer of Simply Beautiful Art provides some details on why things work in general. $\endgroup$ – prosinac Jul 11 at 21:15
3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

As a more general thing, whenever the coefficients are "symmetric", then it will have "symmetric" factors. See that you have $(6x^3-19x^2+19x-6)(x^2+4x+1)$. If there were no rational factors, writing it as a multiple of these "symmetric polynomials" may allow further factoring.

It is also notable that for every root $x$, $1/x$ will also be a root, which halves the amount of roots to check if you plan on using the rational roots theorem.

Lastly, in the latter half of prosinac's answer, the degree involved is reduced by a factor of 2 by the substitution $y=x+\frac1x$. In general, if the degree is even, the substitution $y=x\pm\frac1x$ will allow you to halve the power of the problem you are working on, choosing the sign based on how the signs work out. If the degree is odd, then either $1$ or $-1$ will be a factor, allowing you to reduce this to an even degree "symmetric polynomial" and then halve the power. This means any "symmetric polynomial" of degree less than 10 can be factored algebraically, as you can always reduce the degree below 5.


Edit: thanks to prosinac for pointing this out:

Apparently these are called self-reciprocal or palindromic polynomials. Some more properties are provided on the Wikipedia link.

$\endgroup$
  • $\begingroup$ $\left\{\left\{x\to \frac{2}{3}\right\},\{x\to 1\},\left\{x\to \frac{3}{2}\right\},\left\{x\to -2-\sqrt{3}\right\},\left\{x\to \sqrt{3}-2\right\}\right\}$ $\endgroup$ – David G. Stork Jul 10 at 1:34
  • 2
    $\begingroup$ I'm really not sure what you're getting at. $\endgroup$ – Simply Beautiful Art Jul 10 at 1:37
  • $\begingroup$ @SimplyBeautifulArt Why is it that "for every root 𝑥, 1/𝑥 will also be a root"? Does this rule apply to all polynomials when using the rational root theorem? $\endgroup$ – Alex Jul 11 at 4:44
  • $\begingroup$ @Alex what happens when you divide your equation by $-x^5$ and rewrite everything in terms of $1/x$? $\endgroup$ – Simply Beautiful Art Jul 12 at 1:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.