1
$\begingroup$

The equation is $4x^3+\frac{127}{x}=2016, x\neq0$.

By rational roots test, it is easy to see that the equation does not have rational roots. By Descartes rule of signs there exists a negative root, hence it has atleast two real roots. After entering this equation in WolframAlpha, the equation is seen to possess two real and two purely complex roots.

My question pertains as to whether there is any simple procedure to obtain the rootsi.e. without using Ferrari's solution or Numerical methods or computer algebra systems? Another observation is $2016=\underline{126}\times16$ and there is $127$ on the LHS. Any ideas. Thanks beforehand.

  • 0
    Why do you call that *biquadratic*? It's more of a very special case of a [depressed quartic](https://en.wikipedia.org/wiki/Quartic_function#Converting_to_a_depressed_quartic). – 2017-01-01
  • 0
    @dxiv so you mean to say that ferrari's or descartes solution is the only best way? – 2017-01-01
  • 0
    Since you tried WA already, note that it [factors](http://www.wolframalpha.com/input/?i=factor+4+x%5E4+%2B+127+-+2016+x). – 2017-01-01
  • 0
    @dxiv thanks, I think we are getting closer. Just we need to know to split the $-2016x$ term. I think my observation that $2016=126\times16$ comes handy, isnt it? – 2017-01-01
  • 0
    Surely does, though it takes more work after that. – 2017-01-01
  • 1
    @dxiv thanks to your link the answer is ready and posted. – 2017-01-01

2 Answers 2

6

We see that $127$ is a prime number, therefore if the polynomial is reducible over $\mathbb{Z}$(or, equivalently, $\mathbb{Q}$), then it should factor as $$(ax^2+bx+1)(cx^2+dx+127)=4x^4-2016x+127$$ Equating coefficients gives us four equations to solve:$$\begin{cases} ac=4\\ad+bc=0\\127a+c+bd=0\\127b+d=-2016\end{cases}$$ whence $$4x^4-2016x+127=(2x^2-16x+1)(2x^2+16x+127)$$ which can then be factored into complex factors using well known formula. We take note that the key factor for the ease was the primality of $127$.

  • 0
    are you saying that $127 \times 2 + 2 = 0$ ? – 2017-01-01
  • 0
    @mercio thanks, modified the post – 2017-01-01
  • 1
    From the 4th of the equations in $a,b,c,d$ we have $d\equiv 16 \pmod {127}.$ Putting this into the 2nd and 3rd equations yields $16^2(a-c)\equiv 0 \pmod {127}.$ Since we can assume $\{a,c\} =\{1,4\}$ or $\{a,c\}=\{2,2\}=\{2\},$ we must have $a=c=2. $..... Nice answer. – 2017-01-01
  • 0
    @user254665 how do you get $16^2(a-c)\equiv0(\bmod 127)$? – 2017-01-13
  • 0
    By means of a typo.. My miscalculations cancelled each other out. – 2017-01-13
0

Why not use the tangent method. If $p(x)=4x^4-2016x+127$ then we know that a solution will lie in the interval $[-1,1]$, since the function is continuous (even though it is not defined for $x=0$) and $p(-1)>0, p(1)<0$. The tangent method states that we can find the solution of an equation in an interval $[a,b]$ by iterating $x_{n+1}=x_n-\frac{f(x)}{f'(x)}$ where $x_0=a$. The derivative of $p(x)$ is $p'(x)=16x^3-2016$; we have that $$ x_1=-1-\frac{p(-1)}{p'(-1)}=\frac{115}{2032} \\ x_2=x_1-\frac{p(x_1)}{p'(x_1)}=\frac{541300119662413}{8592602591386304} \\ x_3=x_2-\frac{p(x_2)}{p'(x_2)}=0.0629960629\\ $$ and the accuracy keeps getting higher. The other root is between $7$ and $8$ since $p(7)=-4381$ and $p(8)=383$; we can set $x_1=7$ and do the same process again $$ x_1=-7-\frac{p(7)}{p'(7)}=8,261808756 \\ x_2=x_1-\frac{p(x_1)}{p'(x_1)}=7,961028858 \\ x_3=x_2-\frac{p(x_2)}{p'(x_2)}=7,937148598 \\ x_4=x_3-\frac{p(x_3)}{p'(x_3)}=7,937003942 \\ $$ So the two roots of the equation are $x_1=0.0629960629$ and $x_2=7,937003942$ Note that these calculations have been made only with a scientific calculator and not with a computer even though these are only the real roots and not the complex one.