# Factorisation of quartic equation.

I'm given a logarithm problem which is $$\log_{2} (x^3+1)-2\log_{2}x=\log_{2}(x^2-x+1)-2$$

I'm stuck in the step of $x^4-5x^3+x^2-4=0$

By many times of trial and error, I got $(x^2-4x-4)(x^2-x+1)=0$

My question, is there any standard way to factorise the equation without trials and errors? Thanks in advance.

• Hint: $\;(x^3+1)=(x+1)(x^2-x+1)$
– dxiv
Commented Mar 21, 2017 at 16:45
• In general, no. There exists a quartic formula, much like there exists a quadratic formula, but you rarely see it written in full. A quick search turned up this: curtisbright.com/quartic/quartic-png.html Commented Mar 21, 2017 at 16:46
• @dxiv can you write a answer for me? Thanks in advance. Commented Mar 21, 2017 at 16:59
• It is sometimes useful to look for integer solutions of such a polynomial (even though they might not exist, as in this case): check if any of the divisors of the last term is a solution. In your case, you'd be looking into the divisors of $4$ which are $+-1;+-2+-4$. Commented Mar 21, 2017 at 21:43

Hint: the quartic can be avoided altogether by noting that $x^3+1=(x+1)(x^2-x+1)\,$. Given that $x^2-x+1 \gt 0$ on $\mathbb{R}$ it follows that $x^3+1$ and $x+1$ have the same sign, which must be positive for the $\log_2$ to be defined. Then the equation simplifies to:
$$\require{cancel} \log_{2} (x+1)+\cancel{\log_{2} (x^2-x+1)}-2\log_{2}x=\cancel{\log_{2}(x^2-x+1)}-2 \\ \iff\quad \log_2{}\frac{x+1}{x^2}=-2 \quad\iff\quad \frac{x+1}{x^2}=\frac{1}{4}$$