Find solutions to the polynomial $x^4+x^3-x^2-5x+4$

Question

This question is part of a calculus problem, involving the limit of a rational function. I modified it a little bit to make it more interesting. The question is:

$x^4+x^3-x^2-5x+4=0$, find all the solutions (real and complex) to this polynomial

This is my attempt, please do let me know if this is correct, if it can be improved and please share your approach to!

Since this is a quartic polynomial, factoring it is going to be challenging (if not impossible in some cases). However, I'm going to attempt to factor it into quadratics:

$$x^4+x^3-x^2-5x+4=0$$

$$x^2(x^2-1)+x^3-1-5x+5=0$$

$$x^2(x-1)(x+1)+(x-1)(x^2+x+1)-5(x-1)=0$$

$$(x-1)[(x^3+x^2+x^2+x-4)]=0$$

$$(x-1)(x^3+2x^2+x-4)=0$$

$$(x-1)(x^3-1+2x^2+x-3)=0$$

$$(x-1)[(x-1)(x^2+x+1)+x(2x+3)-1(2x+3)]=0$$

$$(x-1)^2(x^2+3x+4)=0$$

Therefore, $x=1$ and $x=\frac{-3\pm\iota\sqrt{7}}{2}$ are the solutions to this polynomial.

Answer 1

If you look carefully, you'll notice that the sum of the coefficients of the polynomial is equal to $0$. Then, it is obvious that $x = 1$ is a root of the polynomial. Now, let $p(x)$ be our polynomial. By doing $\frac{p(x)}{x-1}$ you'll get $x^3+2x^2+x-4$. The sum of the coefficients is $0$ again. Now divide it by $x-1$, obtaining $x^2 + 3x + 4$, which is clearly irreducible over $\mathbb R$. You can compute the solutions of this last polynomial using the well-known formula for quadratic equations. Hope this helps you somehow.

Answer 2

You have asked for an alternative solutions. Either you can use the rational root theorem, or you can do an observation about the sum of coefficients, or just guess that $1$ is a root, and then do a polynomial division. The sum of coefficients appears to be "ad hoc", or a little bit like a trick. The rational root theorem is a general theorem, but of course it requires to check several possibilities.

Another way is to consider the quartic modulo $2$, then it is $$ x^4+x^3-x^2-5x+4=x^4+x^3+x^2+x= x(x-1)^3, $$ which gives $x=1$ as a candidate also over $\Bbb Q$. Of course the other candidate $x=0$ does not work.