Solving a algebraic equation I am to solve the following equation third order equation
$x^3+x^2+x+1=0$
What I've tried so far is writing the equation as
$ x \cdot (x^2+x+1)+1=0$ 
but that didn't lead anywhere. How do I solve this without using a computer?
 A: Here's another way: Summing the geometric series, we get
$$(x^3+x^2+x+1)(x-1)=x^4-1$$
The roots of $x^4-1$ are $\pm 1$ and $\pm i$, so the roots of $x^3+x^2+x+1$ are $-1, i, -i$.
A: Factor by grouping:
$$
x^3 + x^2 + x + 1 = x^2(x+1)+(x+1) = (x+1)(x^2+1) = 0.
$$
Then $x + 1 = 0$ gives $x = -1$ and $x^2 + 1 = 0$ gives $x = \pm i$ (assuming you want to solve over the complex numbers.
A: Two approaches to finding a root: (We can confirm a real root exists, by the rational root theorem).
$$x^3+x^2+x+1=0$$


*

*We can recognize that $x = -1$ is a zero by using a quick check at $x = \pm 1$ (again, by the rational root theorem, those are possible zeros). Then we know that since $x = -1$ is a zero, $(x - (-1)) = (x + 1)$ is a factor, and we can just use polynomial long division to obtain the remaining factor of $(x^2 + 1)$.

*We can "group" the equation as follows: $$x^3+x^2+x+1=0 \iff x^2(x + 1) + (x + 1) = (x+1)(x^2 + 1) = 0$$
Finally, we know that the remaining factor in $(x+1)(x^2 + 1)=0$, namely $(x^2 + 1),\;$ has no real root (and hence, cannot be factored) by evaluating the discriminant $$\Delta = b^2 - 4ac = 0 - 4 = -4 < 0$$
(Recall the discriminant of a quadratic equation $ax^2 + bx + c = 0$ is given by the $\color{blue}{\bf radicand}$ of the quadratic formula $$\frac{-b \pm  \sqrt{\color{blue}{\bf b^2 - 4ac}}}{2a}$$
A: You could use long division to factorize of $x^3+x^2+x+1$ once you recognize that $x=-1$ is root .
A: We generally use hit and trial method for 3rd degree equations. We get -1 as one of the factor and now diving the whole equation by (x-1) you would get a quadratic which can be easily solved by the quadratic formula.
