Solving a algebraic equation

Question

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?

Answer 1

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 $-1, i, -i$.

Answer 2

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.

Answer 3

Two approaches to finding a root: (We can confirm a real root exists, by the rational root theorem).

$$x^3+x^2+x+1=0$$

1. 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 $(x^2 + 1)$.

2. 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 $$\frac{-b \pm \sqrt{\color{blue}{\bf b^2 - 4ac}}}{2a}$$

Answer 4

You could use long division to factorize of $x^3+x^2+x+1$ once you recognize that $x=-1$ is root .

Answer 5

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.