# 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?

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Try grouping: $x^3 + x^2 + x + 1 = x^2(x + 1) + (x+1) = (x+1)(x^2+1)$ –  Suugaku Jun 3 '13 at 14:17
That did it thanks! –  chase.furlong Jun 3 '13 at 14:21
No problem! Glad to help. –  Suugaku Jun 3 '13 at 14:22

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 $-1, i, -i$.

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This is the best solution, IMO. –  Parth Kohli Jun 3 '13 at 14:30

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.

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I haven't learned complex numbers yet, but that's the answer –  chase.furlong Jun 3 '13 at 14:22
Then don't worry about the $x^2 + 1$ term! hah –  Suugaku Jun 3 '13 at 14:22

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

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I always like to know the mathematical vocabulary in English. –  Sami Ben Romdhane May 29 '14 at 19:52

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

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Nice trick, yeah that works, thanks! –  chase.furlong Jun 3 '13 at 14:20

You can move things around to get a nonzero term on the right hand side, then divide.

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

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

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

Now divide and factor:

$$\dfrac{x^2+1}{x^3+x}=-1$$

$$\implies \dfrac{x^2+1}{x*(x^2+1)}=-1$$

$$\implies \dfrac{1}{x}=-1$$

$$\implies 1=-x$$

$$\implies x=-1$$

Here's how I figured it out: Just by looking at the equation, we can see that we are adding a positive term $(1)$ to a bunch of things in terms of $x$ to get something less than $1\text{ (0)}$. So we know some of those terms have to be nonzero and negative. Therefore $x$ should be negative. Then $x^3$ is negative and $x^2$ is positive. I decided to try and move things around so that both sides of the equation were a positive value, and got the above results.

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what if $x^2+1 = 0$? it would be cleaner just to write $x^3+x^2+x+1 = (x^3+x)+(x^2+1)$. –  John Joy May 30 '14 at 16:26

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.

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