Solve X ^ 3 + X ^ 2 + X + 1 = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$X^{3}+X^{2}+X+1=0$$

Solve for X

$X=-1$

Solution Steps

By Rational Root Theorem, all rational roots of a polynomial are in the form $\frac{p}{q}$, where $p$ divides the constant term $1$ and $q$ divides the leading coefficient $1$. List all candidates $\frac{p}{q}$.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$X=-1$$

By Factor theorem, $X-k$ is a factor of the polynomial for each root $k$. Divide $X^{3}+X^{2}+X+1$ by $X+1$ to get $X^{2}+1$. Solve the equation where the result equals to $0$.

$$X^{2}+1=0$$

All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. Substitute $1$ for $a$, $0$ for $b$, and $1$ for $c$ in the quadratic formula.

$$X=\frac{0±\sqrt{0^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$X=\frac{0±\sqrt{-4}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$X\in \emptyset$$

List all found solutions.

$$X=-1$$

Solve for X (complex solution)

$X=-i$
$X=-1$
$X=i$