Solve for x x^7+x^4+x^3+1=0

$x^{7} + x^{4} + x^{3} + 1 = 0$

Step 1

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{7} + x^{4}) + x^{3} + 1 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{4} (x^{3} + 1) + 1 (x^{3} + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $x^{3} + 1$ .

$(x^{3} + 1) (x^{4} + 1) = 0$

Rewrite $1$ as $1^{3}$ .

$(x^{3} + 1^{3}) (x^{4} + 1) = 0$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 1$ .

$(x + 1) (x^{2} - x \cdot 1 + 1^{2}) (x^{4} + 1) = 0$

Simplify.

Multiply $- 1$ by $1$ .

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

One to any power is one.

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

Step 2

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x + 1 = 0$

$x^{2} - x + 1 = 0$

$x^{4} + 1 = 0$

Step 3

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 4

Set $x^{2} - x + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} - x + 1$ equal to $0$ .

$x^{2} - x + 1 = 0$

Solve $x^{2} - x + 1 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 1$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 1$ .

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4}}{2 \cdot 1}$

Subtract $4$ from $1$ .

$x = \frac{1 \pm \sqrt{- 3}}{2 \cdot 1}$

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{1 \pm \sqrt{- 1 \cdot 3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (3)}$ as $\sqrt{- 1} \cdot \sqrt{3}$ .

$x = \frac{1 \pm \sqrt{- 1} \cdot \sqrt{3}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{1 \pm i \sqrt{3}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm i \sqrt{3}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}$

Step 5

Set $x^{4} + 1$ equal to $0$ and solve for $x$ .

Set $x^{4} + 1$ equal to $0$ .

$x^{4} + 1 = 0$

Solve $x^{4} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{4} = - 1$

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{- 1}$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x = \sqrt[4]{- 1}$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - \sqrt[4]{- 1}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \sqrt[4]{- 1}, - \sqrt[4]{- 1}$

Step 6

The final solution is all the values that make $(x + 1) (x^{2} - x + 1) (x^{4} + 1) = 0$ true.

$x = - 1, \frac{1 + i \sqrt{3}}{2}, \frac{1 - i \sqrt{3}}{2}, \sqrt[4]{- 1}, - \sqrt[4]{- 1}$

Solve for x x^7+x^4+x^3+1=0

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