Algebra Examples

$x^{5} + x^{3} + 8 x^{2} + 8$

Step 1

Set $x^{5} + x^{3} + 8 x^{2} + 8$ equal to $0$ .

$x^{5} + x^{3} + 8 x^{2} + 8 = 0$

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

Factor the left side of the equation.

Tap for more steps...

Step 2.1.1

Factor out the greatest common factor from each group.

Tap for more steps...

Step 2.1.1.1

Group the first two terms and the last two terms.

$(x^{5} + x^{3}) + 8 x^{2} + 8 = 0$

Step 2.1.1.2

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

$x^{3} (x^{2} + 1) + 8 (x^{2} + 1) = 0$

Step 2.1.2

Factor the polynomial by factoring out the greatest common factor, $x^{2} + 1$ .

$(x^{2} + 1) (x^{3} + 8) = 0$

Step 2.1.3

Rewrite $8$ as $2^{3}$ .

$(x^{2} + 1) (x^{3} + 2^{3}) = 0$

Step 2.1.4

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 $b = 2$ .

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

Step 2.1.5

Factor.

Tap for more steps...

Step 2.1.5.1

Simplify.

Tap for more steps...

Step 2.1.5.1.1

Multiply $2$ by $- 1$ .

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

Step 2.1.5.1.2

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

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

Step 2.1.5.2

Remove unnecessary parentheses.

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

Step 2.2

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

$x^{2} + 1 = 0$

$x + 2 = 0$

$x^{2} - 2 x + 4 = 0$

Step 2.3

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

Tap for more steps...

Step 2.3.1

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

$x^{2} + 1 = 0$

Step 2.3.2

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

Tap for more steps...

Step 2.3.2.1

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

Step 2.3.2.2

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

$x = \pm \sqrt{- 1}$

Step 2.3.2.3

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

$x = \pm i$

Step 2.3.2.4

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

Tap for more steps...

Step 2.3.2.4.1

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

$x = i$

Step 2.3.2.4.2

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

$x = - i$

Step 2.3.2.4.3

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

$x = i, - i$

Step 2.4

Set $x + 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.4.1

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Step 2.4.2

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 2.5

Set $x^{2} - 2 x + 4$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.5.1

Set $x^{2} - 2 x + 4$ equal to $0$ .

$x^{2} - 2 x + 4 = 0$

Step 2.5.2

Solve $x^{2} - 2 x + 4 = 0$ for $x$ .

Tap for more steps...

Step 2.5.2.1

Use the quadratic formula to find the solutions.

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

Step 2.5.2.2

Substitute the values $a = 1$ , $b = - 2$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Step 2.5.2.3

Simplify.

Tap for more steps...

Step 2.5.2.3.1

Simplify the numerator.

Tap for more steps...

Step 2.5.2.3.1.1

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

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

Step 2.5.2.3.1.2

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

Tap for more steps...

Step 2.5.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 2.5.2.3.1.2.2

Multiply $- 4$ by $4$ .

$x = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 1}$

Step 2.5.2.3.1.3

Subtract $16$ from $4$ .

$x = \frac{2 \pm \sqrt{- 12}}{2 \cdot 1}$

Step 2.5.2.3.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 1}$

Step 2.5.2.3.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.2.3.1.6

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

$x = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 1}$

Step 2.5.2.3.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

Tap for more steps...

Step 2.5.2.3.1.7.1

Factor $4$ out of $12$ .

$x = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 1}$

Step 2.5.2.3.1.7.2

Rewrite $4$ as $2^{2}$ .

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

Step 2.5.2.3.1.8

Pull terms out from under the radical.

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

Step 2.5.2.3.1.9

Move $2$ to the left of $i$ .

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

Step 2.5.2.3.2

Multiply $2$ by $1$ .

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

Step 2.5.2.3.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{2}$ .

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

Step 2.5.2.4

The final answer is the combination of both solutions.

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

Step 2.6

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

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

Step 3