Algebra Examples

$x^{2} \cdot 4 x + 4 = 0$

Step 1

Simplify the left side.

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Step 1.1

Simplify each term.

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Step 1.1.1

Rewrite using the commutative property of multiplication.

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

Step 1.1.2

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 1.1.2.1

Move $x$ .

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

Step 1.1.2.2

Multiply $x$ by $x^{2}$ .

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Step 1.1.2.2.1

Raise $x$ to the power of $1$ .

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

Step 1.1.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$4 x^{1 + 2} + 4 = 0$

Step 1.1.2.3

Add $1$ and $2$ .

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

Step 2

Subtract $4$ from both sides of the equation.

$4 x^{3} = - 4$

Step 3

Add $4$ to both sides of the equation.

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

Step 4

Factor the left side of the equation.

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Step 4.1

Factor $4$ out of $4 x^{3} + 4$ .

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Step 4.1.1

Factor $4$ out of $4 x^{3}$ .

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

Step 4.1.2

Factor $4$ out of $4$ .

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

Step 4.1.3

Factor $4$ out of $4 (x^{3}) + 4 (1)$ .

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

Step 4.2

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

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

Step 4.3

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 $b = 1$ .

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

Step 4.4

Factor.

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Step 4.4.1

Simplify.

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Step 4.4.1.1

Multiply $- 1$ by $1$ .

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

Step 4.4.1.2

One to any power is one.

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

Step 4.4.2

Remove unnecessary parentheses.

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

Step 5

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

$x + 1 = 0$

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

Step 6

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

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Step 6.1

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

$x + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 7

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

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Step 7.1

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

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

Step 7.2

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

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Step 7.2.1

Use the quadratic formula to find the solutions.

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

Step 7.2.2

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

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

Step 7.2.3

Simplify.

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Step 7.2.3.1

Simplify the numerator.

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Step 7.2.3.1.1

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

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

Step 7.2.3.1.2

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

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Step 7.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.2.3.1.2.2

Multiply $- 4$ by $1$ .

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

Step 7.2.3.1.3

Subtract $4$ from $1$ .

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

Step 7.2.3.1.4

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

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

Step 7.2.3.1.5

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

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

Step 7.2.3.1.6

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

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

Step 7.2.3.2

Multiply $2$ by $1$ .

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

Step 7.2.4

The final answer is the combination of both solutions.

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

Step 8

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

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