Algebra Examples

$9 x^{5} - 72 x^{2} = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Factor $9 x^{2}$ out of $9 x^{5} - 72 x^{2}$ .

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Step 1.1.1

Factor $9 x^{2}$ out of $9 x^{5}$ .

$9 x^{2} (x^{3}) - 72 x^{2} = 0$

Step 1.1.2

Factor $9 x^{2}$ out of $- 72 x^{2}$ .

$9 x^{2} (x^{3}) + 9 x^{2} (- 8) = 0$

Step 1.1.3

Factor $9 x^{2}$ out of $9 x^{2} (x^{3}) + 9 x^{2} (- 8)$ .

$9 x^{2} (x^{3} - 8) = 0$

Step 1.2

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

$9 x^{2} (x^{3} - 2^{3}) = 0$

Step 1.3

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

$9 x^{2} ((x - 2) (x^{2} + x \cdot 2 + 2^{2})) = 0$

Step 1.4

Factor.

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Step 1.4.1

Simplify.

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Step 1.4.1.1

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

$9 x^{2} ((x - 2) (x^{2} + 2 x + 2^{2})) = 0$

Step 1.4.1.2

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

$9 x^{2} ((x - 2) (x^{2} + 2 x + 4)) = 0$

Step 1.4.2

Remove unnecessary parentheses.

$9 x^{2} (x - 2) (x^{2} + 2 x + 4) = 0$

Step 2

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

$x^{2} = 0$

$x - 2 = 0$

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

Step 3

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

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Step 3.1

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

$x^{2} = 0$

Step 3.2

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

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Step 3.2.1

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

$x = \pm \sqrt{0}$

Step 3.2.2

Simplify $\pm \sqrt{0}$ .

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Step 3.2.2.1

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Step 3.2.2.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Step 3.2.2.3

Plus or minus $0$ is $0$ .

$x = 0$

Step 4

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

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Step 4.1

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

$x - 2 = 0$

Step 4.2

Add $2$ to both sides of the equation.

$x = 2$

Step 5

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

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Step 5.1

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

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

Step 5.2

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

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Step 5.2.1

Use the quadratic formula to find the solutions.

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

Step 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 5.2.3

Simplify.

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Step 5.2.3.1

Simplify the numerator.

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Step 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 5.2.3.1.2

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

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Step 5.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 5.2.3.1.2.2

Multiply $- 4$ by $4$ .

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

Step 5.2.3.1.3

Subtract $16$ from $4$ .

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

Step 5.2.3.1.4

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

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

Step 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 5.2.3.1.6

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

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

Step 5.2.3.1.7

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

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Step 5.2.3.1.7.1

Factor $4$ out of $12$ .

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

Step 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 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 5.2.3.1.9

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

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

Step 5.2.3.2

Multiply $2$ by $1$ .

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

Step 5.2.3.3

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

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

Step 5.2.4

The final answer is the combination of both solutions.

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

Step 6

The final solution is all the values that make $9 x^{2} (x - 2) (x^{2} + 2 x + 4) = 0$ true.

$x = 0, 2, - 1 + i \sqrt{3}, - 1 - i \sqrt{3}$