Solve by Factoring cube root of x^2+12x+5=9

$\sqrt[3]{x^{2} + 12 x} + 5 = 9$

Step 1

Subtract $9$ from both sides of the equation.

$\sqrt[3]{x^{2} + 12 x} + 5 - 9 = 0$

Step 2

Simplify $\sqrt[3]{x^{2} + 12 x} + 5 - 9$ .

Factor $x$ out of $x^{2} + 12 x$ .

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

$\sqrt[3]{x \cdot x + 12 x} + 5 - 9 = 0$

Factor $x$ out of $12 x$ .

$\sqrt[3]{x \cdot x + x \cdot 12} + 5 - 9 = 0$

Factor $x$ out of $x \cdot x + x \cdot 12$ .

$\sqrt[3]{x (x + 12)} + 5 - 9 = 0$

Subtract $9$ from $5$ .

$\sqrt[3]{x (x + 12)} - 4 = 0$

Step 3

Add $4$ to both sides of the equation.

$\sqrt[3]{x (x + 12)} = 4$

Step 4

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{x (x + 12)}}^{3} = 4^{3}$

Step 5

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{x (x + 12)}$ as ${(x (x + 12))}^{\frac{1}{3}}$ .

${({(x (x + 12))}^{\frac{1}{3}})}^{3} = 4^{3}$

Simplify the left side.

Simplify ${({(x (x + 12))}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(x (x + 12))}^{\frac{1}{3}})}^{3}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x (x + 12))}^{\frac{1}{3} \cdot 3} = 4^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x (x + 12))}^{\frac{1}{3} \cdot 3} = 4^{3}$

Rewrite the expression.

${(x (x + 12))}^{1} = 4^{3}$

Apply the distributive property.

${(x \cdot x + x \cdot 12)}^{1} = 4^{3}$

Simplify the expression.

Multiply $x$ by $x$ .

${(x^{2} + x \cdot 12)}^{1} = 4^{3}$

Move $12$ to the left of $x$ .

$x^{2} + 12 x = 4^{3}$

Simplify the right side.

Raise $4$ to the power of $3$ .

$x^{2} + 12 x = 64$

Step 6

Solve for $x$ .

Subtract $64$ from both sides of the equation.

$x^{2} + 12 x - 64 = 0$

Factor $x^{2} + 12 x - 64$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 64$ and whose sum is $12$ .

$- 4, 16$

Write the factored form using these integers.

$(x - 4) (x + 16) = 0$

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

$x - 4 = 0$

$x + 16 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 16$ equal to $0$ and solve for $x$ .

Set $x + 16$ equal to $0$ .

$x + 16 = 0$

Subtract $16$ from both sides of the equation.

$x = - 16$

The final solution is all the values that make $(x - 4) (x + 16) = 0$ true.

$x = 4, - 16$

Solve by Factoring cube root of x^2+12x+5=9

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