Solve by Factoring 3x^3+27x=3x(x^2+9)

$3 x^{3} + 27 x = 3 x (x^{2} + 9)$

Step 1

Subtract $3 x (x^{2} + 9)$ from both sides of the equation.

$3 x^{3} + 27 x - 3 x (x^{2} + 9) = 0$

Step 2

Simplify $3 x^{3} + 27 x - 3 x (x^{2} + 9)$ .

Simplify each term.

Apply the distributive property.

$3 x^{3} + 27 x - 3 x \cdot x^{2} - 3 x \cdot 9 = 0$

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

Move $x^{2}$ .

$3 x^{3} + 27 x - 3 (x^{2} x) - 3 x \cdot 9 = 0$

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

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

$3 x^{3} + 27 x - 3 (x^{2} x^{1}) - 3 x \cdot 9 = 0$

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

$3 x^{3} + 27 x - 3 x^{2 + 1} - 3 x \cdot 9 = 0$

Add $2$ and $1$ .

$3 x^{3} + 27 x - 3 x^{3} - 3 x \cdot 9 = 0$

Multiply $9$ by $- 3$ .

$3 x^{3} + 27 x - 3 x^{3} - 27 x = 0$

Combine the opposite terms in $3 x^{3} + 27 x - 3 x^{3} - 27 x$ .

Subtract $3 x^{3}$ from $3 x^{3}$ .

$27 x + 0 - 27 x = 0$

Add $27 x$ and $0$ .

$27 x - 27 x = 0$

Subtract $27 x$ from $27 x$ .

$0 = 0$

Step 3

Since $0 = 0$ , the equation will always be true.

Always true