Solve for x x^4+12x=4x^2+3x^3

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

Step 1

Move all terms containing $x$ to the left side of the equation.

Subtract $4 x^{2}$ from both sides of the equation.

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

Subtract $3 x^{3}$ from both sides of the equation.

$x^{4} + 12 x - 4 x^{2} - 3 x^{3} = 0$

Step 2

Factor the left side of the equation.

Factor $x$ out of $x^{4} + 12 x - 4 x^{2} - 3 x^{3}$ .

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

$x \cdot x^{3} + 12 x - 4 x^{2} - 3 x^{3} = 0$

Factor $x$ out of $12 x$ .

$x \cdot x^{3} + x \cdot 12 - 4 x^{2} - 3 x^{3} = 0$

Factor $x$ out of $- 4 x^{2}$ .

$x \cdot x^{3} + x \cdot 12 + x (- 4 x) - 3 x^{3} = 0$

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

$x \cdot x^{3} + x \cdot 12 + x (- 4 x) + x (- 3 x^{2}) = 0$

Factor $x$ out of $x \cdot x^{3} + x \cdot 12$ .

$x (x^{3} + 12) + x (- 4 x) + x (- 3 x^{2}) = 0$

Factor $x$ out of $x (x^{3} + 12) + x (- 4 x)$ .

$x (x^{3} + 12 - 4 x) + x (- 3 x^{2}) = 0$

Factor $x$ out of $x (x^{3} + 12 - 4 x) + x (- 3 x^{2})$ .

$x (x^{3} + 12 - 4 x - 3 x^{2}) = 0$

Reorder terms.

$x (x^{3} - 3 x^{2} - 4 x + 12) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$x ((x^{3} - 3 x^{2}) - 4 x + 12) = 0$

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

$x (x^{2} (x - 3) - 4 (x - 3)) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 3$ .

$x ((x - 3) (x^{2} - 4)) = 0$

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

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

Factor.

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

$x ((x - 3) ((x + 2) (x - 2))) = 0$

Remove unnecessary parentheses.

$x ((x - 3) (x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$x (x - 3) (x + 2) (x - 2) = 0$

Step 3

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

$x = 0$

$x - 3 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 6

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Step 7

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 8

The final solution is all the values that make $x (x - 3) (x + 2) (x - 2) = 0$ true.

$x = 0, 3, - 2, 2$

Solve for x x^4+12x=4x^2+3x^3

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