Solve by Factoring 32x-16=2x^3-x^2

$32 x - 16 = 2 x^{3} - x^{2}$

Step 1

Move all the expressions to the left side of the equation.

Subtract $2 x^{3}$ from both sides of the equation.

$32 x - 16 - 2 x^{3} = - x^{2}$

Add $x^{2}$ to both sides of the equation.

$32 x - 16 - 2 x^{3} + x^{2} = 0$

Step 2

Reorder terms.

$- 2 x^{3} + x^{2} + 32 x - 16 = 0$

Step 3

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(- 2 x^{3} + x^{2}) + 32 x - 16 = 0$

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

$x^{2} (- 2 x + 1) - 16 (- 2 x + 1) = 0$

Step 4

Factor the polynomial by factoring out the greatest common factor, $- 2 x + 1$ .

$(- 2 x + 1) (x^{2} - 16) = 0$

Step 5

Rewrite $16$ as $4^{2}$ .

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

Step 6

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 $b = 4$ .

$(- 2 x + 1) ((x + 4) (x - 4)) = 0$

Remove unnecessary parentheses.

$(- 2 x + 1) (x + 4) (x - 4) = 0$

Step 7

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

$- 2 x + 1 = 0$

$x + 4 = 0$

$x - 4 = 0$

Step 8

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

Set $- 2 x + 1$ equal to $0$ .

$- 2 x + 1 = 0$

Solve $- 2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$- 2 x = - 1$

Divide each term in $- 2 x = - 1$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 1$ by $- 2$ .

$\frac{- 2 x}{- 2} = \frac{- 1}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} = \frac{- 1}{- 2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{1}{2}$

Step 9

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

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

$x + 4 = 0$

Subtract $4$ from both sides of the equation.

$x = - 4$

Step 10

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.

$x = 4$

Step 11

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

$x = \frac{1}{2}, - 4, 4$

Solve by Factoring 32x-16=2x^3-x^2

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