Find All Complex Solutions 8x^3+x^2-32x-4=0

$8 x^{3} + x^{2} - 32 x - 4 = 0$

Step 1

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(8 x^{3} + x^{2}) - 32 x - 4 = 0$

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

$x^{2} (8 x + 1) - 4 (8 x + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $8 x + 1$ .

$(8 x + 1) (x^{2} - 4) = 0$

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

$(8 x + 1) (x^{2} - 2^{2}) = 0$

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 $b = 2$ .

$(8 x + 1) ((x + 2) (x - 2)) = 0$

Remove unnecessary parentheses.

$(8 x + 1) (x + 2) (x - 2) = 0$

Step 2

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

$8 x + 1 = 0$

$x + 2 = 0$

$x - 2 = 0$

Step 3

Set $8 x + 1$ equal to $0$ and solve for $x$ .

Set $8 x + 1$ equal to $0$ .

$8 x + 1 = 0$

Solve $8 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$8 x = - 1$

Divide each term in $8 x = - 1$ by $8$ and simplify.

Divide each term in $8 x = - 1$ by $8$ .

$\frac{8 x}{8} = \frac{- 1}{8}$

Simplify the left side.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{8 x}{8} = \frac{- 1}{8}$

Divide $x$ by $1$ .

$x = \frac{- 1}{8}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{8}$

Step 4

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 5

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 6

The final solution is all the values that make $(8 x + 1) (x + 2) (x - 2) = 0$ true.

$x = - \frac{1}{8}, - 2, 2$

Find All Complex Solutions 8x^3+x^2-32x-4=0

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