Solve by Factoring (3n-2)(4n+1)=0

$(3 n - 2) (4 n + 1) = 0$

Step 1

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

$3 n - 2 = 0$

$4 n + 1 = 0$

Step 2

Set $3 n - 2$ equal to $0$ and solve for $n$ .

Set $3 n - 2$ equal to $0$ .

$3 n - 2 = 0$

Solve $3 n - 2 = 0$ for $n$ .

Add $2$ to both sides of the equation.

$3 n = 2$

Divide each term in $3 n = 2$ by $3$ and simplify.

Divide each term in $3 n = 2$ by $3$ .

$\frac{3 n}{3} = \frac{2}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 n}{3} = \frac{2}{3}$

Divide $n$ by $1$ .

$n = \frac{2}{3}$

Step 3

Set $4 n + 1$ equal to $0$ and solve for $n$ .

Set $4 n + 1$ equal to $0$ .

$4 n + 1 = 0$

Solve $4 n + 1 = 0$ for $n$ .

Subtract $1$ from both sides of the equation.

$4 n = - 1$

Divide each term in $4 n = - 1$ by $4$ and simplify.

Divide each term in $4 n = - 1$ by $4$ .

$\frac{4 n}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 n}{4} = \frac{- 1}{4}$

Divide $n$ by $1$ .

$n = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$n = - \frac{1}{4}$

Step 4

The final solution is all the values that make $(3 n - 2) (4 n + 1) = 0$ true.

$n = \frac{2}{3}, - \frac{1}{4}$