Solve by Factoring 12(p^2-1)=7p

$12 (p^{2} - 1) = 7 p$

Subtract $7 p$ from both sides of the equation.

$12 (p^{2} - 1) - 7 p = 0$

Simplify each term.

Apply the distributive property.

$12 p^{2} + 12 \cdot - 1 - 7 p = 0$

Multiply $12$ by $- 1$ .

$12 p^{2} - 12 - 7 p = 0$

Factor by grouping.

Reorder terms.

$12 p^{2} - 7 p - 12 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 12 \cdot - 12 = - 144$ and whose sum is $b = - 7$ .

Factor $- 7$ out of $- 7 p$ .

$12 p^{2} - 7 p - 12 = 0$

Rewrite $- 7$ as $9$ plus $- 16$

$12 p^{2} + (9 - 16) p - 12 = 0$

Apply the distributive property.

$12 p^{2} + 9 p - 16 p - 12 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(12 p^{2} + 9 p) - 16 p - 12 = 0$

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

$3 p (4 p + 3) - 4 (4 p + 3) = 0$

Factor the polynomial by factoring out the greatest common factor, $4 p + 3$ .

$(4 p + 3) (3 p - 4) = 0$

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

$4 p + 3 = 0$

$3 p - 4 = 0$

Set $4 p + 3$ equal to $0$ and solve for $p$ .

Set $4 p + 3$ equal to $0$ .

$4 p + 3 = 0$

Solve $4 p + 3 = 0$ for $p$ .

Subtract $3$ from both sides of the equation.

$4 p = - 3$

Divide each term in $4 p = - 3$ by $4$ and simplify.

Divide each term in $4 p = - 3$ by $4$ .

$\frac{4 p}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 p}{4} = \frac{- 3}{4}$

Divide $p$ by $1$ .

$p = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$p = - \frac{3}{4}$

Set $3 p - 4$ equal to $0$ and solve for $p$ .

Set $3 p - 4$ equal to $0$ .

$3 p - 4 = 0$

Solve $3 p - 4 = 0$ for $p$ .

Add $4$ to both sides of the equation.

$3 p = 4$

Divide each term in $3 p = 4$ by $3$ and simplify.

Divide each term in $3 p = 4$ by $3$ .

$\frac{3 p}{3} = \frac{4}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 p}{3} = \frac{4}{3}$

Divide $p$ by $1$ .

$p = \frac{4}{3}$

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

$p = - \frac{3}{4}, \frac{4}{3}$

Solve by Factoring 12(p^2-1)=7p

Related Questions

Support LCMGCF.com by disabling your adblocker.