Solve by Factoring 5p^2-9p-2=0

$5 p^{2} - 9 p - 2 = 0$

Step 1

Factor by grouping.

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 $a \cdot c = 5 \cdot - 2 = - 10$ and whose sum is $b = - 9$ .

Factor $- 9$ out of $- 9 p$ .

$5 p^{2} - 9 p - 2 = 0$

Rewrite $- 9$ as $1$ plus $- 10$

$5 p^{2} + (1 - 10) p - 2 = 0$

Apply the distributive property.

$5 p^{2} + 1 p - 10 p - 2 = 0$

Multiply $p$ by $1$ .

$5 p^{2} + p - 10 p - 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 p^{2} + p) - 10 p - 2 = 0$

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

$p (5 p + 1) - 2 (5 p + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 p + 1$ .

$(5 p + 1) (p - 2) = 0$

Step 2

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

$5 p + 1 = 0$

$p - 2 = 0$

Step 3

Set $5 p + 1$ equal to $0$ and solve for $p$ .

Set $5 p + 1$ equal to $0$ .

$5 p + 1 = 0$

Solve $5 p + 1 = 0$ for $p$ .

Subtract $1$ from both sides of the equation.

$5 p = - 1$

Divide each term in $5 p = - 1$ by $5$ and simplify.

Divide each term in $5 p = - 1$ by $5$ .

$\frac{5 p}{5} = \frac{- 1}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 p}{5} = \frac{- 1}{5}$

Divide $p$ by $1$ .

$p = \frac{- 1}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$p = - \frac{1}{5}$

Step 4

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

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

$p - 2 = 0$

Add $2$ to both sides of the equation.

$p = 2$

Step 5

The final solution is all the values that make $(5 p + 1) (p - 2) = 0$ true.

$p = - \frac{1}{5}, 2$

Solve by Factoring 5p^2-9p-2=0

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