Solve by Factoring 2x^2-x-15=0

$2 x^{2} - x - 15 = 0$

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 $a \cdot c = 2 \cdot - 15 = - 30$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$2 x^{2} - (x) - 15 = 0$

Rewrite $- 1$ as $5$ plus $- 6$

$2 x^{2} + (5 - 6) x - 15 = 0$

Apply the distributive property.

$2 x^{2} + 5 x - 6 x - 15 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 x^{2} + 5 x) - 6 x - 15 = 0$

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

$x (2 x + 5) - 3 (2 x + 5) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 x + 5$ .

$(2 x + 5) (x - 3) = 0$

Step 2

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

$2 x + 5 = 0$

$x - 3 = 0$

Step 3

Set $2 x + 5$ equal to $0$ and solve for $x$ .

Set $2 x + 5$ equal to $0$ .

$2 x + 5 = 0$

Solve $2 x + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Divide each term in $2 x = - 5$ by $2$ and simplify.

Divide each term in $2 x = - 5$ by $2$ .

$\frac{2 x}{2} = \frac{- 5}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 5}{2}$

Divide $x$ by $1$ .

$x = \frac{- 5}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{5}{2}$

Step 4

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 5

The final solution is all the values that make $(2 x + 5) (x - 3) = 0$ true.

$x = - \frac{5}{2}, 3$

Solve by Factoring 2x^2-x-15=0

Related Questions

Support LCMGCF.com by disabling your adblocker.