Pre-Algebra Examples

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

Step 1

Factor by grouping.

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Step 1.1

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 $b = 1$ .

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Step 1.1.1

Multiply by $1$ .

$2 x^{2} + 1 x - 15 = 0$

Step 1.1.2

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

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

Step 1.1.3

Apply the distributive property.

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

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

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

Step 1.2.2

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

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

Step 1.3

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 $x$ .

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Step 3.1

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

$2 x - 5 = 0$

Step 3.2

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

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Step 3.2.1

Add $5$ to both sides of the equation.

$2 x = 5$

Step 3.2.2

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

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Step 3.2.2.1

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

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

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

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

Step 3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 4

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

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Step 4.1

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

$x + 3 = 0$

Step 4.2

Subtract $3$ from 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.

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