Precalculus Examples

$f (x) = 2 x^{2} - 7 x - 30$

Step 1

Set $2 x^{2} - 7 x - 30$ equal to $0$ .

$2 x^{2} - 7 x - 30 = 0$

Step 2

Solve for $x$ .

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Step 2.1

Factor by grouping.

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Step 2.1.1

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 $a \cdot c = 2 \cdot - 30 = - 60$ and whose sum is $b = - 7$ .

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Step 2.1.1.1

Factor $- 7$ out of $- 7 x$ .

$2 x^{2} - 7 x - 30 = 0$

Step 2.1.1.2

Rewrite $- 7$ as $5$ plus $- 12$

$2 x^{2} + (5 - 12) x - 30 = 0$

Step 2.1.1.3

Apply the distributive property.

$2 x^{2} + 5 x - 12 x - 30 = 0$

Step 2.1.2

Factor out the greatest common factor from each group.

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Step 2.1.2.1

Group the first two terms and the last two terms.

$(2 x^{2} + 5 x) - 12 x - 30 = 0$

Step 2.1.2.2

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

$x (2 x + 5) - 6 (2 x + 5) = 0$

Step 2.1.3

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

$(2 x + 5) (x - 6) = 0$

Step 2.2

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

$2 x + 5 = 0$

$x - 6 = 0$

Step 2.3

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

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Step 2.3.1

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

$2 x + 5 = 0$

Step 2.3.2

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

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Step 2.3.2.1

Subtract $5$ from both sides of the equation.

$2 x = - 5$

Step 2.3.2.2

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

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Step 2.3.2.2.1

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

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

Step 2.3.2.2.2

Simplify the left side.

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Step 2.3.2.2.2.1

Cancel the common factor of $2$ .

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Step 2.3.2.2.2.1.1

Cancel the common factor.

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

Step 2.3.2.2.2.1.2

Divide $x$ by $1$ .

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

Step 2.3.2.2.3

Simplify the right side.

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Step 2.3.2.2.3.1

Move the negative in front of the fraction.

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

Step 2.4

Set $x - 6$ equal to $0$ and solve for $x$ .

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Step 2.4.1

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Step 2.4.2

Add $6$ to both sides of the equation.

$x = 6$

Step 2.5

The final solution is all the values that make $(2 x + 5) (x - 6) = 0$ true.

$x = - \frac{5}{2}, 6$

Step 3