Solve the Inequality for x (x+5)(-2x+6)<0

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

Step 1

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

$x + 5 = 0$

$- 2 x + 6 = 0$

Step 2

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

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

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 3

Set $- 2 x + 6$ equal to $0$ and solve for $x$ .

Set $- 2 x + 6$ equal to $0$ .

$- 2 x + 6 = 0$

Solve $- 2 x + 6 = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$- 2 x = - 6$

Divide each term in $- 2 x = - 6$ by $- 2$ and simplify.

Divide each term in $- 2 x = - 6$ by $- 2$ .

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

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 6$ by $- 2$ .

$x = 3$

Step 4

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

$x = - 5, 3$

Step 5

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 3$

$x > 3$

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Test a value on the interval $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$((- 8) + 5) (- 2 \cdot - 8 + 6) < 0$

The left side $- 66$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

Replace $x$ with $0$ in the original inequality.

$((0) + 5) (- 2 \cdot 0 + 6) < 0$

The left side $30$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 3$ to see if it makes the inequality true.

Choose a value on the interval $x > 3$ and see if this value makes the original inequality true.

$x = 6$

Replace $x$ with $6$ in the original inequality.

$((6) + 5) (- 2 \cdot 6 + 6) < 0$

The left side $- 66$ is less than the right side $0$ , which means that the given statement is always true.

True

Compare the intervals to determine which ones satisfy the original inequality.

$x < - 5$ True

$- 5 < x < 3$ False

$x > 3$ True

$x < - 5$ True

$- 5 < x < 3$ False

$x > 3$ True

Step 7

The solution consists of all of the true intervals.

$x < - 5$ or $x > 3$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < - 5 or x > 3$

Interval Notation:

$(- \infty, - 5) \cup (3, \infty)$

Step 9

Solve the Inequality for x (x+5)(-2x+6)<0

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