Solve for x x^2-3x-40<0

$x^{2} - 3 x - 40 < 0$

Step 1

Convert the inequality to an equation.

$x^{2} - 3 x - 40 = 0$

Step 2

Factor $x^{2} - 3 x - 40$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 40$ and whose sum is $- 3$ .

$- 8, 5$

Write the factored form using these integers.

$(x - 8) (x + 5) = 0$

Step 3

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

$x - 8 = 0$

$x + 5 = 0$

Step 4

Set $x - 8$ equal to $0$ and solve for $x$ .

Set $x - 8$ equal to $0$ .

$x - 8 = 0$

Add $8$ to both sides of the equation.

$x = 8$

Step 5

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 6

The final solution is all the values that make $(x - 8) (x + 5) = 0$ true.

$x = 8, - 5$

Step 7

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 8$

$x > 8$

Step 8

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.

${(- 8)}^{2} - 3 \cdot - 8 - 40 < 0$

The left side $48$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < 8$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 8$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} - 3 \cdot 0 - 40 < 0$

The left side $- 40$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 8$ to see if it makes the inequality true.

Choose a value on the interval $x > 8$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

${(10)}^{2} - 3 \cdot 10 - 40 < 0$

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

False

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

$x < - 5$ False

$- 5 < x < 8$ True

$x > 8$ False

$x < - 5$ False

$- 5 < x < 8$ True

$x > 8$ False

Step 9

The solution consists of all of the true intervals.

$- 5 < x < 8$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$- 5 < x < 8$

Interval Notation:

$(- 5, 8)$

Step 11

Solve for x x^2-3x-40<0

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