Convert to Interval Notation x^2-13x+40>0

$x^{2} - 13 x + 40 > 0$

Step 1

Convert the inequality to an equation.

$x^{2} - 13 x + 40 = 0$

Step 2

Factor $x^{2} - 13 x + 40$ 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 $- 13$ .

$- 8, - 5$

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 $0$ .

$x - 5 = 0$

Add $5$ to 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.

$x = 0$

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

${(0)}^{2} - 13 \cdot 0 + 40 > 0$

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

True

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.

$x = 6$

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

${(6)}^{2} - 13 \cdot 6 + 40 > 0$

The left side $- 2$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

${(10)}^{2} - 13 \cdot 10 + 40 > 0$

The left side $10$ is greater 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

$5 < x < 8$ False

$x > 8$ True

$x < 5$ True

$5 < x < 8$ False

$x > 8$ True

Step 9

The solution consists of all of the true intervals.

$x < 5$ or $x > 8$

Step 10

Convert the inequality to interval notation.

$(- \infty, 5) \cup (8, \infty)$

Step 11

Convert to Interval Notation x^2-13x+40>0

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