Solve the Inequality for x (x+1)(x^2-5x+6)<=0

$(x + 1) (x^{2} - 5 x + 6) \leq 0$

Step 1

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

$x + 1 = 0$

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

Step 2

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3

Set $x^{2} - 5 x + 6$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5 x + 6$ equal to $0$ .

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

Solve $x^{2} - 5 x + 6 = 0$ for $x$ .

Factor $x^{2} - 5 x + 6$ 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 $6$ and whose sum is $- 5$ .

$- 3, - 2$

Write the factored form using these integers.

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

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

$x - 3 = 0$

$x - 2 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

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

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

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

The final solution is all the values that make $(x - 3) (x - 2) = 0$ true.

$x = 3, 2$

Step 4

The final solution is all the values that make $(x + 1) (x^{2} - 5 x + 6) \leq 0$ true.

$x = - 1, 3, 2$

Step 5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 2$

$2 < x < 3$

$x > 3$

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 $x < - 1$ to see if it makes the inequality true.

Choose a value on the interval $x < - 1$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

$((- 4) + 1) ({(- 4)}^{2} - 5 \cdot - 4 + 6) \leq 0$

The left side $- 126$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- 1 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $- 1 < x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

$((0) + 1) ({(0)}^{2} - 5 \cdot 0 + 6) \leq 0$

The left side $6$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 3$ and see if this value makes the original inequality true.

$x = 2.5$

Replace $x$ with $2.5$ in the original inequality.

$((2.5) + 1) ({(2.5)}^{2} - 5 \cdot 2.5 + 6) \leq 0$

The left side $- 0.875$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

$((6) + 1) ({(6)}^{2} - 5 \cdot 6 + 6) \leq 0$

The left side $84$ is greater 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.

$x < - 1$ True

$- 1 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

$x < - 1$ True

$- 1 < x < 2$ False

$2 < x < 3$ True

$x > 3$ False

Step 7

The solution consists of all of the true intervals.

$x \leq - 1$ or $2 \leq x \leq 3$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 1 or 2 \leq x \leq 3$

Interval Notation:

$(- \infty, - 1] \cup [2, 3]$

Step 9

Solve the Inequality for x (x+1)(x^2-5x+6)<=0

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