Solve for x (x+1)(x^2-3x+2)<0

$(x + 1) (x^{2} - 3 x + 2) < 0$

Step 1

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

$x + 1 = 0$

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

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 $x^{2} - 3 x + 2$ equal to $0$ and solve for $x$ .

Set $x^{2} - 3 x + 2$ equal to $0$ .

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

Solve $x^{2} - 3 x + 2 = 0$ for $x$ .

Factor $x^{2} - 3 x + 2$ 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 $2$ and whose sum is $- 3$ .

$- 2, - 1$

Write the factored form using these integers.

$(x - 2) (x - 1) = 0$

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

$x - 2 = 0$

$x - 1 = 0$

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.

$x = 2$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(x - 2) (x - 1) = 0$ true.

$x = 2, 1$

Step 4

The final solution is all the values that make $(x + 1) (x^{2} - 3 x + 2) < 0$ true.

$x = - 1, 2, 1$

Step 5

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 2$

$x > 2$

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.

$((- 4) + 1) ({(- 4)}^{2} - 3 \cdot - 4 + 2) < 0$

The left side $- 90$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 0$

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

$((0) + 1) ({(0)}^{2} - 3 \cdot 0 + 2) < 0$

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

False

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.

$x = 1.5$

Replace $x$ with $1.5$ in the original inequality.

$((1.5) + 1) ({(1.5)}^{2} - 3 \cdot 1.5 + 2) < 0$

The left side $- 0.625$ is less than the right side $0$ , which means that the given statement is always true.

True

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

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

$x = 4$

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

$((4) + 1) ({(4)}^{2} - 3 \cdot 4 + 2) < 0$

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.

$x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 2$ True

$x > 2$ False

$x < - 1$ True

$- 1 < x < 1$ False

$1 < x < 2$ True

$x > 2$ False

Step 7

The solution consists of all of the true intervals.

$x < - 1$ or $1 < x < 2$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$x < - 1 or 1 < x < 2$

Interval Notation:

$(- \infty, - 1) \cup (1, 2)$

Step 9

Solve for x (x+1)(x^2-3x+2)<0

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