Solve the Inequality for x (x-1)(x+2)(2x-7)^2<=0

$(x - 1) (x + 2) {(2 x - 7)}^{2} \leq 0$

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

$x - 1 = 0$

$x + 2 = 0$

${(2 x - 7)}^{2} = 0$

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.

$x = 1$

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

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

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

Set ${(2 x - 7)}^{2}$ equal to $0$ and solve for $x$ .

Set ${(2 x - 7)}^{2}$ equal to $0$ .

${(2 x - 7)}^{2} = 0$

Solve ${(2 x - 7)}^{2} = 0$ for $x$ .

Set the $2 x - 7$ equal to $0$ .

$2 x - 7 = 0$

Solve for $x$ .

Add $7$ to both sides of the equation.

$2 x = 7$

Divide each term in $2 x = 7$ by $2$ and simplify.

Divide each term in $2 x = 7$ by $2$ .

$\frac{2 x}{2} = \frac{7}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{7}{2}$

Divide $x$ by $1$ .

$x = \frac{7}{2}$

The final solution is all the values that make $(x - 1) (x + 2) {(2 x - 7)}^{2} \leq 0$ true.

$x = 1, - 2, \frac{7}{2}$

Use each root to create test intervals.

$x < - 2$

$- 2 < x < 1$

$1 < x < \frac{7}{2}$

$x > \frac{7}{2}$

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 $x < - 2$ 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.

$((- 4) - 1) ((- 4) + 2) {(2 (- 4) - 7)}^{2} \leq 0$

The left side $2250$ is greater than the right side $0$ , which means that the given statement is false.

False

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

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

$x = 0$

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

$((0) - 1) ((0) + 2) {(2 (0) - 7)}^{2} \leq 0$

The left side $- 98$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < \frac{7}{2}$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < \frac{7}{2}$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$((2) - 1) ((2) + 2) {(2 (2) - 7)}^{2} \leq 0$

The left side $36$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > \frac{7}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > \frac{7}{2}$ and see if this value makes the original inequality true.

$x = 6$

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

$((6) - 1) ((6) + 2) {(2 (6) - 7)}^{2} \leq 0$

The left side $1000$ 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.

$x < - 2$ False

$- 2 < x < 1$ True

$1 < x < \frac{7}{2}$ False

$x > \frac{7}{2}$ False

$x < - 2$ False

$- 2 < x < 1$ True

$1 < x < \frac{7}{2}$ False

$x > \frac{7}{2}$ False

The solution consists of all of the true intervals.

$- 2 \leq x \leq 1$ or $x = \frac{7}{2}$

The result can be shown in multiple forms.

Inequality Form:

$- 2 \leq x \leq 1 or x = \frac{7}{2}$

Interval Notation:

$[- 2, 1] \cup [\frac{7}{2}, \frac{7}{2}]$

Solve the Inequality for x (x-1)(x+2)(2x-7)^2<=0

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