Solve the Inequality for x x^3-4x^2-x+4>0

$x^{3} - 4 x^{2} - x + 4 > 0$

Convert the inequality to an equation.

$x^{3} - 4 x^{2} - x + 4 = 0$

Factor the left side of the equation.

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(x^{3} - 4 x^{2}) - x + 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$x^{2} (x - 4) - (x - 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $x - 4$ .

$(x - 4) (x^{2} - 1) = 0$

Rewrite $1$ as $1^{2}$ .

$(x - 4) (x^{2} - 1^{2}) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$(x - 4) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x - 4) (x + 1) (x - 1) = 0$

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

$x - 4 = 0$

$x + 1 = 0$

$x - 1 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

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.

$x = - 1$

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 $(x - 4) (x + 1) (x - 1) = 0$ true.

$x = 4, - 1, 1$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 1$

$1 < x < 4$

$x > 4$

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.

${(- 4)}^{3} - 4 {(- 4)}^{2} - (- 4) + 4 > 0$

The left side $- 120$ is not greater than the right side $0$ , which means that the given statement is false.

False

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.

${(0)}^{3} - 4 {(0)}^{2} - (0) + 4 > 0$

The left side $4$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

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

${(2)}^{3} - 4 {(2)}^{2} - (2) + 4 > 0$

The left side $- 6$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

${(6)}^{3} - 4 {(6)}^{2} - (6) + 4 > 0$

The left side $70$ 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.

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

$x < - 1$ False

$- 1 < x < 1$ True

$1 < x < 4$ False

$x > 4$ True

The solution consists of all of the true intervals.

$- 1 < x < 1$ or $x > 4$

The result can be shown in multiple forms.

Inequality Form:

$- 1 < x < 1 or x > 4$

Interval Notation:

$(- 1, 1) \cup (4, \infty)$

Solve the Inequality for x x^3-4x^2-x+4>0

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