Solve by Factoring x^4+x^3-6x^2<=0

$x^{4} + x^{3} - 6 x^{2} \leq 0$

Factor $x^{2}$ out of $x^{4} + x^{3} - 6 x^{2}$ .

Factor $x^{2}$ out of $x^{4}$ .

$x^{2} x^{2} + x^{3} - 6 x^{2} \leq 0$

Factor $x^{2}$ out of $x^{3}$ .

$x^{2} x^{2} + x^{2} x - 6 x^{2} \leq 0$

Factor $x^{2}$ out of $- 6 x^{2}$ .

$x^{2} x^{2} + x^{2} x + x^{2} \cdot - 6 \leq 0$

Factor $x^{2}$ out of $x^{2} x^{2} + x^{2} x$ .

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

Factor $x^{2}$ out of $x^{2} (x^{2} + x) + x^{2} \cdot - 6$ .

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

Factor.

Factor $x^{2} + x - 6$ 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 $1$ .

$- 2, 3$

Write the factored form using these integers.

$x^{2} ((x - 2) (x + 3)) \leq 0$

Remove unnecessary parentheses.

$x^{2} (x - 2) (x + 3) \leq 0$

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

$x^{2} = 0$

$x - 2 = 0$

$x + 3 = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

Take the square root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

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 $x + 3$ equal to $0$ and solve for $x$ .

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

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

$x = 0, 2, - 3$

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 0$

$0 < x < 2$

$x > 2$

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 $x < - 3$ 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.

${(- 6)}^{2} ((- 6) - 2) ((- 6) + 3) \leq 0$

The left side $864$ is greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 3 < x < 0$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 0$ and see if this value makes the original inequality true.

$x = - 2$

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

${(- 2)}^{2} ((- 2) - 2) ((- 2) + 3) \leq 0$

The left side $- 16$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $0 < x < 2$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 2$ and see if this value makes the original inequality true.

$x = 1$

Replace $x$ with $1$ in the original inequality.

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

The left side $- 4$ 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.

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

The left side $224$ 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.

$x < - 3$ False

$- 3 < x < 0$ True

$0 < x < 2$ True

$x > 2$ False

$x < - 3$ False

$- 3 < x < 0$ True

$0 < x < 2$ True

$x > 2$ False

The solution consists of all of the true intervals.

$- 3 \leq x \leq 0$ or $0 \leq x \leq 2$

Combine the intervals.

$- 3 \leq x \leq 2$

Solve by Factoring x^4+x^3-6x^2<=0

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