Solve the Inequality for x x^3-4x^2+x+6<=0

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

Convert the inequality to an equation.

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

Factor the left side of the equation.

Factor $x^{3} - 4 x^{2} + x + 6$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 6, \pm 2, \pm 3 q = \pm 1$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 6, \pm 2, \pm 3$

Substitute $- 1$ and simplify the expression. In this case, the expression is equal to $0$ so $- 1$ is a root of the polynomial.

Substitute $- 1$ into the polynomial.

${(- 1)}^{3} - 4 {(- 1)}^{2} - 1 + 6$

Raise $- 1$ to the power of $3$ .

$- 1 - 4 {(- 1)}^{2} - 1 + 6$

Raise $- 1$ to the power of $2$ .

$- 1 - 4 \cdot 1 - 1 + 6$

Multiply $- 4$ by $1$ .

$- 1 - 4 - 1 + 6$

Subtract $4$ from $- 1$ .

$- 5 - 1 + 6$

Subtract $1$ from $- 5$ .

$- 6 + 6$

Add $- 6$ and $6$ .

$0$

Since $- 1$ is a known root, divide the polynomial by $x + 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{x^{3} - 4 x^{2} + x + 6}{x + 1}$

Divide $x^{3} - 4 x^{2} + x + 6$ by $x + 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$x$

$1$

$x^{3}$

$4 x^{2}$

$x$

$6$

Divide the highest order term in the dividend $x^{3}$ by the highest order term in divisor $x$ .

					$x^{2}$
	$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			+	$x^{3}$	+	$x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $x^{3} + x^{2}$

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$

Divide the highest order term in the dividend $- 5 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$5 x$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					-	$5 x^{2}$	-	$5 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 5 x^{2} - 5 x$

				$x^{2}$	-	$5 x$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	+	$5 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$5 x$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	+	$5 x$
							+	$6 x$

Pull the next terms from the original dividend down into the current dividend.

				$x^{2}$	-	$5 x$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	+	$5 x$
							+	$6 x$	+	$6$

Divide the highest order term in the dividend $6 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$5 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	+	$5 x$
							+	$6 x$	+	$6$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$5 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	+	$5 x$
							+	$6 x$	+	$6$
							+	$6 x$	+	$6$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x + 6$

				$x^{2}$	-	$5 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	+	$5 x$
							+	$6 x$	+	$6$
							-	$6 x$	-	$6$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$x^{2}$	-	$5 x$	+	$6$
$x$	+	$1$		$x^{3}$	-	$4 x^{2}$	+	$x$	+	$6$
			-	$x^{3}$	-	$x^{2}$
					-	$5 x^{2}$	+	$x$
					+	$5 x^{2}$	+	$5 x$
							+	$6 x$	+	$6$
							-	$6 x$	-	$6$
										$0$

Since the remander is $0$ , the final answer is the quotient.

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

Write $x^{3} - 4 x^{2} + x + 6$ as a set of factors.

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

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.

$(x + 1) ((x - 3) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x + 1) (x - 3) (x - 2) = 0$

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

$x + 1 = 0$

$x - 3 = 0$

$x - 2 = 0$

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 $x - 3$ 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 $(x + 1) (x - 3) (x - 2) = 0$ true.

$x = - 1, 3, 2$

Use each root to create test intervals.

$x < - 1$

$- 1 < x < 2$

$2 < x < 3$

$x > 3$

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.

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

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.

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

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.

${(2.5)}^{3} - 4 {(2.5)}^{2} + 2.5 + 6 \leq 0$

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.

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

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

The solution consists of all of the true intervals.

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

The result can be shown in multiple forms.

Inequality Form:

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

Interval Notation:

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

Solve the Inequality for x x^3-4x^2+x+6<=0

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