Solve the Inequality for x x^3-7x^2+14x-8<0

$x^{3} - 7 x^{2} + 14 x - 8 < 0$

Step 1

Convert the inequality to an equation.

$x^{3} - 7 x^{2} + 14 x - 8 = 0$

Step 2

Factor the left side of the equation.

Factor $x^{3} - 7 x^{2} + 14 x - 8$ 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.

$p = \pm 1, \pm 8, \pm 2, \pm 4 q = \pm 1$

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

$\pm 1, \pm 8, \pm 2, \pm 4$

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.

$1^{3} - 7 \cdot 1^{2} + 14 \cdot 1 - 8$

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

$1 - 7 \cdot 1^{2} + 14 \cdot 1 - 8$

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

$1 - 7 \cdot 1 + 14 \cdot 1 - 8$

Multiply $- 7$ by $1$ .

$1 - 7 + 14 \cdot 1 - 8$

Subtract $7$ from $1$ .

$- 6 + 14 \cdot 1 - 8$

Multiply $14$ by $1$ .

$- 6 + 14 - 8$

Add $- 6$ and $14$ .

$8 - 8$

Subtract $8$ from $8$ .

$0$

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.

$\frac{x^{3} - 7 x^{2} + 14 x - 8}{x - 1}$

Divide $x^{3} - 7 x^{2} + 14 x - 8$ 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 $0$ .

$x$

$1$

$x^{3}$

$7 x^{2}$

$14 x$

$8$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$

Divide the highest order term in the dividend $- 6 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					-	$6 x^{2}$	+	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 6 x^{2} + 6 x$

				$x^{2}$	-	$6 x$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					+	$6 x^{2}$	-	$6 x$

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

				$x^{2}$	-	$6 x$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					+	$6 x^{2}$	-	$6 x$
							+	$8 x$

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

				$x^{2}$	-	$6 x$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					+	$6 x^{2}$	-	$6 x$
							+	$8 x$	-	$8$

Divide the highest order term in the dividend $8 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$6 x$	+	$8$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					+	$6 x^{2}$	-	$6 x$
							+	$8 x$	-	$8$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$6 x$	+	$8$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					+	$6 x^{2}$	-	$6 x$
							+	$8 x$	-	$8$
							+	$8 x$	-	$8$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x - 8$

				$x^{2}$	-	$6 x$	+	$8$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					+	$6 x^{2}$	-	$6 x$
							+	$8 x$	-	$8$
							-	$8 x$	+	$8$

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

				$x^{2}$	-	$6 x$	+	$8$
$x$	-	$1$		$x^{3}$	-	$7 x^{2}$	+	$14 x$	-	$8$
			-	$x^{3}$	+	$x^{2}$
					-	$6 x^{2}$	+	$14 x$
					+	$6 x^{2}$	-	$6 x$
							+	$8 x$	-	$8$
							-	$8 x$	+	$8$
										$0$

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

$x^{2} - 6 x + 8$

Write $x^{3} - 7 x^{2} + 14 x - 8$ as a set of factors.

$(x - 1) (x^{2} - 6 x + 8) = 0$

Factor $x^{2} - 6 x + 8$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$(x - 1) ((x - 4) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x - 1) (x - 4) (x - 2) = 0$

Step 3

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

$x - 1 = 0$

$x - 4 = 0$

$x - 2 = 0$

Step 4

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.

$x = 1$

Step 5

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.

$x = 4$

Step 6

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.

$x = 2$

Step 7

The final solution is all the values that make $(x - 1) (x - 4) (x - 2) = 0$ true.

$x = 1, 4, 2$

Step 8

Use each root to create test intervals.

$x < 1$

$1 < x < 2$

$2 < x < 4$

$x > 4$

Step 9

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.

$x = 0$

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

${(0)}^{3} - 7 {(0)}^{2} + 14 (0) - 8 < 0$

The left side $- 8$ 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.

$x = 1.5$

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

${(1.5)}^{3} - 7 {(1.5)}^{2} + 14 (1.5) - 8 < 0$

The left side $0.625$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 4$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

${(3)}^{3} - 7 {(3)}^{2} + 14 (3) - 8 < 0$

The left side $- 2$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

${(6)}^{3} - 7 {(6)}^{2} + 14 (6) - 8 < 0$

The left side $40$ 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

$1 < x < 2$ False

$2 < x < 4$ True

$x > 4$ False

$x < 1$ True

$1 < x < 2$ False

$2 < x < 4$ True

$x > 4$ False

Step 10

The solution consists of all of the true intervals.

$x < 1$ or $2 < x < 4$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x < 1 or 2 < x < 4$

Interval Notation:

$(- \infty, 1) \cup (2, 4)$

Step 12

Solve the Inequality for x x^3-7x^2+14x-8<0

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