Solve for x x^3-8x^2+17x-10<0

$x^{3} - 8 x^{2} + 17 x - 10 < 0$

Step 1

Convert the inequality to an equation.

$x^{3} - 8 x^{2} + 17 x - 10 = 0$

Step 2

Factor the left side of the equation.

Factor $x^{3} - 8 x^{2} + 17 x - 10$ 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.

$p = \pm 1, \pm 10, \pm 2, \pm 5 q = \pm 1$

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

$\pm 1, \pm 10, \pm 2, \pm 5$

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.

$1^{3} - 8 \cdot 1^{2} + 17 \cdot 1 - 10$

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

$1 - 8 \cdot 1^{2} + 17 \cdot 1 - 10$

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

$1 - 8 \cdot 1 + 17 \cdot 1 - 10$

Multiply $- 8$ by $1$ .

$1 - 8 + 17 \cdot 1 - 10$

Subtract $8$ from $1$ .

$- 7 + 17 \cdot 1 - 10$

Multiply $17$ by $1$ .

$- 7 + 17 - 10$

Add $- 7$ and $17$ .

$10 - 10$

Subtract $10$ from $10$ .

$0$

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.

$\frac{x^{3} - 8 x^{2} + 17 x - 10}{x - 1}$

Divide $x^{3} - 8 x^{2} + 17 x - 10$ 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 $0$ .

$x$

$1$

$x^{3}$

$8 x^{2}$

$17 x$

$10$

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

					$x^{2}$
	$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			+	$x^{3}$	-	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$

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

				$x^{2}$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$

Divide the highest order term in the dividend $- 7 x^{2}$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					-	$7 x^{2}$	+	$7 x$

The expression needs to be subtracted from the dividend, so change all the signs in $- 7 x^{2} + 7 x$

				$x^{2}$	-	$7 x$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					+	$7 x^{2}$	-	$7 x$

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

				$x^{2}$	-	$7 x$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					+	$7 x^{2}$	-	$7 x$
							+	$10 x$

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

				$x^{2}$	-	$7 x$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					+	$7 x^{2}$	-	$7 x$
							+	$10 x$	-	$10$

Divide the highest order term in the dividend $10 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$7 x$	+	$10$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					+	$7 x^{2}$	-	$7 x$
							+	$10 x$	-	$10$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$7 x$	+	$10$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					+	$7 x^{2}$	-	$7 x$
							+	$10 x$	-	$10$
							+	$10 x$	-	$10$

The expression needs to be subtracted from the dividend, so change all the signs in $10 x - 10$

				$x^{2}$	-	$7 x$	+	$10$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					+	$7 x^{2}$	-	$7 x$
							+	$10 x$	-	$10$
							-	$10 x$	+	$10$

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

				$x^{2}$	-	$7 x$	+	$10$
$x$	-	$1$		$x^{3}$	-	$8 x^{2}$	+	$17 x$	-	$10$
			-	$x^{3}$	+	$x^{2}$
					-	$7 x^{2}$	+	$17 x$
					+	$7 x^{2}$	-	$7 x$
							+	$10 x$	-	$10$
							-	$10 x$	+	$10$
										$0$

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

$x^{2} - 7 x + 10$

Write $x^{3} - 8 x^{2} + 17 x - 10$ as a set of factors.

$(x - 1) (x^{2} - 7 x + 10) = 0$

Factor $x^{2} - 7 x + 10$ 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 $10$ and whose sum is $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

$(x - 1) ((x - 5) (x - 2)) = 0$

Remove unnecessary parentheses.

$(x - 1) (x - 5) (x - 2) = 0$

Step 3

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

$x - 1 = 0$

$x - 5 = 0$

$x - 2 = 0$

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 $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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 $(x - 1) (x - 5) (x - 2) = 0$ true.

$x = 1, 5, 2$

Step 8

Use each root to create test intervals.

$x < 1$

$1 < x < 2$

$2 < x < 5$

$x > 5$

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.

${(0)}^{3} - 8 {(0)}^{2} + 17 (0) - 10 < 0$

The left side $- 10$ 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.

${(1.5)}^{3} - 8 {(1.5)}^{2} + 17 (1.5) - 10 < 0$

The left side $0.875$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $2 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

Replace $x$ with $4$ in the original inequality.

${(4)}^{3} - 8 {(4)}^{2} + 17 (4) - 10 < 0$

The left side $- 6$ is less than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

${(8)}^{3} - 8 {(8)}^{2} + 17 (8) - 10 < 0$

The left side $126$ 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

$2 < x < 5$ True

$x > 5$ False

$x < 1$ True

$1 < x < 2$ False

$2 < x < 5$ True

$x > 5$ False

Step 10

The solution consists of all of the true intervals.

$x < 1$ or $2 < x < 5$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x < 1 or 2 < x < 5$

Interval Notation:

$(- \infty, 1) \cup (2, 5)$

Step 12

Solve for x x^3-8x^2+17x-10<0

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