Convert to Set Notation h^3+2h^2-16<=0

$h^{3} + 2 h^{2} - 16 \leq 0$

Step 1

Solve $h \leq 2$ .

Convert the inequality to an equation.

$h^{3} + 2 h^{2} - 16 = 0$

Factor $h^{3} + 2 h^{2} - 16$ 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.

$p = \pm 1, \pm 16, \pm 2, \pm 8, \pm 4 q = \pm 1$

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

$\pm 1, \pm 16, \pm 2, \pm 8, \pm 4$

Substitute $2$ and simplify the expression. In this case, the expression is equal to $0$ so $2$ is a root of the polynomial.

Substitute $2$ into the polynomial.

$2^{3} + 2 \cdot 2^{2} - 16$

Raise $2$ to the power of $3$ .

$8 + 2 \cdot 2^{2} - 16$

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

$8 + 2 \cdot 4 - 16$

Multiply $2$ by $4$ .

$8 + 8 - 16$

Add $8$ and $8$ .

$16 - 16$

Subtract $16$ from $16$ .

$0$

Since $2$ is a known root, divide the polynomial by $h - 2$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{h^{3} + 2 h^{2} - 16}{h - 2}$

Divide $h^{3} + 2 h^{2} - 16$ by $h - 2$ .

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

$h$

$2$

$h^{3}$

$2 h^{2}$

$0 h$

$16$

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

					$h^{2}$
	$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$

Multiply the new quotient term by the divisor.

				$h^{2}$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			+	$h^{3}$	-	$2 h^{2}$

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

				$h^{2}$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$

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

				$h^{2}$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$

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

				$h^{2}$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$

Divide the highest order term in the dividend $4 h^{2}$ by the highest order term in divisor $h$ .

				$h^{2}$	+	$4 h$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$

Multiply the new quotient term by the divisor.

				$h^{2}$	+	$4 h$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					+	$4 h^{2}$	-	$8 h$

The expression needs to be subtracted from the dividend, so change all the signs in $4 h^{2} - 8 h$

				$h^{2}$	+	$4 h$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					-	$4 h^{2}$	+	$8 h$

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

				$h^{2}$	+	$4 h$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					-	$4 h^{2}$	+	$8 h$
							+	$8 h$

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

				$h^{2}$	+	$4 h$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					-	$4 h^{2}$	+	$8 h$
							+	$8 h$	-	$16$

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

				$h^{2}$	+	$4 h$	+	$8$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					-	$4 h^{2}$	+	$8 h$
							+	$8 h$	-	$16$

Multiply the new quotient term by the divisor.

				$h^{2}$	+	$4 h$	+	$8$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					-	$4 h^{2}$	+	$8 h$
							+	$8 h$	-	$16$
							+	$8 h$	-	$16$

The expression needs to be subtracted from the dividend, so change all the signs in $8 h - 16$

				$h^{2}$	+	$4 h$	+	$8$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					-	$4 h^{2}$	+	$8 h$
							+	$8 h$	-	$16$
							-	$8 h$	+	$16$

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

				$h^{2}$	+	$4 h$	+	$8$
$h$	-	$2$		$h^{3}$	+	$2 h^{2}$	+	$0 h$	-	$16$
			-	$h^{3}$	+	$2 h^{2}$
					+	$4 h^{2}$	+	$0 h$
					-	$4 h^{2}$	+	$8 h$
							+	$8 h$	-	$16$
							-	$8 h$	+	$16$
										$0$

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

$h^{2} + 4 h + 8$

Write $h^{3} + 2 h^{2} - 16$ as a set of factors.

$(h - 2) (h^{2} + 4 h + 8) = 0$

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

$h - 2 = 0$

$h^{2} + 4 h + 8 = 0$

Set $h - 2$ equal to $0$ and solve for $h$ .

Set $h - 2$ equal to $0$ .

$h - 2 = 0$

Add $2$ to both sides of the equation.

$h = 2$

Set $h^{2} + 4 h + 8$ equal to $0$ and solve for $h$ .

Set $h^{2} + 4 h + 8$ equal to $0$ .

$h^{2} + 4 h + 8 = 0$

Solve $h^{2} + 4 h + 8 = 0$ for $h$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 4$ , and $c = 8$ into the quadratic formula and solve for $h$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 8)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $4$ to the power of $2$ .

$h = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 8$ .

Multiply $- 4$ by $1$ .

$h = \frac{- 4 \pm \sqrt{16 - 4 \cdot 8}}{2 \cdot 1}$

Multiply $- 4$ by $8$ .

$h = \frac{- 4 \pm \sqrt{16 - 32}}{2 \cdot 1}$

Subtract $32$ from $16$ .

$h = \frac{- 4 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$h = \frac{- 4 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$h = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$h = \frac{- 4 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$h = \frac{- 4 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

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

$h = \frac{- 4 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$h = \frac{- 4 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$h = \frac{- 4 \pm 4 i}{2}$

Simplify $\frac{- 4 \pm 4 i}{2}$ .

$h = - 2 \pm 2 i$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$h = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 8$ .

Multiply $- 4$ by $1$ .

$h = \frac{- 4 \pm \sqrt{16 - 4 \cdot 8}}{2 \cdot 1}$

Multiply $- 4$ by $8$ .

$h = \frac{- 4 \pm \sqrt{16 - 32}}{2 \cdot 1}$

Subtract $32$ from $16$ .

$h = \frac{- 4 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$h = \frac{- 4 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$h = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$h = \frac{- 4 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$h = \frac{- 4 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

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

$h = \frac{- 4 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$h = \frac{- 4 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$h = \frac{- 4 \pm 4 i}{2}$

Simplify $\frac{- 4 \pm 4 i}{2}$ .

$h = - 2 \pm 2 i$

Change the $\pm$ to $+$ .

$h = - 2 + 2 i$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Simplify the numerator.

Raise $4$ to the power of $2$ .

$h = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 8}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 8$ .

Multiply $- 4$ by $1$ .

$h = \frac{- 4 \pm \sqrt{16 - 4 \cdot 8}}{2 \cdot 1}$

Multiply $- 4$ by $8$ .

$h = \frac{- 4 \pm \sqrt{16 - 32}}{2 \cdot 1}$

Subtract $32$ from $16$ .

$h = \frac{- 4 \pm \sqrt{- 16}}{2 \cdot 1}$

Rewrite $- 16$ as $- 1 (16)$ .

$h = \frac{- 4 \pm \sqrt{- 1 \cdot 16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$h = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$h = \frac{- 4 \pm i \cdot \sqrt{16}}{2 \cdot 1}$

Rewrite $16$ as $4^{2}$ .

$h = \frac{- 4 \pm i \cdot \sqrt{4^{2}}}{2 \cdot 1}$

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

$h = \frac{- 4 \pm i \cdot 4}{2 \cdot 1}$

Move $4$ to the left of $i$ .

$h = \frac{- 4 \pm 4 i}{2 \cdot 1}$

Multiply $2$ by $1$ .

$h = \frac{- 4 \pm 4 i}{2}$

Simplify $\frac{- 4 \pm 4 i}{2}$ .

$h = - 2 \pm 2 i$

Change the $\pm$ to $-$ .

$h = - 2 - 2 i$

The final answer is the combination of both solutions.

$h = - 2 + 2 i, - 2 - 2 i$

The final solution is all the values that make $(h - 2) (h^{2} + 4 h + 8) = 0$ true.

$h = 2, - 2 + 2 i, - 2 - 2 i$

The solution consists of all of the true intervals.

$h \leq 2$

Step 2

Use the inequality $h \leq 2$ to build the set notation.

${h | h \leq 2}$

Step 3

Convert to Set Notation h^3+2h^2-16<=0

Related Questions

Support LCMGCF.com by disabling your adblocker.