Solve for x 1/8x^3+1 1/2x=3/4x^2+1

$\frac{1}{8} x^{3} + 1 \frac{1}{2} x = \frac{3}{4} x^{2} + 1$

Simplify the left side.

Simplify $\frac{1}{8} x^{3} + 1 \frac{1}{2} x$ .

Convert $1 \frac{1}{2}$ to an improper fraction.

A mixed number is an addition of its whole and fractional parts.

$\frac{1}{8} x^{3} + (1 + \frac{1}{2}) x = \frac{3}{4} x^{2} + 1$

Add $1$ and $\frac{1}{2}$ .

Write $1$ as a fraction with a common denominator.

$\frac{1}{8} x^{3} + (\frac{2}{2} + \frac{1}{2}) x = \frac{3}{4} x^{2} + 1$

Combine the numerators over the common denominator.

$\frac{1}{8} x^{3} + \frac{2 + 1}{2} x = \frac{3}{4} x^{2} + 1$

Add $2$ and $1$ .

$\frac{1}{8} x^{3} + \frac{3}{2} x = \frac{3}{4} x^{2} + 1$

Simplify each term.

Combine $\frac{1}{8}$ and $x^{3}$ .

$\frac{x^{3}}{8} + \frac{3}{2} x = \frac{3}{4} x^{2} + 1$

Combine $\frac{3}{2}$ and $x$ .

$\frac{x^{3}}{8} + \frac{3 x}{2} = \frac{3}{4} x^{2} + 1$

Simplify the right side.

Combine $\frac{3}{4}$ and $x^{2}$ .

$\frac{x^{3}}{8} + \frac{3 x}{2} = \frac{3 x^{2}}{4} + 1$

Subtract $\frac{3 x^{2}}{4}$ from both sides of the equation.

$\frac{x^{3}}{8} + \frac{3 x}{2} - \frac{3 x^{2}}{4} = 1$

Multiply each term in $\frac{x^{3}}{8} + \frac{3 x}{2} - \frac{3 x^{2}}{4} = 1$ by $8$ to eliminate the fractions.

Multiply each term in $\frac{x^{3}}{8} + \frac{3 x}{2} - \frac{3 x^{2}}{4} = 1$ by $8$ .

$\frac{x^{3}}{8} \cdot 8 + \frac{3 x}{2} \cdot 8 - \frac{3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Simplify the left side.

Simplify each term.

Cancel the common factor of $8$ .

Cancel the common factor.

$\frac{x^{3}}{8} \cdot 8 + \frac{3 x}{2} \cdot 8 - \frac{3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Rewrite the expression.

$x^{3} + \frac{3 x}{2} \cdot 8 - \frac{3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Cancel the common factor of $2$ .

Factor $2$ out of $8$ .

$x^{3} + \frac{3 x}{2} \cdot (2 (4)) - \frac{3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Cancel the common factor.

$x^{3} + \frac{3 x}{2} \cdot (2 \cdot 4) - \frac{3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Rewrite the expression.

$x^{3} + 3 x \cdot 4 - \frac{3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Multiply $4$ by $3$ .

$x^{3} + 12 x - \frac{3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Cancel the common factor of $4$ .

Move the leading negative in $- \frac{3 x^{2}}{4}$ into the numerator.

$x^{3} + 12 x + \frac{- 3 x^{2}}{4} \cdot 8 = 1 \cdot 8$

Factor $4$ out of $8$ .

$x^{3} + 12 x + \frac{- 3 x^{2}}{4} \cdot (4 (2)) = 1 \cdot 8$

Cancel the common factor.

$x^{3} + 12 x + \frac{- 3 x^{2}}{4} \cdot (4 \cdot 2) = 1 \cdot 8$

Rewrite the expression.

$x^{3} + 12 x - 3 x^{2} \cdot 2 = 1 \cdot 8$

Multiply $2$ by $- 3$ .

$x^{3} + 12 x - 6 x^{2} = 1 \cdot 8$

Simplify the right side.

Multiply $8$ by $1$ .

$x^{3} + 12 x - 6 x^{2} = 8$

Subtract $8$ from both sides of the equation.

$x^{3} + 12 x - 6 x^{2} - 8 = 0$

Factor the left side of the equation.

Reorder terms.

$x^{3} - 6 x^{2} + 12 x - 8 = 0$

Factor $x^{3} - 6 x^{2} + 12 x - 8$ 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 $2$ 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.

$2^{3} - 6 \cdot 2^{2} + 12 \cdot 2 - 8$

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

$8 - 6 \cdot 2^{2} + 12 \cdot 2 - 8$

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

$8 - 6 \cdot 4 + 12 \cdot 2 - 8$

Multiply $- 6$ by $4$ .

$8 - 24 + 12 \cdot 2 - 8$

Subtract $24$ from $8$ .

$- 16 + 12 \cdot 2 - 8$

Multiply $12$ by $2$ .

$- 16 + 24 - 8$

Add $- 16$ and $24$ .

$8 - 8$

Subtract $8$ from $8$ .

$0$

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

$\frac{x^{3} - 6 x^{2} + 12 x - 8}{x - 2}$

Divide $x^{3} - 6 x^{2} + 12 x - 8$ by $x - 2$ .

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

$x$

$2$

$x^{3}$

$6 x^{2}$

$12 x$

$8$

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

					$x^{2}$
	$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$

Multiply the new quotient term by the divisor.

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			+	$x^{3}$	-	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$

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

				$x^{2}$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					-	$4 x^{2}$	+	$8 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$

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

				$x^{2}$	-	$4 x$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$

Divide the highest order term in the dividend $4 x$ by the highest order term in divisor $x$ .

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$

Multiply the new quotient term by the divisor.

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$
							+	$4 x$	-	$8$

The expression needs to be subtracted from the dividend, so change all the signs in $4 x - 8$

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$
							-	$4 x$	+	$8$

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

				$x^{2}$	-	$4 x$	+	$4$
$x$	-	$2$		$x^{3}$	-	$6 x^{2}$	+	$12 x$	-	$8$
			-	$x^{3}$	+	$2 x^{2}$
					-	$4 x^{2}$	+	$12 x$
					+	$4 x^{2}$	-	$8 x$
							+	$4 x$	-	$8$
							-	$4 x$	+	$8$
										$0$

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

$x^{2} - 4 x + 4$

Write $x^{3} - 6 x^{2} + 12 x - 8$ as a set of factors.

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

Factor using the perfect square rule.

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

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

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

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

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

$(x - 2) {(x - 2)}^{2} = 0$

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

$x - 2 = 0$

${(x - 2)}^{2} = 0$

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.

$x = 2$

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

$x = 2$

Solve for x 1/8x^3+1 1/2x=3/4x^2+1

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