Solve 3/4x^7-5/8x^5-3/2x^4+frac{7/6x^6}{-frac{5}{10}{x}^3}

$$\frac{ \frac{ 3 }{ 4 } { x }^{ 7 } - \frac{ 5 }{ 8 } { x }^{ 5 } - \frac{ 3 }{ 2 } { x }^{ 4 } + \frac{ 7 }{ 6 } { x }^{ 6 } }{ - \frac{ 5 }{ 10 } { x }^{ 3 } }$$

Evaluate

$$-\frac{3x^{4}}{2}-\frac{7x^{3}}{3}+\frac{5x^{2}}{4}+3x$$

Short Solution Steps

Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.

$$\frac{\frac{3}{4}x^{7}-\frac{5}{8}x^{5}-\frac{3}{2}x^{4}+\frac{7}{6}x^{6}}{-\frac{1}{2}x^{3}}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{24}\left(18x^{3}+28x^{2}-15x-36\right)x^{4}}{-\frac{1}{2}x^{3}}$$

Cancel out x^{3} in both numerator and denominator.

$$\frac{\frac{1}{24}x\left(18x^{3}+28x^{2}-15x-36\right)}{-\frac{1}{2}}$$

Expand the expression.

$$\frac{\frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x}{-\frac{1}{2}}$$

Divide \frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x by -\frac{1}{2} by multiplying \frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x by the reciprocal of -\frac{1}{2}.

$$\frac{\left(\frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x\right)\times 2}{-1}$$

Anything divided by -1 gives its opposite.

$$-\left(\frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x\right)\times 2$$

Use the distributive property to multiply \frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x by 2.

$$-\left(\frac{3}{2}x^{4}+\frac{7}{3}x^{3}-\frac{5}{4}x^{2}-3x\right)$$

To find the opposite of \frac{3}{2}x^{4}+\frac{7}{3}x^{3}-\frac{5}{4}x^{2}-3x, find the opposite of each term.

$$-\frac{3}{2}x^{4}-\frac{7}{3}x^{3}+\frac{5}{4}x^{2}+3x$$

Expand

$$-\frac{3x^{4}}{2}-\frac{7x^{3}}{3}+\frac{5x^{2}}{4}+3x$$

Short Solution Steps

Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.

$$\frac{\frac{3}{4}x^{7}-\frac{5}{8}x^{5}-\frac{3}{2}x^{4}+\frac{7}{6}x^{6}}{-\frac{1}{2}x^{3}}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{24}\left(18x^{3}+28x^{2}-15x-36\right)x^{4}}{-\frac{1}{2}x^{3}}$$

Cancel out x^{3} in both numerator and denominator.

$$\frac{\frac{1}{24}x\left(18x^{3}+28x^{2}-15x-36\right)}{-\frac{1}{2}}$$

Expand the expression.

$$\frac{\frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x}{-\frac{1}{2}}$$

Divide \frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x by -\frac{1}{2} by multiplying \frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x by the reciprocal of -\frac{1}{2}.

$$\frac{\left(\frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x\right)\times 2}{-1}$$

Anything divided by -1 gives its opposite.

$$-\left(\frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x\right)\times 2$$

Use the distributive property to multiply \frac{3}{4}x^{4}+\frac{7}{6}x^{3}-\frac{5}{8}x^{2}-\frac{3}{2}x by 2.

$$-\left(\frac{3}{2}x^{4}+\frac{7}{3}x^{3}-\frac{5}{4}x^{2}-3x\right)$$

To find the opposite of \frac{3}{2}x^{4}+\frac{7}{3}x^{3}-\frac{5}{4}x^{2}-3x, find the opposite of each term.

$$-\frac{3}{2}x^{4}-\frac{7}{3}x^{3}+\frac{5}{4}x^{2}+3x$$