Solve 6m+mn/4mdivn^2-36

$$\frac { 6 m + m n } { 4 m } \div n ^ { 2 } - 36$$

Evaluate

$$-36+\frac{1}{4n}+\frac{3}{2n^{2}}$$

Short Solution Steps

Express \frac{\frac{6m+mn}{4m}}{n^{2}} as a single fraction.

$$\frac{6m+mn}{4mn^{2}}-36$$

Factor the expressions that are not already factored in \frac{6m+mn}{4mn^{2}}.

$$\frac{m\left(n+6\right)}{4mn^{2}}-36$$

Cancel out m in both numerator and denominator.

$$\frac{n+6}{4n^{2}}-36$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 36 times \frac{4n^{2}}{4n^{2}}.

$$\frac{n+6}{4n^{2}}-\frac{36\times 4n^{2}}{4n^{2}}$$

Since \frac{n+6}{4n^{2}} and \frac{36\times 4n^{2}}{4n^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{n+6-36\times 4n^{2}}{4n^{2}}$$

Do the multiplications in n+6-36\times 4n^{2}.

$$\frac{n+6-144n^{2}}{4n^{2}}$$

Factor the expressions that are not already factored in \frac{n+6-144n^{2}}{4n^{2}}.

$$\frac{-144\left(n-\left(-\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)\left(n-\left(\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)}{4n^{2}}$$

Cancel out 4 in both numerator and denominator.

$$\frac{-36\left(n-\left(-\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)\left(n-\left(\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)}{n^{2}}$$

To find the opposite of -\frac{1}{288}\sqrt{3457}+\frac{1}{288}, find the opposite of each term.

$$\frac{-36\left(n+\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)\left(n-\left(\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)}{n^{2}}$$

To find the opposite of \frac{1}{288}\sqrt{3457}+\frac{1}{288}, find the opposite of each term.

$$\frac{-36\left(n+\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)\left(n-\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)}{n^{2}}$$

Use the distributive property to multiply -36 by n+\frac{1}{288}\sqrt{3457}-\frac{1}{288}.

$$\frac{\left(-36n-\frac{1}{8}\sqrt{3457}+\frac{1}{8}\right)\left(n-\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)}{n^{2}}$$

Use the distributive property to multiply -36n-\frac{1}{8}\sqrt{3457}+\frac{1}{8} by n-\frac{1}{288}\sqrt{3457}-\frac{1}{288} and combine like terms.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{1}{2304}\left(\sqrt{3457}\right)^{2}-\frac{1}{2304}}{n^{2}}$$

The square of \sqrt{3457} is 3457.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{1}{2304}\times 3457-\frac{1}{2304}}{n^{2}}$$

Multiply \frac{1}{2304} and 3457 to get \frac{3457}{2304}.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{3457}{2304}-\frac{1}{2304}}{n^{2}}$$

Subtract \frac{1}{2304} from \frac{3457}{2304} to get \frac{3}{2}.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{3}{2}}{n^{2}}$$

Expand

$$-36+\frac{1}{4n}+\frac{3}{2n^{2}}$$

Short Solution Steps

Express \frac{\frac{6m+mn}{4m}}{n^{2}} as a single fraction.

$$\frac{6m+mn}{4mn^{2}}-36$$

Factor the expressions that are not already factored in \frac{6m+mn}{4mn^{2}}.

$$\frac{m\left(n+6\right)}{4mn^{2}}-36$$

Cancel out m in both numerator and denominator.

$$\frac{n+6}{4n^{2}}-36$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 36 times \frac{4n^{2}}{4n^{2}}.

$$\frac{n+6}{4n^{2}}-\frac{36\times 4n^{2}}{4n^{2}}$$

Since \frac{n+6}{4n^{2}} and \frac{36\times 4n^{2}}{4n^{2}} have the same denominator, subtract them by subtracting their numerators.

$$\frac{n+6-36\times 4n^{2}}{4n^{2}}$$

Do the multiplications in n+6-36\times 4n^{2}.

$$\frac{n+6-144n^{2}}{4n^{2}}$$

Factor the expressions that are not already factored in \frac{n+6-144n^{2}}{4n^{2}}.

$$\frac{-144\left(n-\left(-\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)\left(n-\left(\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)}{4n^{2}}$$

Cancel out 4 in both numerator and denominator.

$$\frac{-36\left(n-\left(-\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)\left(n-\left(\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)}{n^{2}}$$

To find the opposite of -\frac{1}{288}\sqrt{3457}+\frac{1}{288}, find the opposite of each term.

$$\frac{-36\left(n+\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)\left(n-\left(\frac{1}{288}\sqrt{3457}+\frac{1}{288}\right)\right)}{n^{2}}$$

To find the opposite of \frac{1}{288}\sqrt{3457}+\frac{1}{288}, find the opposite of each term.

$$\frac{-36\left(n+\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)\left(n-\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)}{n^{2}}$$

Use the distributive property to multiply -36 by n+\frac{1}{288}\sqrt{3457}-\frac{1}{288}.

$$\frac{\left(-36n-\frac{1}{8}\sqrt{3457}+\frac{1}{8}\right)\left(n-\frac{1}{288}\sqrt{3457}-\frac{1}{288}\right)}{n^{2}}$$

Use the distributive property to multiply -36n-\frac{1}{8}\sqrt{3457}+\frac{1}{8} by n-\frac{1}{288}\sqrt{3457}-\frac{1}{288} and combine like terms.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{1}{2304}\left(\sqrt{3457}\right)^{2}-\frac{1}{2304}}{n^{2}}$$

The square of \sqrt{3457} is 3457.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{1}{2304}\times 3457-\frac{1}{2304}}{n^{2}}$$

Multiply \frac{1}{2304} and 3457 to get \frac{3457}{2304}.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{3457}{2304}-\frac{1}{2304}}{n^{2}}$$

Subtract \frac{1}{2304} from \frac{3457}{2304} to get \frac{3}{2}.

$$\frac{-36n^{2}+\frac{1}{4}n+\frac{3}{2}}{n^{2}}$$