Solve 1/4+1/4n+2+1/2left(2n+4right)-1/2n+1

$$\frac{ 1 }{ 4 } + \frac{ 1 }{ 4n+2 } + \frac{ 1 }{ 2 \left( 2n+4 \right) } - \frac{ 1 }{ 2n+1 }$$

Evaluate

$$\frac{2n^{2}+5n-1}{4\left(n+2\right)\left(2n+1\right)}$$

Short Solution Steps

Factor 4n+2.

$$\frac{1}{4}+\frac{1}{2\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2\left(2n+1\right) is 4\left(2n+1\right). Multiply \frac{1}{4} times \frac{2n+1}{2n+1}. Multiply \frac{1}{2\left(2n+1\right)} times \frac{2}{2}.

$$\frac{2n+1}{4\left(2n+1\right)}+\frac{2}{4\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

Since \frac{2n+1}{4\left(2n+1\right)} and \frac{2}{4\left(2n+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2n+1+2}{4\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

Combine like terms in 2n+1+2.

$$\frac{2n+3}{4\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

Factor 2\left(2n+4\right).

$$\frac{2n+3}{4\left(2n+1\right)}+\frac{1}{2^{2}\left(n+2\right)}-\frac{1}{2n+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(2n+1\right) and 2^{2}\left(n+2\right) is 4\left(n+2\right)\left(2n+1\right). Multiply \frac{2n+3}{4\left(2n+1\right)} times \frac{n+2}{n+2}. Multiply \frac{1}{2^{2}\left(n+2\right)} times \frac{2n+1}{2n+1}.

$$\frac{\left(2n+3\right)\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)}+\frac{2n+1}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

Since \frac{\left(2n+3\right)\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)} and \frac{2n+1}{4\left(n+2\right)\left(2n+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(2n+3\right)\left(n+2\right)+2n+1}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

Do the multiplications in \left(2n+3\right)\left(n+2\right)+2n+1.

$$\frac{2n^{2}+4n+3n+6+2n+1}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

Combine like terms in 2n^{2}+4n+3n+6+2n+1.

$$\frac{2n^{2}+9n+7}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(n+2\right)\left(2n+1\right) and 2n+1 is 4\left(n+2\right)\left(2n+1\right). Multiply \frac{1}{2n+1} times \frac{4\left(n+2\right)}{4\left(n+2\right)}.

$$\frac{2n^{2}+9n+7}{4\left(n+2\right)\left(2n+1\right)}-\frac{4\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)}$$

Since \frac{2n^{2}+9n+7}{4\left(n+2\right)\left(2n+1\right)} and \frac{4\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2n^{2}+9n+7-4\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)}$$

Do the multiplications in 2n^{2}+9n+7-4\left(n+2\right).

$$\frac{2n^{2}+9n+7-4n-8}{4\left(n+2\right)\left(2n+1\right)}$$

Combine like terms in 2n^{2}+9n+7-4n-8.

$$\frac{2n^{2}+5n-1}{4\left(n+2\right)\left(2n+1\right)}$$

Factor the expressions that are not already factored in \frac{2n^{2}+5n-1}{4\left(n+2\right)\left(2n+1\right)}.

$$\frac{2\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4\left(n+2\right)\left(2n+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{2\left(n+2\right)\left(2n+1\right)}$$

Expand 2\left(n+2\right)\left(2n+1\right).

$$\frac{\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

To find the opposite of -\frac{1}{4}\sqrt{33}-\frac{5}{4}, find the opposite of each term.

$$\frac{\left(n-\left(-\frac{1}{4}\sqrt{33}\right)-\left(-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

The opposite of -\frac{1}{4}\sqrt{33} is \frac{1}{4}\sqrt{33}.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}-\left(-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

The opposite of -\frac{5}{4} is \frac{5}{4}.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

To find the opposite of \frac{1}{4}\sqrt{33}-\frac{5}{4}, find the opposite of each term.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)\left(n-\frac{1}{4}\sqrt{33}-\left(-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

The opposite of -\frac{5}{4} is \frac{5}{4}.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)\left(n-\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)}{4n^{2}+10n+4}$$

Apply the distributive property by multiplying each term of n+\frac{1}{4}\sqrt{33}+\frac{5}{4} by each term of n-\frac{1}{4}\sqrt{33}+\frac{5}{4}.

$$\frac{n^{2}+n\left(-\frac{1}{4}\right)\sqrt{33}+n\times \frac{5}{4}+\frac{1}{4}\sqrt{33}n+\frac{1}{4}\sqrt{33}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \sqrt{33} and \sqrt{33} to get 33.

$$\frac{n^{2}+n\left(-\frac{1}{4}\right)\sqrt{33}+n\times \frac{5}{4}+\frac{1}{4}\sqrt{33}n+\frac{1}{4}\times 33\left(-\frac{1}{4}\right)+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Combine n\left(-\frac{1}{4}\right)\sqrt{33} and \frac{1}{4}\sqrt{33}n to get 0.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{1}{4}\times 33\left(-\frac{1}{4}\right)+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{1}{4} and 33 to get \frac{33}{4}.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{33}{4}\left(-\frac{1}{4}\right)+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{33}{4} times -\frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{33\left(-1\right)}{4\times 4}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{33\left(-1\right)}{4\times 4}.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{-33}{16}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Fraction \frac{-33}{16} can be rewritten as -\frac{33}{16} by extracting the negative sign.

$$\frac{n^{2}+n\times \frac{5}{4}-\frac{33}{16}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{1}{4} times \frac{5}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+n\times \frac{5}{4}-\frac{33}{16}+\frac{1\times 5}{4\times 4}\sqrt{33}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{1\times 5}{4\times 4}.

$$\frac{n^{2}+n\times \frac{5}{4}-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Combine n\times \frac{5}{4} and \frac{5}{4}n to get \frac{5}{2}n.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{5}{4} times -\frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{5\left(-1\right)}{4\times 4}\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{5\left(-1\right)}{4\times 4}.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{-5}{16}\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Fraction \frac{-5}{16} can be rewritten as -\frac{5}{16} by extracting the negative sign.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}-\frac{5}{16}\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Combine \frac{5}{16}\sqrt{33} and -\frac{5}{16}\sqrt{33} to get 0.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{5}{4} times \frac{5}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5\times 5}{4\times 4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{5\times 5}{4\times 4}.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{25}{16}}{4n^{2}+10n+4}$$

Since -\frac{33}{16} and \frac{25}{16} have the same denominator, add them by adding their numerators.

$$\frac{n^{2}+\frac{5}{2}n+\frac{-33+25}{16}}{4n^{2}+10n+4}$$

Add -33 and 25 to get -8.

$$\frac{n^{2}+\frac{5}{2}n+\frac{-8}{16}}{4n^{2}+10n+4}$$

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$\frac{n^{2}+\frac{5}{2}n-\frac{1}{2}}{4n^{2}+10n+4}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{2}\times 2\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{2\left(n+2\right)\left(2n+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{\frac{1}{2}\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{\left(n+2\right)\left(2n+1\right)}$$

Expand the expression.

$$\frac{\frac{1}{2}n^{2}+\frac{5}{4}n-\frac{1}{4}}{2n^{2}+5n+2}$$

Expand

$$\frac{2n^{2}+5n-1}{4\left(n+2\right)\left(2n+1\right)}$$

Short Solution Steps

Factor 4n+2.

$$\frac{1}{4}+\frac{1}{2\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4 and 2\left(2n+1\right) is 4\left(2n+1\right). Multiply \frac{1}{4} times \frac{2n+1}{2n+1}. Multiply \frac{1}{2\left(2n+1\right)} times \frac{2}{2}.

$$\frac{2n+1}{4\left(2n+1\right)}+\frac{2}{4\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

Since \frac{2n+1}{4\left(2n+1\right)} and \frac{2}{4\left(2n+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{2n+1+2}{4\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

Combine like terms in 2n+1+2.

$$\frac{2n+3}{4\left(2n+1\right)}+\frac{1}{2\left(2n+4\right)}-\frac{1}{2n+1}$$

Factor 2\left(2n+4\right).

$$\frac{2n+3}{4\left(2n+1\right)}+\frac{1}{2^{2}\left(n+2\right)}-\frac{1}{2n+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(2n+1\right) and 2^{2}\left(n+2\right) is 4\left(n+2\right)\left(2n+1\right). Multiply \frac{2n+3}{4\left(2n+1\right)} times \frac{n+2}{n+2}. Multiply \frac{1}{2^{2}\left(n+2\right)} times \frac{2n+1}{2n+1}.

$$\frac{\left(2n+3\right)\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)}+\frac{2n+1}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

Since \frac{\left(2n+3\right)\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)} and \frac{2n+1}{4\left(n+2\right)\left(2n+1\right)} have the same denominator, add them by adding their numerators.

$$\frac{\left(2n+3\right)\left(n+2\right)+2n+1}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

Do the multiplications in \left(2n+3\right)\left(n+2\right)+2n+1.

$$\frac{2n^{2}+4n+3n+6+2n+1}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

Combine like terms in 2n^{2}+4n+3n+6+2n+1.

$$\frac{2n^{2}+9n+7}{4\left(n+2\right)\left(2n+1\right)}-\frac{1}{2n+1}$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(n+2\right)\left(2n+1\right) and 2n+1 is 4\left(n+2\right)\left(2n+1\right). Multiply \frac{1}{2n+1} times \frac{4\left(n+2\right)}{4\left(n+2\right)}.

$$\frac{2n^{2}+9n+7}{4\left(n+2\right)\left(2n+1\right)}-\frac{4\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)}$$

Since \frac{2n^{2}+9n+7}{4\left(n+2\right)\left(2n+1\right)} and \frac{4\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{2n^{2}+9n+7-4\left(n+2\right)}{4\left(n+2\right)\left(2n+1\right)}$$

Do the multiplications in 2n^{2}+9n+7-4\left(n+2\right).

$$\frac{2n^{2}+9n+7-4n-8}{4\left(n+2\right)\left(2n+1\right)}$$

Combine like terms in 2n^{2}+9n+7-4n-8.

$$\frac{2n^{2}+5n-1}{4\left(n+2\right)\left(2n+1\right)}$$

Factor the expressions that are not already factored in \frac{2n^{2}+5n-1}{4\left(n+2\right)\left(2n+1\right)}.

$$\frac{2\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4\left(n+2\right)\left(2n+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{2\left(n+2\right)\left(2n+1\right)}$$

Expand 2\left(n+2\right)\left(2n+1\right).

$$\frac{\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

To find the opposite of -\frac{1}{4}\sqrt{33}-\frac{5}{4}, find the opposite of each term.

$$\frac{\left(n-\left(-\frac{1}{4}\sqrt{33}\right)-\left(-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

The opposite of -\frac{1}{4}\sqrt{33} is \frac{1}{4}\sqrt{33}.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}-\left(-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

The opposite of -\frac{5}{4} is \frac{5}{4}.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

To find the opposite of \frac{1}{4}\sqrt{33}-\frac{5}{4}, find the opposite of each term.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)\left(n-\frac{1}{4}\sqrt{33}-\left(-\frac{5}{4}\right)\right)}{4n^{2}+10n+4}$$

The opposite of -\frac{5}{4} is \frac{5}{4}.

$$\frac{\left(n+\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)\left(n-\frac{1}{4}\sqrt{33}+\frac{5}{4}\right)}{4n^{2}+10n+4}$$

Apply the distributive property by multiplying each term of n+\frac{1}{4}\sqrt{33}+\frac{5}{4} by each term of n-\frac{1}{4}\sqrt{33}+\frac{5}{4}.

$$\frac{n^{2}+n\left(-\frac{1}{4}\right)\sqrt{33}+n\times \frac{5}{4}+\frac{1}{4}\sqrt{33}n+\frac{1}{4}\sqrt{33}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \sqrt{33} and \sqrt{33} to get 33.

$$\frac{n^{2}+n\left(-\frac{1}{4}\right)\sqrt{33}+n\times \frac{5}{4}+\frac{1}{4}\sqrt{33}n+\frac{1}{4}\times 33\left(-\frac{1}{4}\right)+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Combine n\left(-\frac{1}{4}\right)\sqrt{33} and \frac{1}{4}\sqrt{33}n to get 0.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{1}{4}\times 33\left(-\frac{1}{4}\right)+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{1}{4} and 33 to get \frac{33}{4}.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{33}{4}\left(-\frac{1}{4}\right)+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{33}{4} times -\frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{33\left(-1\right)}{4\times 4}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{33\left(-1\right)}{4\times 4}.

$$\frac{n^{2}+n\times \frac{5}{4}+\frac{-33}{16}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Fraction \frac{-33}{16} can be rewritten as -\frac{33}{16} by extracting the negative sign.

$$\frac{n^{2}+n\times \frac{5}{4}-\frac{33}{16}+\frac{1}{4}\sqrt{33}\times \frac{5}{4}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{1}{4} times \frac{5}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+n\times \frac{5}{4}-\frac{33}{16}+\frac{1\times 5}{4\times 4}\sqrt{33}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{1\times 5}{4\times 4}.

$$\frac{n^{2}+n\times \frac{5}{4}-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{5}{4}n+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Combine n\times \frac{5}{4} and \frac{5}{4}n to get \frac{5}{2}n.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{5}{4}\left(-\frac{1}{4}\right)\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{5}{4} times -\frac{1}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{5\left(-1\right)}{4\times 4}\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{5\left(-1\right)}{4\times 4}.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}+\frac{-5}{16}\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Fraction \frac{-5}{16} can be rewritten as -\frac{5}{16} by extracting the negative sign.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{16}\sqrt{33}-\frac{5}{16}\sqrt{33}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Combine \frac{5}{16}\sqrt{33} and -\frac{5}{16}\sqrt{33} to get 0.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5}{4}\times \frac{5}{4}}{4n^{2}+10n+4}$$

Multiply \frac{5}{4} times \frac{5}{4} by multiplying numerator times numerator and denominator times denominator.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{5\times 5}{4\times 4}}{4n^{2}+10n+4}$$

Do the multiplications in the fraction \frac{5\times 5}{4\times 4}.

$$\frac{n^{2}+\frac{5}{2}n-\frac{33}{16}+\frac{25}{16}}{4n^{2}+10n+4}$$

Since -\frac{33}{16} and \frac{25}{16} have the same denominator, add them by adding their numerators.

$$\frac{n^{2}+\frac{5}{2}n+\frac{-33+25}{16}}{4n^{2}+10n+4}$$

Add -33 and 25 to get -8.

$$\frac{n^{2}+\frac{5}{2}n+\frac{-8}{16}}{4n^{2}+10n+4}$$

Reduce the fraction \frac{-8}{16} to lowest terms by extracting and canceling out 8.

$$\frac{n^{2}+\frac{5}{2}n-\frac{1}{2}}{4n^{2}+10n+4}$$

Factor the expressions that are not already factored.

$$\frac{\frac{1}{2}\times 2\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{2\left(n+2\right)\left(2n+1\right)}$$

Cancel out 2 in both numerator and denominator.

$$\frac{\frac{1}{2}\left(n-\left(-\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)\left(n-\left(\frac{1}{4}\sqrt{33}-\frac{5}{4}\right)\right)}{\left(n+2\right)\left(2n+1\right)}$$

Expand the expression.

$$\frac{\frac{1}{2}n^{2}+\frac{5}{4}n-\frac{1}{4}}{2n^{2}+5n+2}$$