Solve 1/(n-1)^2+5(n-1)+4-1/n^2+5n+4

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

Evaluate

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

Short Solution Steps

Factor \left(n-1\right)^{2}+5\left(n-1\right)+4. Factor n^{2}+5n+4.

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

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

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

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

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

Do the multiplications in \left(n+1\right)\left(n+4\right)-n\left(n+3\right).

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

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

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

Expand n\left(n+1\right)\left(n+3\right)\left(n+4\right).

$$\frac{2n+4}{n^{4}+8n^{3}+19n^{2}+12n}$$

Expand

$$\frac{2\left(n+2\right)}{\left(n^{2}+3n\right)\left(n^{2}+5n+4\right)}$$

Short Solution Steps

Factor \left(n-1\right)^{2}+5\left(n-1\right)+4. Factor n^{2}+5n+4.

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

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

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

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

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

Do the multiplications in \left(n+1\right)\left(n+4\right)-n\left(n+3\right).

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

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

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

Expand n\left(n+1\right)\left(n+3\right)\left(n+4\right).

$$\frac{2n+4}{n^{4}+8n^{3}+19n^{2}+12n}$$