Solve p+1/p^2-2p-3-1/p^2+p-3/p^2-3p

$$\frac { p + 1 } { p ^ { 2 } - 2 p - 3 } - \frac { 1 } { p ^ { 2 } + p } - \frac { 3 } { p ^ { 2 } - 3 p }$$

Evaluate

$$\frac{1}{p+1}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{p+1}{p^{2}-2p-3}.

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

Cancel out p+1 in both numerator and denominator.

$$\frac{1}{p-3}-\frac{1}{p^{2}+p}-\frac{3}{p^{2}-3p}$$

Factor p^{2}+p.

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

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

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

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

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

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

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

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

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

Factor p^{2}-3p.

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

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

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

Since \frac{p^{2}+3}{p\left(p-3\right)\left(p+1\right)} and \frac{3\left(p+1\right)}{p\left(p-3\right)\left(p+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in p^{2}+3-3\left(p+1\right).

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

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

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

Factor the expressions that are not already factored in \frac{p^{2}-3p}{p\left(p-3\right)\left(p+1\right)}.

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

Cancel out p\left(p-3\right) in both numerator and denominator.

$$\frac{1}{p+1}$$

Expand

$$\frac{1}{p+1}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{p+1}{p^{2}-2p-3}.

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

Cancel out p+1 in both numerator and denominator.

$$\frac{1}{p-3}-\frac{1}{p^{2}+p}-\frac{3}{p^{2}-3p}$$

Factor p^{2}+p.

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

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

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

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

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

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

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

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

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

Factor p^{2}-3p.

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

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

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

Since \frac{p^{2}+3}{p\left(p-3\right)\left(p+1\right)} and \frac{3\left(p+1\right)}{p\left(p-3\right)\left(p+1\right)} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in p^{2}+3-3\left(p+1\right).

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

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

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

Factor the expressions that are not already factored in \frac{p^{2}-3p}{p\left(p-3\right)\left(p+1\right)}.

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

Cancel out p\left(p-3\right) in both numerator and denominator.

$$\frac{1}{p+1}$$