Solve k(3k+5)/4(k+1)(k+2)+1/(k+1)(k+3)

$$\frac { k ( 3 k + 5 ) } { 4 ( k + 1 ) ( k + 2 ) } + \frac { 1 } { ( k + 1 ) ( k + 3 ) }$$

Evaluate

$$\frac{\left(k+1\right)\left(3k+8\right)}{4\left(k+2\right)\left(k+3\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(k+1\right)\left(k+2\right) and \left(k+1\right)\left(k+3\right) is 4\left(k+1\right)\left(k+2\right)\left(k+3\right). Multiply \frac{k\left(3k+5\right)}{4\left(k+1\right)\left(k+2\right)} times \frac{k+3}{k+3}. Multiply \frac{1}{\left(k+1\right)\left(k+3\right)} times \frac{4\left(k+2\right)}{4\left(k+2\right)}.

$$\frac{k\left(3k+5\right)\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}+\frac{4\left(k+2\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Since \frac{k\left(3k+5\right)\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)} and \frac{4\left(k+2\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{k\left(3k+5\right)\left(k+3\right)+4\left(k+2\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Do the multiplications in k\left(3k+5\right)\left(k+3\right)+4\left(k+2\right).

$$\frac{3k^{3}+9k^{2}+5k^{2}+15k+4k+8}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Combine like terms in 3k^{3}+9k^{2}+5k^{2}+15k+4k+8.

$$\frac{3k^{3}+14k^{2}+19k+8}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Factor the expressions that are not already factored in \frac{3k^{3}+14k^{2}+19k+8}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}.

$$\frac{\left(3k+8\right)\left(k+1\right)^{2}}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Cancel out k+1 in both numerator and denominator.

$$\frac{\left(k+1\right)\left(3k+8\right)}{4\left(k+2\right)\left(k+3\right)}$$

Expand 4\left(k+2\right)\left(k+3\right).

$$\frac{\left(k+1\right)\left(3k+8\right)}{4k^{2}+20k+24}$$

Apply the distributive property by multiplying each term of k+1 by each term of 3k+8.

$$\frac{3k^{2}+8k+3k+8}{4k^{2}+20k+24}$$

Combine 8k and 3k to get 11k.

$$\frac{3k^{2}+11k+8}{4k^{2}+20k+24}$$

Expand

$$\frac{3k^{2}+11k+8}{4\left(k+2\right)\left(k+3\right)}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 4\left(k+1\right)\left(k+2\right) and \left(k+1\right)\left(k+3\right) is 4\left(k+1\right)\left(k+2\right)\left(k+3\right). Multiply \frac{k\left(3k+5\right)}{4\left(k+1\right)\left(k+2\right)} times \frac{k+3}{k+3}. Multiply \frac{1}{\left(k+1\right)\left(k+3\right)} times \frac{4\left(k+2\right)}{4\left(k+2\right)}.

$$\frac{k\left(3k+5\right)\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}+\frac{4\left(k+2\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Since \frac{k\left(3k+5\right)\left(k+3\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)} and \frac{4\left(k+2\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{k\left(3k+5\right)\left(k+3\right)+4\left(k+2\right)}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Do the multiplications in k\left(3k+5\right)\left(k+3\right)+4\left(k+2\right).

$$\frac{3k^{3}+9k^{2}+5k^{2}+15k+4k+8}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Combine like terms in 3k^{3}+9k^{2}+5k^{2}+15k+4k+8.

$$\frac{3k^{3}+14k^{2}+19k+8}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Factor the expressions that are not already factored in \frac{3k^{3}+14k^{2}+19k+8}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}.

$$\frac{\left(3k+8\right)\left(k+1\right)^{2}}{4\left(k+1\right)\left(k+2\right)\left(k+3\right)}$$

Cancel out k+1 in both numerator and denominator.

$$\frac{\left(k+1\right)\left(3k+8\right)}{4\left(k+2\right)\left(k+3\right)}$$

Expand 4\left(k+2\right)\left(k+3\right).

$$\frac{\left(k+1\right)\left(3k+8\right)}{4k^{2}+20k+24}$$

Apply the distributive property by multiplying each term of k+1 by each term of 3k+8.

$$\frac{3k^{2}+8k+3k+8}{4k^{2}+20k+24}$$

Combine 8k and 3k to get 11k.

$$\frac{3k^{2}+11k+8}{4k^{2}+20k+24}$$