Solve 2x+15x^2+39x+33/x^2+5x+6-(2x^3+10x^2+12x)

$$\frac { 2 x + 15 x ^ { 2 } + 39 x + 33 } { x ^ { 2 } + 5 x + 6 } - ( 2 x ^ { 3 } + 10 x ^ { 2 } + 12 x )$$

Evaluate

$$\frac{33-31x-105x^{2}-74x^{3}-20x^{4}-2x^{5}}{\left(x+2\right)\left(x+3\right)}$$

Short Solution Steps

Combine 2x and 39x to get 41x.

$$\frac{41x+15x^{2}+33}{x^{2}+5x+6}-\left(2x^{3}+10x^{2}+12x\right)$$

To find the opposite of 2x^{3}+10x^{2}+12x, find the opposite of each term.

$$\frac{41x+15x^{2}+33}{x^{2}+5x+6}-2x^{3}-10x^{2}-12x$$

Factor x^{2}+5x+6.

$$\frac{41x+15x^{2}+33}{\left(x+2\right)\left(x+3\right)}-2x^{3}-10x^{2}-12x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x^{3}-10x^{2}-12x times \frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}.

$$\frac{41x+15x^{2}+33}{\left(x+2\right)\left(x+3\right)}+\frac{\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}$$

Since \frac{41x+15x^{2}+33}{\left(x+2\right)\left(x+3\right)} and \frac{\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{41x+15x^{2}+33+\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}$$

Do the multiplications in 41x+15x^{2}+33+\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right).

$$\frac{41x+15x^{2}+33-2x^{5}-10x^{4}-12x^{3}-10x^{4}-50x^{3}-60x^{2}-12x^{3}-60x^{2}-72x}{\left(x+2\right)\left(x+3\right)}$$

Combine like terms in 41x+15x^{2}+33-2x^{5}-10x^{4}-12x^{3}-10x^{4}-50x^{3}-60x^{2}-12x^{3}-60x^{2}-72x.

$$\frac{-31x-105x^{2}+33-2x^{5}-20x^{4}-74x^{3}}{\left(x+2\right)\left(x+3\right)}$$

Expand \left(x+2\right)\left(x+3\right).

$$\frac{-31x-105x^{2}+33-2x^{5}-20x^{4}-74x^{3}}{x^{2}+5x+6}$$

Expand

$$\frac{33-31x-105x^{2}-74x^{3}-20x^{4}-2x^{5}}{x^{2}+5x+6}$$

Short Solution Steps

Combine 2x and 39x to get 41x.

$$\frac{41x+15x^{2}+33}{x^{2}+5x+6}-\left(2x^{3}+10x^{2}+12x\right)$$

To find the opposite of 2x^{3}+10x^{2}+12x, find the opposite of each term.

$$\frac{41x+15x^{2}+33}{x^{2}+5x+6}-2x^{3}-10x^{2}-12x$$

Factor x^{2}+5x+6.

$$\frac{41x+15x^{2}+33}{\left(x+2\right)\left(x+3\right)}-2x^{3}-10x^{2}-12x$$

To add or subtract expressions, expand them to make their denominators the same. Multiply -2x^{3}-10x^{2}-12x times \frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}.

$$\frac{41x+15x^{2}+33}{\left(x+2\right)\left(x+3\right)}+\frac{\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}$$

Since \frac{41x+15x^{2}+33}{\left(x+2\right)\left(x+3\right)} and \frac{\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)} have the same denominator, add them by adding their numerators.

$$\frac{41x+15x^{2}+33+\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}$$

Do the multiplications in 41x+15x^{2}+33+\left(-2x^{3}-10x^{2}-12x\right)\left(x+2\right)\left(x+3\right).

$$\frac{41x+15x^{2}+33-2x^{5}-10x^{4}-12x^{3}-10x^{4}-50x^{3}-60x^{2}-12x^{3}-60x^{2}-72x}{\left(x+2\right)\left(x+3\right)}$$

Combine like terms in 41x+15x^{2}+33-2x^{5}-10x^{4}-12x^{3}-10x^{4}-50x^{3}-60x^{2}-12x^{3}-60x^{2}-72x.

$$\frac{-31x-105x^{2}+33-2x^{5}-20x^{4}-74x^{3}}{\left(x+2\right)\left(x+3\right)}$$

Expand \left(x+2\right)\left(x+3\right).

$$\frac{-31x-105x^{2}+33-2x^{5}-20x^{4}-74x^{3}}{x^{2}+5x+6}$$