Solve E(x)=[x^4+2x^3+x^2/(x+1)^2-9]:(3x-9)-x/3

$$E ( x ) = [ \frac { x ^ { 4 } + 2 x ^ { 3 } + x ^ { 2 } } { ( x + 1 ) ^ { 2 } } - 9 ] : ( 3 x - 9 ) - \frac { x } { 3 }$$

Evaluate

$$1$$

Short Solution Steps

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

$$\frac{\frac{x^{2}\left(x+1\right)^{2}}{\left(x+1\right)^{2}}-9}{3x-9}-\frac{x}{3}$$

Cancel out \left(x+1\right)^{2} in both numerator and denominator.

$$\frac{x^{2}-9}{3x-9}-\frac{x}{3}$$

Factor the expressions that are not already factored in \frac{x^{2}-9}{3x-9}.

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

Cancel out x-3 in both numerator and denominator.

$$\frac{x+3}{3}-\frac{x}{3}$$

Since \frac{x+3}{3} and \frac{x}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{x+3-x}{3}$$

Combine like terms in x+3-x.

$$\frac{3}{3}$$

Divide 3 by 3 to get 1.

$$1$$

Factor

$$1$$