Solve (2n+2)(2n+1)-(2n+2)*(2n+1)/n+1

$$( 2 n + 2 ) ( 2 n + 1 ) - \frac { ( 2 n + 2 ) \cdot ( 2 n + 1 ) } { n + 1 }$$

Evaluate

$$2n\left(2n+1\right)$$

Short Solution Steps

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

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

Cancel out n+1 in both numerator and denominator.

$$\left(2n+2\right)\left(2n+1\right)-2\left(2n+1\right)$$

Expand the expression.

$$\left(2n+2\right)\left(2n+1\right)-\left(4n+2\right)$$

To find the opposite of 4n+2, find the opposite of each term.

$$\left(2n+2\right)\left(2n+1\right)-4n-2$$

Apply the distributive property by multiplying each term of 2n+2 by each term of 2n+1.

$$4n^{2}+2n+4n+2-4n-2$$

Combine 2n and 4n to get 6n.

$$4n^{2}+6n+2-4n-2$$

Combine 6n and -4n to get 2n.

$$4n^{2}+2n+2-2$$

Subtract 2 from 2 to get 0.

$$4n^{2}+2n$$

Expand

$$4n^{2}+2n$$

Short Solution Steps

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

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

Cancel out n+1 in both numerator and denominator.

$$\left(2n+2\right)\left(2n+1\right)-2\left(2n+1\right)$$

Expand the expression.

$$\left(2n+2\right)\left(2n+1\right)-\left(4n+2\right)$$

To find the opposite of 4n+2, find the opposite of each term.

$$\left(2n+2\right)\left(2n+1\right)-4n-2$$

Apply the distributive property by multiplying each term of 2n+2 by each term of 2n+1.

$$4n^{2}+2n+4n+2-4n-2$$

Combine 2n and 4n to get 6n.

$$4n^{2}+6n+2-4n-2$$

Combine 6n and -4n to get 2n.

$$4n^{2}+2n+2-2$$

Subtract 2 from 2 to get 0.

$$4n^{2}+2n$$