Solve (m+n)^2-(2m+n)(-m+n)

$$( m + n ) ^ { 2 } - ( 2 m + n ) ( - m + n )$$

Evaluate

$$m\left(3m+n\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+n\right)^{2}.

$$m^{2}+2mn+n^{2}-\left(2m+n\right)\left(-m+n\right)$$

Use the distributive property to multiply 2m+n by -m+n.

$$m^{2}+2mn+n^{2}-\left(2m\left(-m\right)+2mn+n\left(-m\right)+n^{2}\right)$$

To find the opposite of 2m\left(-m\right)+2mn+n\left(-m\right)+n^{2}, find the opposite of each term.

$$m^{2}+2mn+n^{2}-2m\left(-m\right)-2mn-n\left(-m\right)-n^{2}$$

Multiply -2 and -1 to get 2.

$$m^{2}+2mn+n^{2}+2mm-2mn-n\left(-m\right)-n^{2}$$

Multiply m and m to get m^{2}.

$$m^{2}+2mn+n^{2}+2m^{2}-2mn-n\left(-m\right)-n^{2}$$

Combine m^{2} and 2m^{2} to get 3m^{2}.

$$3m^{2}+2mn+n^{2}-2mn-n\left(-m\right)-n^{2}$$

Combine 2mn and -2mn to get 0.

$$3m^{2}+n^{2}-n\left(-m\right)-n^{2}$$

Multiply -1 and -1 to get 1.

$$3m^{2}+n^{2}+nm-n^{2}$$

Combine n^{2} and -n^{2} to get 0.

$$3m^{2}+nm$$

Expand

$$3m^{2}+mn$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(m+n\right)^{2}.

$$m^{2}+2mn+n^{2}-\left(2m+n\right)\left(-m+n\right)$$

Use the distributive property to multiply 2m+n by -m+n.

$$m^{2}+2mn+n^{2}-\left(2m\left(-m\right)+2mn+n\left(-m\right)+n^{2}\right)$$

To find the opposite of 2m\left(-m\right)+2mn+n\left(-m\right)+n^{2}, find the opposite of each term.

$$m^{2}+2mn+n^{2}-2m\left(-m\right)-2mn-n\left(-m\right)-n^{2}$$

Multiply -2 and -1 to get 2.

$$m^{2}+2mn+n^{2}+2mm-2mn-n\left(-m\right)-n^{2}$$

Multiply m and m to get m^{2}.

$$m^{2}+2mn+n^{2}+2m^{2}-2mn-n\left(-m\right)-n^{2}$$

Combine m^{2} and 2m^{2} to get 3m^{2}.

$$3m^{2}+2mn+n^{2}-2mn-n\left(-m\right)-n^{2}$$

Combine 2mn and -2mn to get 0.

$$3m^{2}+n^{2}-n\left(-m\right)-n^{2}$$

Multiply -1 and -1 to get 1.

$$3m^{2}+n^{2}+nm-n^{2}$$

Combine n^{2} and -n^{2} to get 0.

$$3m^{2}+nm$$