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

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

Evaluate

$$3m\left(m+2n\right)$$

Solution Steps

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

$$4m^{2}+4mn+n^{2}-\left(m-n\right)^{2}$$

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

$$4m^{2}+4mn+n^{2}-\left(m^{2}-2mn+n^{2}\right)$$

To find the opposite of m^{2}-2mn+n^{2}, find the opposite of each term.

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

Combine 4m^{2} and -m^{2} to get 3m^{2}.

$$3m^{2}+4mn+n^{2}+2mn-n^{2}$$

Combine 4mn and 2mn to get 6mn.

$$3m^{2}+6mn+n^{2}-n^{2}$$

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

$$3m^{2}+6mn$$

Expand

$$3m^{2}+6mn$$

Solution Steps

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

$$4m^{2}+4mn+n^{2}-\left(m-n\right)^{2}$$

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

$$4m^{2}+4mn+n^{2}-\left(m^{2}-2mn+n^{2}\right)$$

To find the opposite of m^{2}-2mn+n^{2}, find the opposite of each term.

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

Combine 4m^{2} and -m^{2} to get 3m^{2}.

$$3m^{2}+4mn+n^{2}+2mn-n^{2}$$

Combine 4mn and 2mn to get 6mn.

$$3m^{2}+6mn+n^{2}-n^{2}$$

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

$$3m^{2}+6mn$$