Solve (2m+n)^2-9(3m-n)^2

$$( 2 m + n ) ^ { 2 } - 9 ( 3 m - n ) ^ { 2 }$$

Evaluate

$$\left(4n-7m\right)\left(11m-2n\right)$$

Solution Steps

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

$$4m^{2}+4mn+n^{2}-9\left(3m-n\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3m-n\right)^{2}.

$$4m^{2}+4mn+n^{2}-9\left(9m^{2}-6mn+n^{2}\right)$$

Use the distributive property to multiply -9 by 9m^{2}-6mn+n^{2}.

$$4m^{2}+4mn+n^{2}-81m^{2}+54mn-9n^{2}$$

Combine 4m^{2} and -81m^{2} to get -77m^{2}.

$$-77m^{2}+4mn+n^{2}+54mn-9n^{2}$$

Combine 4mn and 54mn to get 58mn.

$$-77m^{2}+58mn+n^{2}-9n^{2}$$

Combine n^{2} and -9n^{2} to get -8n^{2}.

$$-77m^{2}+58mn-8n^{2}$$

Expand

$$-77m^{2}+58mn-8n^{2}$$

Solution Steps

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

$$4m^{2}+4mn+n^{2}-9\left(3m-n\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3m-n\right)^{2}.

$$4m^{2}+4mn+n^{2}-9\left(9m^{2}-6mn+n^{2}\right)$$

Use the distributive property to multiply -9 by 9m^{2}-6mn+n^{2}.

$$4m^{2}+4mn+n^{2}-81m^{2}+54mn-9n^{2}$$

Combine 4m^{2} and -81m^{2} to get -77m^{2}.

$$-77m^{2}+4mn+n^{2}+54mn-9n^{2}$$

Combine 4mn and 54mn to get 58mn.

$$-77m^{2}+58mn+n^{2}-9n^{2}$$

Combine n^{2} and -9n^{2} to get -8n^{2}.

$$-77m^{2}+58mn-8n^{2}$$