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

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

Evaluate

$$m\left(9m-4n\right)$$

Solution Steps

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

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

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

$$9m^{2}-6mn+n^{2}+2nm-n^{2}$$

Combine -6mn and 2nm to get -4mn.

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

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

$$9m^{2}-4mn$$

Expand

$$9m^{2}-4mn$$

Solution Steps

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

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

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

$$9m^{2}-6mn+n^{2}+2nm-n^{2}$$

Combine -6mn and 2nm to get -4mn.

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

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

$$9m^{2}-4mn$$