Solve (m+n)(m-n)+(m-n)^2-4m(m+n)rfloordiv(2m)

$$( m + n ) ( m - n ) + ( m - n ) ^ { 2 } - 4 m ( m + n ) \rfloor \div ( 2 m )$$

Evaluate

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

Solution Steps

Consider \left(m+n\right)\left(m-n\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

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

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

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

Use the distributive property to multiply -4m by m+n.

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

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

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

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

$$-2m^{2}-6mn$$

Expand

$$-2m^{2}-6mn$$

Solution Steps

Consider \left(m+n\right)\left(m-n\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

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

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

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

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

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

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

Use the distributive property to multiply -4m by m+n.

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

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

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

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

$$-2m^{2}-6mn$$