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

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

Evaluate

$$m^{4}-16$$

Solution Steps

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Expand \left(2n\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Use the distributive property to multiply 2n-4 by 4+2n and combine like terms.

$$m^{4}-4n^{2}+4n^{2}-16$$

Combine -4n^{2} and 4n^{2} to get 0.

$$m^{4}-16$$

Expand

$$m^{4}-16$$

Solution Steps

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

Expand \left(2n\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Use the distributive property to multiply 2n-4 by 4+2n and combine like terms.

$$m^{4}-4n^{2}+4n^{2}-16$$

Combine -4n^{2} and 4n^{2} to get 0.

$$m^{4}-16$$