Solve (b+2)*(b-2)-(b+2)^2=

$$( b + 2 ) \cdot ( b - 2 ) - ( b + 2 ) ^ { 2 } =$$

Evaluate

$$-4b-8$$

Solution Steps

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

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(b+2\right)^{2}.

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

To find the opposite of b^{2}+4b+4, find the opposite of each term.

$$b^{2}-4-b^{2}-4b-4$$

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

$$-4-4b-4$$

Subtract 4 from -4 to get -8.

$$-8-4b$$

Expand

$$-4b-8$$

Solution Steps

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

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(b+2\right)^{2}.

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

To find the opposite of b^{2}+4b+4, find the opposite of each term.

$$b^{2}-4-b^{2}-4b-4$$

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

$$-4-4b-4$$

Subtract 4 from -4 to get -8.

$$-8-4b$$