Solve (p-2)^2-(p-3)^2+(p-4)(p+4)

$$( p - 2 ) ^ { 2 } - ( p - 3 ) ^ { 2 } + ( p - 4 ) ( p + 4 )$$

Evaluate

$$p^{2}+2p-21$$

Solution Steps

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

$$p^{2}-4p+4-\left(p-3\right)^{2}+\left(p-4\right)\left(p+4\right)$$

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

$$p^{2}-4p+4-\left(p^{2}-6p+9\right)+\left(p-4\right)\left(p+4\right)$$

To find the opposite of p^{2}-6p+9, find the opposite of each term.

$$p^{2}-4p+4-p^{2}+6p-9+\left(p-4\right)\left(p+4\right)$$

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

$$-4p+4+6p-9+\left(p-4\right)\left(p+4\right)$$

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

$$2p+4-9+\left(p-4\right)\left(p+4\right)$$

Subtract 9 from 4 to get -5.

$$2p-5+\left(p-4\right)\left(p+4\right)$$

Consider \left(p-4\right)\left(p+4\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 4.

$$2p-5+p^{2}-16$$

Subtract 16 from -5 to get -21.

$$2p-21+p^{2}$$

Expand

$$p^{2}+2p-21$$

Solution Steps

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

$$p^{2}-4p+4-\left(p-3\right)^{2}+\left(p-4\right)\left(p+4\right)$$

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

$$p^{2}-4p+4-\left(p^{2}-6p+9\right)+\left(p-4\right)\left(p+4\right)$$

To find the opposite of p^{2}-6p+9, find the opposite of each term.

$$p^{2}-4p+4-p^{2}+6p-9+\left(p-4\right)\left(p+4\right)$$

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

$$-4p+4+6p-9+\left(p-4\right)\left(p+4\right)$$

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

$$2p+4-9+\left(p-4\right)\left(p+4\right)$$

Subtract 9 from 4 to get -5.

$$2p-5+\left(p-4\right)\left(p+4\right)$$

Consider \left(p-4\right)\left(p+4\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 4.

$$2p-5+p^{2}-16$$

Subtract 16 from -5 to get -21.

$$2p-21+p^{2}$$