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

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

Evaluate

$$\left(p-8\right)^{2}-39$$

Solution Steps

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

$$4p^{2}-12p+9-\left(p-4\right)\left(p+4\right)-2p\left(p+2\right)$$

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.

$$4p^{2}-12p+9-\left(p^{2}-16\right)-2p\left(p+2\right)$$

To find the opposite of p^{2}-16, find the opposite of each term.

$$4p^{2}-12p+9-p^{2}+16-2p\left(p+2\right)$$

Combine 4p^{2} and -p^{2} to get 3p^{2}.

$$3p^{2}-12p+9+16-2p\left(p+2\right)$$

Add 9 and 16 to get 25.

$$3p^{2}-12p+25-2p\left(p+2\right)$$

Use the distributive property to multiply -2p by p+2.

$$3p^{2}-12p+25-2p^{2}-4p$$

Combine 3p^{2} and -2p^{2} to get p^{2}.

$$p^{2}-12p+25-4p$$

Combine -12p and -4p to get -16p.

$$p^{2}-16p+25$$

Expand

$$p^{2}-16p+25$$

Solution Steps

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

$$4p^{2}-12p+9-\left(p-4\right)\left(p+4\right)-2p\left(p+2\right)$$

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.

$$4p^{2}-12p+9-\left(p^{2}-16\right)-2p\left(p+2\right)$$

To find the opposite of p^{2}-16, find the opposite of each term.

$$4p^{2}-12p+9-p^{2}+16-2p\left(p+2\right)$$

Combine 4p^{2} and -p^{2} to get 3p^{2}.

$$3p^{2}-12p+9+16-2p\left(p+2\right)$$

Add 9 and 16 to get 25.

$$3p^{2}-12p+25-2p\left(p+2\right)$$

Use the distributive property to multiply -2p by p+2.

$$3p^{2}-12p+25-2p^{2}-4p$$

Combine 3p^{2} and -2p^{2} to get p^{2}.

$$p^{2}-12p+25-4p$$

Combine -12p and -4p to get -16p.

$$p^{2}-16p+25$$