Solve 2(a+1)^2-(2a-3)(2a+3)

$$2 ( a + 1 ) ^ { 2 } - ( 2 a - 3 ) ( 2 a + 3 )$$

Evaluate

$$11+4a-2a^{2}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+1\right)^{2}.

$$2\left(a^{2}+2a+1\right)-\left(2a-3\right)\left(2a+3\right)$$

Use the distributive property to multiply 2 by a^{2}+2a+1.

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

Consider \left(2a-3\right)\left(2a+3\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 3.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

To find the opposite of 4a^{2}-9, find the opposite of each term.

$$2a^{2}+4a+2-4a^{2}+9$$

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

$$-2a^{2}+4a+2+9$$

Add 2 and 9 to get 11.

$$-2a^{2}+4a+11$$

Expand

$$11+4a-2a^{2}$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+1\right)^{2}.

$$2\left(a^{2}+2a+1\right)-\left(2a-3\right)\left(2a+3\right)$$

Use the distributive property to multiply 2 by a^{2}+2a+1.

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

Consider \left(2a-3\right)\left(2a+3\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 3.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

To find the opposite of 4a^{2}-9, find the opposite of each term.

$$2a^{2}+4a+2-4a^{2}+9$$

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

$$-2a^{2}+4a+2+9$$

Add 2 and 9 to get 11.

$$-2a^{2}+4a+11$$