Solve (a+3)^2-(a-1)(a+1)-2(2a+4)

$$( a + 3 ) ^ { 2 } - ( a - 1 ) ( a + 1 ) - 2 ( 2 a + 4 )$$

Evaluate

$$2\left(a+1\right)$$

Solution Steps

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

$$a^{2}+6a+9-\left(a-1\right)\left(a+1\right)-2\left(2a+4\right)$$

Consider \left(a-1\right)\left(a+1\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 1.

$$a^{2}+6a+9-\left(a^{2}-1\right)-2\left(2a+4\right)$$

To find the opposite of a^{2}-1, find the opposite of each term.

$$a^{2}+6a+9-a^{2}+1-2\left(2a+4\right)$$

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

$$6a+9+1-2\left(2a+4\right)$$

Add 9 and 1 to get 10.

$$6a+10-2\left(2a+4\right)$$

Use the distributive property to multiply -2 by 2a+4.

$$6a+10-4a-8$$

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

$$2a+10-8$$

Subtract 8 from 10 to get 2.

$$2a+2$$

Expand

$$2a+2$$

Solution Steps

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

$$a^{2}+6a+9-\left(a-1\right)\left(a+1\right)-2\left(2a+4\right)$$

Consider \left(a-1\right)\left(a+1\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 1.

$$a^{2}+6a+9-\left(a^{2}-1\right)-2\left(2a+4\right)$$

To find the opposite of a^{2}-1, find the opposite of each term.

$$a^{2}+6a+9-a^{2}+1-2\left(2a+4\right)$$

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

$$6a+9+1-2\left(2a+4\right)$$

Add 9 and 1 to get 10.

$$6a+10-2\left(2a+4\right)$$

Use the distributive property to multiply -2 by 2a+4.

$$6a+10-4a-8$$

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

$$2a+10-8$$

Subtract 8 from 10 to get 2.

$$2a+2$$