Solve (3a+1)^2-(a+6)^2=

$$( 3 a + 1 ) ^ { 2 } - ( a + 6 ) ^ { 2 } =$$

Evaluate

$$\left(2a-5\right)\left(4a+7\right)$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3a+1\right)^{2}.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+6\right)^{2}.

$$9a^{2}+6a+1-\left(a^{2}+12a+36\right)$$

To find the opposite of a^{2}+12a+36, find the opposite of each term.

$$9a^{2}+6a+1-a^{2}-12a-36$$

Combine 9a^{2} and -a^{2} to get 8a^{2}.

$$8a^{2}+6a+1-12a-36$$

Combine 6a and -12a to get -6a.

$$8a^{2}-6a+1-36$$

Subtract 36 from 1 to get -35.

$$8a^{2}-6a-35$$

Expand

$$8a^{2}-6a-35$$

Solution Steps

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(3a+1\right)^{2}.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(a+6\right)^{2}.

$$9a^{2}+6a+1-\left(a^{2}+12a+36\right)$$

To find the opposite of a^{2}+12a+36, find the opposite of each term.

$$9a^{2}+6a+1-a^{2}-12a-36$$

Combine 9a^{2} and -a^{2} to get 8a^{2}.

$$8a^{2}+6a+1-12a-36$$

Combine 6a and -12a to get -6a.

$$8a^{2}-6a+1-36$$

Subtract 36 from 1 to get -35.

$$8a^{2}-6a-35$$