Solve (a-5)^2-(a+6)^2

$$( a - 5 ) ^ { 2 } - ( a + 6 ) ^ { 2 }$$

Evaluate

$$-22a-11$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-5\right)^{2}.

$$a^{2}-10a+25-\left(a+6\right)^{2}$$

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

$$a^{2}-10a+25-\left(a^{2}+12a+36\right)$$

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

$$a^{2}-10a+25-a^{2}-12a-36$$

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

$$-10a+25-12a-36$$

Combine -10a and -12a to get -22a.

$$-22a+25-36$$

Subtract 36 from 25 to get -11.

$$-22a-11$$

Expand

$$-22a-11$$

Solution Steps

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-5\right)^{2}.

$$a^{2}-10a+25-\left(a+6\right)^{2}$$

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

$$a^{2}-10a+25-\left(a^{2}+12a+36\right)$$

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

$$a^{2}-10a+25-a^{2}-12a-36$$

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

$$-10a+25-12a-36$$

Combine -10a and -12a to get -22a.

$$-22a+25-36$$

Subtract 36 from 25 to get -11.

$$-22a-11$$