Solve (a-2)^2-(a-5)(a+5)-(a-7)^2+2(a^2-10)

$$( a - 2 ) ^ { 2 } - ( a - 5 ) ( a + 5 ) - ( a - 7 ) ^ { 2 } + 2 ( a ^ { 2 } - 10 )$$

Evaluate

$$a^{2}+10a-40$$

Solution Steps

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

$$a^{2}-4a+4-\left(a-5\right)\left(a+5\right)-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

Consider \left(a-5\right)\left(a+5\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 5.

$$a^{2}-4a+4-\left(a^{2}-25\right)-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

To find the opposite of a^{2}-25, find the opposite of each term.

$$a^{2}-4a+4-a^{2}+25-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

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

$$-4a+4+25-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

Add 4 and 25 to get 29.

$$-4a+29-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-7\right)^{2}.

$$-4a+29-\left(a^{2}-14a+49\right)+2\left(a^{2}-10\right)$$

To find the opposite of a^{2}-14a+49, find the opposite of each term.

$$-4a+29-a^{2}+14a-49+2\left(a^{2}-10\right)$$

Combine -4a and 14a to get 10a.

$$10a+29-a^{2}-49+2\left(a^{2}-10\right)$$

Subtract 49 from 29 to get -20.

$$10a-20-a^{2}+2\left(a^{2}-10\right)$$

Use the distributive property to multiply 2 by a^{2}-10.

$$10a-20-a^{2}+2a^{2}-20$$

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

$$10a-20+a^{2}-20$$

Subtract 20 from -20 to get -40.

$$10a-40+a^{2}$$

Expand

$$a^{2}+10a-40$$

Solution Steps

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

$$a^{2}-4a+4-\left(a-5\right)\left(a+5\right)-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

Consider \left(a-5\right)\left(a+5\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 5.

$$a^{2}-4a+4-\left(a^{2}-25\right)-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

To find the opposite of a^{2}-25, find the opposite of each term.

$$a^{2}-4a+4-a^{2}+25-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

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

$$-4a+4+25-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

Add 4 and 25 to get 29.

$$-4a+29-\left(a-7\right)^{2}+2\left(a^{2}-10\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(a-7\right)^{2}.

$$-4a+29-\left(a^{2}-14a+49\right)+2\left(a^{2}-10\right)$$

To find the opposite of a^{2}-14a+49, find the opposite of each term.

$$-4a+29-a^{2}+14a-49+2\left(a^{2}-10\right)$$

Combine -4a and 14a to get 10a.

$$10a+29-a^{2}-49+2\left(a^{2}-10\right)$$

Subtract 49 from 29 to get -20.

$$10a-20-a^{2}+2\left(a^{2}-10\right)$$

Use the distributive property to multiply 2 by a^{2}-10.

$$10a-20-a^{2}+2a^{2}-20$$

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

$$10a-20+a^{2}-20$$

Subtract 20 from -20 to get -40.

$$10a-40+a^{2}$$