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

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

Evaluate

$$50-3a^{2}$$

Solution Steps

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

$$a^{2}+10a+25-10a+\left(5+2a\right)\left(5-2a\right)$$

Combine 10a and -10a to get 0.

$$a^{2}+25+\left(5+2a\right)\left(5-2a\right)$$

Consider \left(5+2a\right)\left(5-2a\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.

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

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

$$a^{2}+25+25-2^{2}a^{2}$$

Calculate 2 to the power of 2 and get 4.

$$a^{2}+25+25-4a^{2}$$

Add 25 and 25 to get 50.

$$a^{2}+50-4a^{2}$$

Combine a^{2} and -4a^{2} to get -3a^{2}.

$$-3a^{2}+50$$

Expand

$$50-3a^{2}$$

Solution Steps

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

$$a^{2}+10a+25-10a+\left(5+2a\right)\left(5-2a\right)$$

Combine 10a and -10a to get 0.

$$a^{2}+25+\left(5+2a\right)\left(5-2a\right)$$

Consider \left(5+2a\right)\left(5-2a\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.

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

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

$$a^{2}+25+25-2^{2}a^{2}$$

Calculate 2 to the power of 2 and get 4.

$$a^{2}+25+25-4a^{2}$$

Add 25 and 25 to get 50.

$$a^{2}+50-4a^{2}$$

Combine a^{2} and -4a^{2} to get -3a^{2}.

$$-3a^{2}+50$$