Solve (a-2)quad(a+2)+3(a+2)^2-6a(a+2)

$$( a - 2 ) \quad ( a + 2 ) + 3 ( a + 2 ) ^ { 2 } - 6 a ( a + 2 )$$

Evaluate

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

Solution Steps

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

$$a^{2}-4+3\left(a+2\right)^{2}-6a\left(a+2\right)$$

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

$$a^{2}-4+3\left(a^{2}+4a+4\right)-6a\left(a+2\right)$$

Use the distributive property to multiply 3 by a^{2}+4a+4.

$$a^{2}-4+3a^{2}+12a+12-6a\left(a+2\right)$$

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

$$4a^{2}-4+12a+12-6a\left(a+2\right)$$

Add -4 and 12 to get 8.

$$4a^{2}+8+12a-6a\left(a+2\right)$$

Use the distributive property to multiply -6a by a+2.

$$4a^{2}+8+12a-6a^{2}-12a$$

Combine 4a^{2} and -6a^{2} to get -2a^{2}.

$$-2a^{2}+8+12a-12a$$

Combine 12a and -12a to get 0.

$$-2a^{2}+8$$

Expand

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

Solution Steps

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

$$a^{2}-4+3\left(a+2\right)^{2}-6a\left(a+2\right)$$

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

$$a^{2}-4+3\left(a^{2}+4a+4\right)-6a\left(a+2\right)$$

Use the distributive property to multiply 3 by a^{2}+4a+4.

$$a^{2}-4+3a^{2}+12a+12-6a\left(a+2\right)$$

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

$$4a^{2}-4+12a+12-6a\left(a+2\right)$$

Add -4 and 12 to get 8.

$$4a^{2}+8+12a-6a\left(a+2\right)$$

Use the distributive property to multiply -6a by a+2.

$$4a^{2}+8+12a-6a^{2}-12a$$

Combine 4a^{2} and -6a^{2} to get -2a^{2}.

$$-2a^{2}+8+12a-12a$$

Combine 12a and -12a to get 0.

$$-2a^{2}+8$$