Solve a^{2}(2a-3)^2-(2a^2+a)^2+a(4a-1)^2-(a-2)(a+2)

$$a ^ { 2 } ( 2 a - 3 ) ^ { 2 } - ( 2 a ^ { 2 } + a ) ^ { 2 } + a ( 4 a - 1 ) ^ { 2 } - ( a - 2 ) ( a + 2 )$$

Evaluate

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

Solution Steps

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

$$a^{2}\left(4a^{2}-12a+9\right)-\left(2a^{2}+a\right)^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Use the distributive property to multiply a^{2} by 4a^{2}-12a+9.

$$4a^{4}-12a^{3}+9a^{2}-\left(2a^{2}+a\right)^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

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

$$4a^{4}-12a^{3}+9a^{2}-\left(4\left(a^{2}\right)^{2}+4a^{2}a+a^{2}\right)+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$4a^{4}-12a^{3}+9a^{2}-\left(4a^{4}+4a^{2}a+a^{2}\right)+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$4a^{4}-12a^{3}+9a^{2}-\left(4a^{4}+4a^{3}+a^{2}\right)+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

To find the opposite of 4a^{4}+4a^{3}+a^{2}, find the opposite of each term.

$$4a^{4}-12a^{3}+9a^{2}-4a^{4}-4a^{3}-a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Combine 4a^{4} and -4a^{4} to get 0.

$$-12a^{3}+9a^{2}-4a^{3}-a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Combine -12a^{3} and -4a^{3} to get -16a^{3}.

$$-16a^{3}+9a^{2}-a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

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

$$-16a^{3}+8a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(4a-1\right)^{2}.

$$-16a^{3}+8a^{2}+a\left(16a^{2}-8a+1\right)-\left(a-2\right)\left(a+2\right)$$

Use the distributive property to multiply a by 16a^{2}-8a+1.

$$-16a^{3}+8a^{2}+16a^{3}-8a^{2}+a-\left(a-2\right)\left(a+2\right)$$

Combine -16a^{3} and 16a^{3} to get 0.

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

Combine 8a^{2} and -8a^{2} to get 0.

$$a-\left(a-2\right)\left(a+2\right)$$

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.

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

To find the opposite of a^{2}-4, find the opposite of each term.

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

Expand

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

Solution Steps

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

$$a^{2}\left(4a^{2}-12a+9\right)-\left(2a^{2}+a\right)^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Use the distributive property to multiply a^{2} by 4a^{2}-12a+9.

$$4a^{4}-12a^{3}+9a^{2}-\left(2a^{2}+a\right)^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

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

$$4a^{4}-12a^{3}+9a^{2}-\left(4\left(a^{2}\right)^{2}+4a^{2}a+a^{2}\right)+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$4a^{4}-12a^{3}+9a^{2}-\left(4a^{4}+4a^{2}a+a^{2}\right)+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$4a^{4}-12a^{3}+9a^{2}-\left(4a^{4}+4a^{3}+a^{2}\right)+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

To find the opposite of 4a^{4}+4a^{3}+a^{2}, find the opposite of each term.

$$4a^{4}-12a^{3}+9a^{2}-4a^{4}-4a^{3}-a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Combine 4a^{4} and -4a^{4} to get 0.

$$-12a^{3}+9a^{2}-4a^{3}-a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Combine -12a^{3} and -4a^{3} to get -16a^{3}.

$$-16a^{3}+9a^{2}-a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

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

$$-16a^{3}+8a^{2}+a\left(4a-1\right)^{2}-\left(a-2\right)\left(a+2\right)$$

Use binomial theorem \left(p-q\right)^{2}=p^{2}-2pq+q^{2} to expand \left(4a-1\right)^{2}.

$$-16a^{3}+8a^{2}+a\left(16a^{2}-8a+1\right)-\left(a-2\right)\left(a+2\right)$$

Use the distributive property to multiply a by 16a^{2}-8a+1.

$$-16a^{3}+8a^{2}+16a^{3}-8a^{2}+a-\left(a-2\right)\left(a+2\right)$$

Combine -16a^{3} and 16a^{3} to get 0.

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

Combine 8a^{2} and -8a^{2} to get 0.

$$a-\left(a-2\right)\left(a+2\right)$$

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.

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

To find the opposite of a^{2}-4, find the opposite of each term.

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