Solve (2a-1)(2a+1)+(2a-1)^2-2a(a-2)

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

Evaluate

$$6a^{2}$$

Solution Steps

Consider \left(2a-1\right)\left(2a+1\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 1.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Combine 4a^{2} and 4a^{2} to get 8a^{2}.

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

Add -1 and 1 to get 0.

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

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

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

Combine 8a^{2} and -2a^{2} to get 6a^{2}.

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

Combine -4a and 4a to get 0.

$$6a^{2}$$

Expand

$$6a^{2}$$

Solution Steps

Consider \left(2a-1\right)\left(2a+1\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 1.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Combine 4a^{2} and 4a^{2} to get 8a^{2}.

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

Add -1 and 1 to get 0.

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

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

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

Combine 8a^{2} and -2a^{2} to get 6a^{2}.

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

Combine -4a and 4a to get 0.

$$6a^{2}$$