Solve (2a+1)(2a-1)-5(a-1)-4(1+a^2)+(a-1)^2

$$( 2 a + 1 ) ( 2 a - 1 ) - 5 ( a - 1 ) - 4 ( 1 + a ^ { 2 } ) + ( a - 1 ) ^ { 2 }$$

Evaluate

$$a^{2}-7a+1$$

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.

$$\left(2a\right)^{2}-1-5\left(a-1\right)-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

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

$$2^{2}a^{2}-1-5\left(a-1\right)-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4a^{2}-1-5\left(a-1\right)-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Use the distributive property to multiply -5 by a-1.

$$4a^{2}-1-5a+5-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Add -1 and 5 to get 4.

$$4a^{2}+4-5a-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Use the distributive property to multiply -4 by 1+a^{2}.

$$4a^{2}+4-5a-4-4a^{2}+\left(a-1\right)^{2}$$

Subtract 4 from 4 to get 0.

$$4a^{2}-5a-4a^{2}+\left(a-1\right)^{2}$$

Combine 4a^{2} and -4a^{2} to get 0.

$$-5a+\left(a-1\right)^{2}$$

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

$$-5a+a^{2}-2a+1$$

Combine -5a and -2a to get -7a.

$$-7a+a^{2}+1$$

Expand

$$a^{2}-7a+1$$

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.

$$\left(2a\right)^{2}-1-5\left(a-1\right)-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

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

$$2^{2}a^{2}-1-5\left(a-1\right)-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4a^{2}-1-5\left(a-1\right)-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Use the distributive property to multiply -5 by a-1.

$$4a^{2}-1-5a+5-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Add -1 and 5 to get 4.

$$4a^{2}+4-5a-4\left(1+a^{2}\right)+\left(a-1\right)^{2}$$

Use the distributive property to multiply -4 by 1+a^{2}.

$$4a^{2}+4-5a-4-4a^{2}+\left(a-1\right)^{2}$$

Subtract 4 from 4 to get 0.

$$4a^{2}-5a-4a^{2}+\left(a-1\right)^{2}$$

Combine 4a^{2} and -4a^{2} to get 0.

$$-5a+\left(a-1\right)^{2}$$

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

$$-5a+a^{2}-2a+1$$

Combine -5a and -2a to get -7a.

$$-7a+a^{2}+1$$