Solve 4(a-1)^2-4(a-1)-3

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

Evaluate

$$\left(2a-5\right)\left(2a-1\right)$$

Solution Steps

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

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

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

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

Use the distributive property to multiply -4 by a-1.

$$4a^{2}-8a+4-4a+4-3$$

Combine -8a and -4a to get -12a.

$$4a^{2}-12a+4+4-3$$

Add 4 and 4 to get 8.

$$4a^{2}-12a+8-3$$

Subtract 3 from 8 to get 5.

$$4a^{2}-12a+5$$

Expand

$$4a^{2}-12a+5$$

Solution Steps

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

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

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

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

Use the distributive property to multiply -4 by a-1.

$$4a^{2}-8a+4-4a+4-3$$

Combine -8a and -4a to get -12a.

$$4a^{2}-12a+4+4-3$$

Add 4 and 4 to get 8.

$$4a^{2}-12a+8-3$$

Subtract 3 from 8 to get 5.

$$4a^{2}-12a+5$$