Solve (c+2)(c-2)-4c(c-1)

$$( c + 2 ) ( c - 2 ) - 4 c ( c - 1 )$$

Evaluate

$$-3c^{2}+4c-4$$

Solution Steps

Consider \left(c+2\right)\left(c-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}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Combine c^{2} and -4c^{2} to get -3c^{2}.

$$-3c^{2}-4+4c$$

Expand

$$-3c^{2}+4c-4$$

Solution Steps

Consider \left(c+2\right)\left(c-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}.

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

Calculate 2 to the power of 2 and get 4.

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

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

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

Combine c^{2} and -4c^{2} to get -3c^{2}.

$$-3c^{2}-4+4c$$