Solve (x+2)(x-2)-(x-4)(x+3)

$$( x + 2 ) ( x - 2 ) - ( x - 4 ) ( x + 3 )$$

Evaluate

$$x+8$$

Solution Steps

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

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

Calculate 2 to the power of 2 and get 4.

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

Apply the distributive property by multiplying each term of x-4 by each term of x+3.

$$x^{2}-4-\left(x^{2}+3x-4x-12\right)$$

Combine 3x and -4x to get -x.

$$x^{2}-4-\left(x^{2}-x-12\right)$$

To find the opposite of x^{2}-x-12, find the opposite of each term.

$$x^{2}-4-x^{2}-\left(-x\right)-\left(-12\right)$$

The opposite of -x is x.

$$x^{2}-4-x^{2}+x-\left(-12\right)$$

The opposite of -12 is 12.

$$x^{2}-4-x^{2}+x+12$$

Combine x^{2} and -x^{2} to get 0.

$$-4+x+12$$

Add -4 and 12 to get 8.

$$8+x$$

Expand

$$x+8$$

Solution Steps

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

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

Calculate 2 to the power of 2 and get 4.

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

Apply the distributive property by multiplying each term of x-4 by each term of x+3.

$$x^{2}-4-\left(x^{2}+3x-4x-12\right)$$

Combine 3x and -4x to get -x.

$$x^{2}-4-\left(x^{2}-x-12\right)$$

To find the opposite of x^{2}-x-12, find the opposite of each term.

$$x^{2}-4-x^{2}-\left(-x\right)-\left(-12\right)$$

The opposite of -x is x.

$$x^{2}-4-x^{2}+x-\left(-12\right)$$

The opposite of -12 is 12.

$$x^{2}-4-x^{2}+x+12$$

Combine x^{2} and -x^{2} to get 0.

$$-4+x+12$$

Add -4 and 12 to get 8.

$$8+x$$