Solve (x+3)(x-2)-(x-1)(x+1)

$$( x + 3 ) ( x - 2 ) - ( x - 1 ) ( x + 1 )$$

Evaluate

$$x-5$$

Solution Steps

Apply the distributive property by multiplying each term of x+3 by each term of x-2.

$$x^{2}-2x+3x-6-\left(x-1\right)\left(x+1\right)$$

Combine -2x and 3x to get x.

$$x^{2}+x-6-\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+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}.

$$x^{2}+x-6-\left(x^{2}-1^{2}\right)$$

Calculate 1 to the power of 2 and get 1.

$$x^{2}+x-6-\left(x^{2}-1\right)$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}+x-6-x^{2}-\left(-1\right)$$

The opposite of -1 is 1.

$$x^{2}+x-6-x^{2}+1$$

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

$$x-6+1$$

Add -6 and 1 to get -5.

$$x-5$$

Expand

$$x-5$$

Solution Steps

Apply the distributive property by multiplying each term of x+3 by each term of x-2.

$$x^{2}-2x+3x-6-\left(x-1\right)\left(x+1\right)$$

Combine -2x and 3x to get x.

$$x^{2}+x-6-\left(x-1\right)\left(x+1\right)$$

Consider \left(x-1\right)\left(x+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}.

$$x^{2}+x-6-\left(x^{2}-1^{2}\right)$$

Calculate 1 to the power of 2 and get 1.

$$x^{2}+x-6-\left(x^{2}-1\right)$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}+x-6-x^{2}-\left(-1\right)$$

The opposite of -1 is 1.

$$x^{2}+x-6-x^{2}+1$$

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

$$x-6+1$$

Add -6 and 1 to get -5.

$$x-5$$