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

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

Evaluate

$$x-7$$

Solution Steps

Apply the distributive property by multiplying each term of 1+x by each term of 2-x.

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

Combine -x and 2x to get x.

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

Consider \left(3-x\right)\left(3+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$2+x-x^{2}-\left(9-x^{2}\right)$$

To find the opposite of 9-x^{2}, find the opposite of each term.

$$2+x-x^{2}-9-\left(-x^{2}\right)$$

The opposite of -x^{2} is x^{2}.

$$2+x-x^{2}-9+x^{2}$$

Subtract 9 from 2 to get -7.

$$-7+x-x^{2}+x^{2}$$

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

$$-7+x$$

Expand

$$x-7$$

Solution Steps

Apply the distributive property by multiplying each term of 1+x by each term of 2-x.

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

Combine -x and 2x to get x.

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

Consider \left(3-x\right)\left(3+x\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Calculate 3 to the power of 2 and get 9.

$$2+x-x^{2}-\left(9-x^{2}\right)$$

To find the opposite of 9-x^{2}, find the opposite of each term.

$$2+x-x^{2}-9-\left(-x^{2}\right)$$

The opposite of -x^{2} is x^{2}.

$$2+x-x^{2}-9+x^{2}$$

Subtract 9 from 2 to get -7.

$$-7+x-x^{2}+x^{2}$$

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

$$-7+x$$