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

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

Evaluate

$$-x^{2}+3x-7$$

Solution Steps

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

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

Calculate 3 to the power of 2 and get 9.

$$x^{2}-9-\left(2x+1\right)\left(x-2\right)$$

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

$$x^{2}-9-\left(2x^{2}-4x+x-2\right)$$

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

$$x^{2}-9-\left(2x^{2}-3x-2\right)$$

To find the opposite of 2x^{2}-3x-2, find the opposite of each term.

$$x^{2}-9-2x^{2}-\left(-3x\right)-\left(-2\right)$$

The opposite of -3x is 3x.

$$x^{2}-9-2x^{2}+3x-\left(-2\right)$$

The opposite of -2 is 2.

$$x^{2}-9-2x^{2}+3x+2$$

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

$$-x^{2}-9+3x+2$$

Add -9 and 2 to get -7.

$$-x^{2}-7+3x$$

Expand

$$-x^{2}+3x-7$$

Solution Steps

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

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

Calculate 3 to the power of 2 and get 9.

$$x^{2}-9-\left(2x+1\right)\left(x-2\right)$$

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

$$x^{2}-9-\left(2x^{2}-4x+x-2\right)$$

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

$$x^{2}-9-\left(2x^{2}-3x-2\right)$$

To find the opposite of 2x^{2}-3x-2, find the opposite of each term.

$$x^{2}-9-2x^{2}-\left(-3x\right)-\left(-2\right)$$

The opposite of -3x is 3x.

$$x^{2}-9-2x^{2}+3x-\left(-2\right)$$

The opposite of -2 is 2.

$$x^{2}-9-2x^{2}+3x+2$$

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

$$-x^{2}-9+3x+2$$

Add -9 and 2 to get -7.

$$-x^{2}-7+3x$$