Solve (x-3)(x+3)-(x-5)(2x+3)

$$( x - 3 ) ( x + 3 ) - ( x - 5 ) ( 2 x + 3 )$$

Evaluate

$$6+7x-x^{2}$$

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}.

$$x^{2}-3^{2}-\left(x-5\right)\left(2x+3\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}-9-\left(x-5\right)\left(2x+3\right)$$

Apply the distributive property by multiplying each term of x-5 by each term of 2x+3.

$$x^{2}-9-\left(2x^{2}+3x-10x-15\right)$$

Combine 3x and -10x to get -7x.

$$x^{2}-9-\left(2x^{2}-7x-15\right)$$

To find the opposite of 2x^{2}-7x-15, find the opposite of each term.

$$x^{2}-9-2x^{2}-\left(-7x\right)-\left(-15\right)$$

The opposite of -7x is 7x.

$$x^{2}-9-2x^{2}+7x-\left(-15\right)$$

The opposite of -15 is 15.

$$x^{2}-9-2x^{2}+7x+15$$

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

$$-x^{2}-9+7x+15$$

Add -9 and 15 to get 6.

$$-x^{2}+6+7x$$

Expand

$$6+7x-x^{2}$$

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}.

$$x^{2}-3^{2}-\left(x-5\right)\left(2x+3\right)$$

Calculate 3 to the power of 2 and get 9.

$$x^{2}-9-\left(x-5\right)\left(2x+3\right)$$

Apply the distributive property by multiplying each term of x-5 by each term of 2x+3.

$$x^{2}-9-\left(2x^{2}+3x-10x-15\right)$$

Combine 3x and -10x to get -7x.

$$x^{2}-9-\left(2x^{2}-7x-15\right)$$

To find the opposite of 2x^{2}-7x-15, find the opposite of each term.

$$x^{2}-9-2x^{2}-\left(-7x\right)-\left(-15\right)$$

The opposite of -7x is 7x.

$$x^{2}-9-2x^{2}+7x-\left(-15\right)$$

The opposite of -15 is 15.

$$x^{2}-9-2x^{2}+7x+15$$

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

$$-x^{2}-9+7x+15$$

Add -9 and 15 to get 6.

$$-x^{2}+6+7x$$