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

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

Evaluate

$$5x^{2}+20x-29$$

Solution Steps

Consider \left(3x-2\right)\left(3x+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}. Square 2.

$$\left(3x\right)^{2}-4-\left(2x-5\right)^{2}$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4-\left(2x-5\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4-\left(2x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$9x^{2}-4-\left(4x^{2}-20x+25\right)$$

To find the opposite of 4x^{2}-20x+25, find the opposite of each term.

$$9x^{2}-4-4x^{2}+20x-25$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-4+20x-25$$

Subtract 25 from -4 to get -29.

$$5x^{2}-29+20x$$

Expand

$$5x^{2}+20x-29$$

Solution Steps

Consider \left(3x-2\right)\left(3x+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}. Square 2.

$$\left(3x\right)^{2}-4-\left(2x-5\right)^{2}$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-4-\left(2x-5\right)^{2}$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-4-\left(2x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$9x^{2}-4-\left(4x^{2}-20x+25\right)$$

To find the opposite of 4x^{2}-20x+25, find the opposite of each term.

$$9x^{2}-4-4x^{2}+20x-25$$

Combine 9x^{2} and -4x^{2} to get 5x^{2}.

$$5x^{2}-4+20x-25$$

Subtract 25 from -4 to get -29.

$$5x^{2}-29+20x$$