Solve (2x+5)(2x-5)+2(4x+3)-4left(x+1right)^2

$$(2x+5)(2x-5)+2(4x+3)-4 { \left(x+1 \right) }^{ 2 }$$

Evaluate

$$-23$$

Solution Steps

Consider \left(2x+5\right)\left(2x-5\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 5.

$$\left(2x\right)^{2}-25+2\left(4x+3\right)-4\left(x+1\right)^{2}$$

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

$$2^{2}x^{2}-25+2\left(4x+3\right)-4\left(x+1\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-25+2\left(4x+3\right)-4\left(x+1\right)^{2}$$

Use the distributive property to multiply 2 by 4x+3.

$$4x^{2}-25+8x+6-4\left(x+1\right)^{2}$$

Add -25 and 6 to get -19.

$$4x^{2}-19+8x-4\left(x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$4x^{2}-19+8x-4\left(x^{2}+2x+1\right)$$

Use the distributive property to multiply -4 by x^{2}+2x+1.

$$4x^{2}-19+8x-4x^{2}-8x-4$$

Combine 4x^{2} and -4x^{2} to get 0.

$$-19+8x-8x-4$$

Combine 8x and -8x to get 0.

$$-19-4$$

Subtract 4 from -19 to get -23.

$$-23$$

Factor

$$-23$$