Solve 4(x+1)^2-(2x-5)(2x+5)

$$4 ( x + 1 ) ^ { 2 } - ( 2 x - 5 ) ( 2 x + 5 )$$

Evaluate

$$8x+29$$

Solution Steps

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

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

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

$$4x^{2}+8x+4-\left(2x-5\right)\left(2x+5\right)$$

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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

$$4x^{2}+8x+4-4x^{2}+25$$

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

$$8x+4+25$$

Add 4 and 25 to get 29.

$$8x+29$$

Expand

$$8x+29$$

Solution Steps

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

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

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

$$4x^{2}+8x+4-\left(2x-5\right)\left(2x+5\right)$$

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.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

$$4x^{2}+8x+4-4x^{2}+25$$

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

$$8x+4+25$$

Add 4 and 25 to get 29.

$$8x+29$$