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

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

Evaluate

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

Solution Steps

Consider \left(x+1\right)\left(x-1\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 1.

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

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

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

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

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

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

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

Add -1 and 4 to get 3.

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

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

$$3+4x^{2}+4x+1$$

Add 3 and 1 to get 4.

$$4+4x^{2}+4x$$

Expand

$$4x^{2}+4x+4$$

Solution Steps

Consider \left(x+1\right)\left(x-1\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 1.

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

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

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

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

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

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

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

Add -1 and 4 to get 3.

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

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

$$3+4x^{2}+4x+1$$

Add 3 and 1 to get 4.

$$4+4x^{2}+4x$$