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

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

Evaluate

$$2\left(x+1\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-\left(x+1\right)\left(x-1\right)-2\left(2x+4\right)$$

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.

$$x^{2}+6x+9-\left(x^{2}-1\right)-2\left(2x+4\right)$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}+6x+9-x^{2}+1-2\left(2x+4\right)$$

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

$$6x+9+1-2\left(2x+4\right)$$

Add 9 and 1 to get 10.

$$6x+10-2\left(2x+4\right)$$

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

$$6x+10-4x-8$$

Combine 6x and -4x to get 2x.

$$2x+10-8$$

Subtract 8 from 10 to get 2.

$$2x+2$$

Expand

$$2x+2$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9-\left(x+1\right)\left(x-1\right)-2\left(2x+4\right)$$

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.

$$x^{2}+6x+9-\left(x^{2}-1\right)-2\left(2x+4\right)$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{2}+6x+9-x^{2}+1-2\left(2x+4\right)$$

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

$$6x+9+1-2\left(2x+4\right)$$

Add 9 and 1 to get 10.

$$6x+10-2\left(2x+4\right)$$

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

$$6x+10-4x-8$$

Combine 6x and -4x to get 2x.

$$2x+10-8$$

Subtract 8 from 10 to get 2.

$$2x+2$$