Solve (x+1)^2-(x-3)(x+3)=

$$( x + 1 ) ^ { 2 } - ( x - 3 ) ( x + 3 ) =$$

Evaluate

$$2\left(x+5\right)$$

Solution Steps

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

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

Consider \left(x-3\right)\left(x+3\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 3.

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

To find the opposite of x^{2}-9, find the opposite of each term.

$$x^{2}+2x+1-x^{2}+9$$

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

$$2x+1+9$$

Add 1 and 9 to get 10.

$$2x+10$$

Expand

$$2x+10$$

Solution Steps

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

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

Consider \left(x-3\right)\left(x+3\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 3.

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

To find the opposite of x^{2}-9, find the opposite of each term.

$$x^{2}+2x+1-x^{2}+9$$

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

$$2x+1+9$$

Add 1 and 9 to get 10.

$$2x+10$$