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

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

Evaluate

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

Solution Steps

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

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

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

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

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

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

Subtract 1 from 1 to get 0.

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

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

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

Expand

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

Solution Steps

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

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

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

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

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

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

Subtract 1 from 1 to get 0.

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

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

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