Solve left(2x-1right)^2+left(1+2xright)^2-8x^2=

$${ \left(2x-1 \right) }^{ 2 } + { \left(1+2x \right) }^{ 2 } -8 { x }^{ 2 } =$$

Evaluate

$$2$$

Solution Steps

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

$$4x^{2}-4x+1+\left(1+2x\right)^{2}-8x^{2}$$

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

$$4x^{2}-4x+1+1+4x+4x^{2}-8x^{2}$$

Add 1 and 1 to get 2.

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

Combine -4x and 4x to get 0.

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

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

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

Combine 8x^{2} and -8x^{2} to get 0.

$$2$$

Factor

$$2$$