Solve 2500(1+x)^2-3600

$$2500 ( 1 + x ) ^ { 2 } - 3600$$

Evaluate

$$2500x^{2}+5000x-1100$$

Solution Steps

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

$$2500\left(1+2x+x^{2}\right)-3600$$

Use the distributive property to multiply 2500 by 1+2x+x^{2}.

$$2500+5000x+2500x^{2}-3600$$

Subtract 3600 from 2500 to get -1100.

$$-1100+5000x+2500x^{2}$$

Expand

$$2500x^{2}+5000x-1100$$

Solution Steps

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

$$2500\left(1+2x+x^{2}\right)-3600$$

Use the distributive property to multiply 2500 by 1+2x+x^{2}.

$$2500+5000x+2500x^{2}-3600$$

Subtract 3600 from 2500 to get -1100.

$$-1100+5000x+2500x^{2}$$