# How do you find the coefficient of x^2 in the expansion of (2+x)^5?

Binomial coefficient are may be found by applyıng the Pascal's rule . Pascal's rule or the Pascal's triangle can easiliy be found in the internet. ${\left(x + y\right)}^{5} = {x}^{5} + 5 {x}^{4} y + 10 {x}^{3} {y}^{2} + 10 {x}^{2} {y}^{3} + 5 x {y}^{4} + {y}^{5}$
the x in the question correspond to y in the formulation. Acoording to this ${\left(2 + x\right)}^{5} = {2}^{5} + {5.2}^{4} x + {10.2}^{3} {x}^{2} + {10.2}^{2} {x}^{3} + 5.2 {x}^{4} + {x}^{5}$
or ${\left(2 + x\right)}^{5} = 32 + 80 x + 80 {x}^{2} + 40 {x}^{3} + 10 {x}^{4} + {x}^{5}$ Therefore the coefficient of ${x}^{2}$ in the expansion of ${\left(2 + x\right)}^{5}$ is 80.