How do you use the binomial theorem to expand #(sqrtx+3)^4#?

1 Answer
Jan 6, 2018

Answer:

#=x^2 + 12xsqrtx + 54x + 108sqrtx + 81#

Explanation:

row in pascal's triangle for powers of #4#:

1 4 6 4 1

for #(x+y)^4#, this means #1(x^4y^0) + 4(x^3y^1) + 6(x^2y^2) + 4(x^1y^3) + 1(x^0y^4)#,

or #x^4+4x^3y+6x^2y^2+4xy^3+y^4#

for #(sqrtx+3)^4:#

#(sqrtx)^4 + 4((sqrtx)^3*3) + 6(x*9) + 4((sqrtx)*27) + 81#

#=x^2 + 12xsqrtx + 54x + 108sqrtx + 81#

or #x^2 + 12x^(3/2) + 54x + 108x^(1/2) + 81#