How do you use the binomial theorem to expand and simplify #(2+x)^5-(2-x)^5# and apply the result to calculate #2.1^5-1.9^5# ?

1 Answer
George C.
Feb 21, 2016

#(2+x)^5-(2-x)^5=160x+80x^3+2x^5#

#2.1^5-1.9^5=16.08002#

Explanation:

Use Pascal's triangle to help expand the #5#th power of the binomials.

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Choose the row #1, 5, 10, 10, 5, 1# to find that for any #a# and #b# we have:

#(a+b)^5 = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5#

Hence:

#(2+x)^5#

#= 2^5+5(2^4)x+10(2^3)x^2+10(2^2)x^3+5(2)x^4+x^5#

#=32+80x+80x^2+40x^3+10x^4+x^5#

#color(white)()#
Similarly:

#(2-x)^5=32-80x+80x^2-40x^3+10x^4-x^5#

#color(white)()#
Hence:

#(2+x)^5-(2-x)^5#

#=(color(red)(cancel(color(black)(32)))+80x+color(red)(cancel(color(black)(80x^2)))+40x^3+color(red)(cancel(color(black)(10x^4)))+x^5)-(color(red)(cancel(color(black)(32)))-80x+color(red)(cancel(color(black)(80x^2)))-40x^3+color(red)(cancel(color(black)(10x^4)))-x^5)#

#=160x+80x^3+2x^5#

#color(white)()#
Use this with #x=0.1# as follows:

#2.1^5-1.9^5#

#=(2+0.1)^5-(2-0.1)^5#

#=160(0.1)+80(0.1)^3+2(0.1)^5#

#=160(0.1)+80(0.001)+2(0.00001)#

#=16+0.08+0.00002#

#=16.08002#

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