How do you find the in binomial expansion of #(x-3)^5 #?

1 Answer
Alan P.
Jun 1, 2015

In general the expansion of (a+b)^n is
#sum_(i=0)^n(nCi*a^(n-i)*b^i)#

where #pCq# (p Choose q)" can be evaluated as #(p!)/((p-q)!(q!))#

Alternately (and probably more simply), you could use Pascal's Triangle to optain the coefficients:

#{: (0, rarr, 1,"-","-","-", "-", "-"), (1, rarr, 1,1,"-","-", "-", "-"), (2, rarr, 1,2,1,"-", "-", "-"), (3, rarr, 1,3,3,1, "-", "-"), ( 4, rarr, 1, 4, 6, 4, 1, "-"), ( 5, rarr, 1, 5, 10, 10, 5, 1) :}#
(etc.)

Giving
#(x-3)^5#
#=(1)(x^5)(-3)^0+5(x^4)(-3)^1+10(x^3)(-3)^2+10(x^2)(-3)^3+5(x^1)(-3)^4+(1)(x^0)(-3)^5#

=#x^5-15x^4+90x^3-270x^2+405x-243#

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