We have the Binomial Theorem:
#(x+a)^n=sum_(k=0)^n((n),(k))x^(n-k)a^k#.
Here, #((n),(k))# is read as "n choose k" and is equal to #(n!)/(k!(n-k)!)#, where #!# is the factorial function.
Here, #x=3, a=-5x, n=14#. Inputting:
#(3-5x)^14=sum_(k=0)^14((14),(k))3^(14-k)(-5x)^k#
We'll do the first three terms:
When #k=0#:
#((14),(0))3^(14-0)(-5x)^0#
#=(14!)/(0!(14-0)!)*3^14#
#=1*4782969#
#=4782969#
When #k=1#:
#((14),(1))3^(14-1)(-5x)^1#
#=(14!)/(1!(14-1)!)*3^13*(-5x)^1#
#=14*1594323*-5x#
#=-111602610x#
When #k=2#:
#((14),(2))3^(14-2)(-5x)^2#
#=(14!)/(2!(14-2)!)*3^12(-5x)^2#
#=91*531441*25x^2#
#=1209028275x^2#
And we keep going on till we get to #k=14#.