How do you multiply #(x-3)^5#?

1 Answer
bp
Aug 18, 2015

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

Explanation:

Either use binomial theorem to get the answer or using the Pascal's triangle, the expansion of #(x+y)^5 = x^5+5x^4y+10x^3y^2 +10x^2y^3 +5xy^4 +y^5#
Now, substitute y= -3 to get,

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

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