How do you use the Binomial Theorem to expand # (x-5)^5#?

1 Answer
May 26, 2018

Answer:

#(-5+x)^5=-3125+3125x -1250x^2+250x^3-25x^4+x^5#

Explanation:

#(a+bx)^n=sum_(r=0)^n((n),(r))a^(n-r)(bx)^r=sum_(r=0)^n(n!)/(r!(n-r)!)a^(n-r)(bx)^r#

#(-5+x)^5=sum_(r=0)^5(5!)/(r!(5-r)!)(-5)^(5-r)x^r#

#(-5+x)^5=(5!)/(0!(5-0)!)(-5)^(5-0)x^0+(5!)/(1!(5-1)!)(-5)^(5-1)x^1+(5!)/(2!(5-2)!)(-5)^(5-2)x^2+(5!)/(3!(5-3)!)(-5)^(5-3)x^3+(5!)/(4!(5-4)!)(-5)^(5-4)x^4+(5!)/(5!(5-5)!)(-5)^(5-5)x^5#

#(-5+x)^5=(5!)/(0!5!)(-5)^5+(5!)/(1!4!)(-5)^4x+(5!)/(2!3!)(-5)^3x^2+(5!)/(3!2!)(-5)^2x^3+(5!)/(4!1!)(-5)x^4+(5!)/(5!0!)x^5#

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

#(-5+x)^5=-3125+3125x -1250x^2+250x^3-25x^4+x^5#