How do you expand #(3x-2y)^5#?

1 Answer
Jul 13, 2017

Answer:

#243x^5-810x^4y+1080x^3y^2-720x^2y^3+240xy^4-32y^5#

Explanation:

#"using the "color(blue)"binomial theorem"#

#•color(white)(x)(x+y)^n=sum_(r=0)^n((n),(r))x^(n-r)y^r#

#"where" ((n),(r))=(n!)/(r!(n-r)!)#

#"here " x=3x" and " y=-2y#

#rArr(3x-2y)^5#

#=((5),(0))(3x)^5(-2y)^0+((5),(1))(3x)^4(-2y)^1+((5),(2))(3x)^3(-2y)^2+((5),(3))(3x)^2(-2y)^3+((5),(4))(3x)^1(-2y)^4+((5),(5))(3x)^0(-2y)^5#

#"we can obtain the binomial coefficients using the appropriate"#
#"row of "color(blue)"Pascal's triangle"#

#"for n = 5 the row of coefficients is"#

#1color(white)(x)5color(white)(x)10color(white)(x)10color(white)(x)5color(white)(x)1#

#=1.243x^5+5.81x^4(-2y)+10.27x^3. 4y^2+10.9x^2(-8y^3)+5.3x.16y^4+1.-32y^5#

#=243x^5-810x^4y+1080x^3y^2-720x^2y^3+240xy^4-32y^5#