Question 407935


here is the rule:

{{{(x+y)^4 =x^4 + 4(x^3)y + 6(x^2)(y^2) + 4x(y^3) + y^4}}}

{{{(x^2-2)^4}}}...you have {{{x=x}}} and {{{y= -2}}}, so substitute {{{-2 }}}for {{{y}}}


{{{(x-2)^4 =x^4 + 4(x^3)(-2) + 6(x^2)((-2)^2) + 4x((-2)^3) + (-2)^4}}}

{{{(x-2)^4 =x^4 -8x^3 + 6(x^2)(4) + 4x(-8) + 16}}}

{{{(x-2)^4 = x^4 -8x^3 + 24x^2 - 32x + 16}}}