Question 153702
{{{(y^5-3y^3+8y^2)/(5y)}}} Start with the given expression



{{{(y^5)/(5y)-(3y^3)/(5y)+(8y^2)/(5y)}}} Break up the fraction



{{{(y^(5-1))/(5)-(3y^(3-1))/(5)+(8y^(2-1))/(5)}}} Divide the variable terms by simply subtracting the exponents. So {{{y^5/y^1=y^(5-1)=y^4}}}. Note: {{{y=y^1}}}



{{{(y^4)/(5)-(3y^2)/(5)+(8y^1)/(5)}}} Subtract



{{{(y^4)/(5)-(3y^2)/(5)+(8y)/(5)}}} Simplify



So {{{(y^5-3y^3+8y^2)/(5y)=(y^4)/(5)-(3y^2)/(5)+(8y)/(5)}}}



Note: you could rearrange the coefficients to get this answer as well


{{{(y^5-3y^3+8y^2)/(5y)=(1/5)y^4-(3/5)y^2+(8/5)y}}}