## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$y^3-2y+4$
Divide $4y^2$ to each term of the numerator to obtain: $=\dfrac{4y^5}{4y^2}-\dfrac{8y^3}{4y^2}+\dfrac{16y^2}{4y^2}$ Simplify by dividing the coefficients and using the Quotient Rule of Exponents $(\frac{a^m}{a^n} = a^{m-n})$ to obtain: $=1y^{5-2} - 2y^{3-2} + 4y^{2-2} \\=y^3-2y+4y^{0}$ Since $a^0=1, a\ne0$, then the expression above simplifies to: $=y^3-2y+4(1) \\=y^3-2y+4$