## College Algebra (6th Edition)

The solutions are $y=2$ and $y=\pm\dfrac{2}{3}$
$9y^{3}+8=4y+18y^{2}$ Take all terms to the left side of the equation: $9y^{3}-18y^{2}-4y+8=0$ Group the first two terms together and the last two terms together: $(9y^{3}-18y^{2})-(4y-8)=0$ Take out common factor $9y^{2}$ from the first group and common factor $4$ from the second group: $9y^{2}(y-2)-4(y-2)=0$ Factor out $(y-2)$: $(y-2)(9y^{2}-4)=0$ Set both factors equal to $0$ and solve each individual equation for $y$: $y-2=0$ $y=2$ $9y^{2}-4=0$ $9y^{2}=4$ $y^{2}=\dfrac{4}{9}$ $\sqrt{y^{2}}=\pm\sqrt{\dfrac{4}{9}}$ $y=\pm\dfrac{2}{3}$ The solutions are $y=2$ and $y=\pm\dfrac{2}{3}$