## Intermediate Algebra (6th Edition)

The solutions are $-3i, 3i,$ and $1$.
Group the first two terms together and the last two terms together to obtain: $(y^3+9y)+(-y^2-9)=0$ Factor out the GCF in each group ($y$ in the first and $-1$) to obtain: $y(y^2+9)+(-1)(y^2+9)$ Factor out the GCF of the expression ($y^2+9$) to obtain: $=(y^2+9)[y+(-1)] \\=(y^2+9)(y-1)$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&y^2+9=0 &\text{or} &y-1=0 \\&y^2=-9 &\text{or} &y=1 \\&y=\pm\sqrt{-9} &\text{or} &y=1 \\&y=\pm 3i &\text{or} &y=1 \end{array} Therefore, the solutions are $-3i, 3i,$ and $1$.