## College Algebra (10th Edition)

Solution set: $\displaystyle \{-\frac{3}{2}\}$
We can factor in pairs here: $x^{2}(2x+3)+(2x+3)=0$ $(2x+3)(x^{2}+1)=0$ $x^{2}+1$ has no real zeros (does not yield zero for any real x). $2x+3=0$ when $x=-\displaystyle \frac{3}{2}$ Solution set: $\displaystyle \{-\frac{3}{2}\}$