## College Algebra (10th Edition)

$\quad x\displaystyle \in\left\{\frac{5}{2}\right\}$
Factoring in pairs is not obvious here, so we try to find rational zeros of $f(x)=2x^{3}-3x^{2}-3x-5$ Possible rational roots: $\displaystyle \frac{p}{q}=\frac{\pm 1,\pm 5}{\pm 1,\pm 2}$ Testing with synthetic division, ... we eventually try $x-\displaystyle \frac{5}{2}$ $\left.\begin{array}{l} 5/2 \ \ |\\ \\ \\ \end{array}\right.\begin{array}{rrrrrr} 2 &-3 &-3 &-5 \\\hline &5 &5 & 5 \\\hline 2& 2 &2 & |\ \ 0 \end{array}$ $f(x)=(x-\displaystyle \frac{5}{2})2( x^{2}+x+1)$ applying the quadratic formula, $b^{2}-4ac=1-4=-3$ the discriminant is negative, $x^{2}+x+1=0\quad$ has no real solutions. $f(x)=0\quad$when$\quad x\displaystyle \in\left\{\frac{5}{2}\right\}$