jmc

Since the coefficients of the polynomial are all real, the four non-real roots must come in two conjugate pairs. Let $z$ and $w$ be the two roots that multiply to $13+i$. Since $13+i$ is not real, $z$ and $w$ cannot be conjugates of each other (since any complex number times its conjugate is a real number). Therefore, the other two roots must be $\overline{z}$ and $\overline{w}$, the conjugates of $z$ and $w$. Therefore, we have $$zw=13+i\text{and}\overline{z}+\overline{w}=3+4i.$$To find $b$, we use Vieta's formulas: $b$ equals the second symmetric sum of the roots, which is $$b=zw+z\overline{z}+z\overline{w}+w\overline{z}+w\overline{w}+\overline{z}\cdot \overline{w}.$$To evaluate this expression, we first recognize the terms $zw$ and $\overline{z}\cdot \overline{w}$. We have $zw=13+i$, so $\overline{z}\cdot \overline{w}=\overline{zw}=13-i$. Thus, $$b=26+(z\overline{z}+z\overline{w}+w\overline{z}+w\overline{w}).$$To finish, we can factor the remaining terms by grouping: $$b=26+(z+w)(\overline{z}+\overline{w}).$$From $\overline{z}+\overline{w}=3+4i$, we get $z+w=3-4i$. Thus, $$b=26+(3-4i)(3+4i)=\boxed{51}.$$