## Precalculus (10th Edition)

$-3$ and $5i$
The Conjugate Pairs Theorem says that if a polynomial has real coefficients, then any complex zeros occur in conjugate pairs. That is, if $a + bi$ is a zero then so is $a – bi$ and vice-versa. Its degree is $3$, hence it has $3$ complex zeros. According to the Conjugate Pair Theorem since $-5i$ is a complex zero, $5i$ is also a complex zero. Assume $r$ is the last zero. Then we know that \begin{align*} (x-5i)(x-(-5i))(x-r)&=x^3+3x^2+25x+75\\(x-5i)(x+5i)(x-r)&=x^3+3x^2+25x+75\\(x^2+25)(x-r)&=x^3+3x^2+25x+75\\x^3-rx^2+25x-25r&=x^3+3x^2+25x+75\end{align*} Thus, $-25r=75\\ r=-3$