## College Algebra (10th Edition)

The remaining zeros of the function are: $\color{blue}{\bf 2i, \frac{1}{2}, -3}$
The polynomial function's degree is four so it has four zeros. RECALL: If $a+bi$ is a zero of a polynomial function with real number coefficients, then its conjugate $a-bi$ is also a zero of the function. Thus, one of the missing zeros of the given function is the conjugate of $-2i$, which is ${\bf 2i}$. Two of the four zeros of the polynomial function with real coefficients are: $-2i$ and $2i$ This means that $(x-2i)$ and $(x+2i)$ are both factors of $P(x)$. Multiply these two to obtain: $(x-2i)(x+2i) = x^2-4i^2 = x^2-4(-1) = x^2+4$ Divide $P(x)$ by $x^2+4$ using long division to obtain $2x^2+5x-3$. (refer to the attached image below for the actual long division) Factor $2x^2+5x-3$ to obtain: $2x^2+5x-3=(2x-1)(x+3)$ Thus, the completely factored form of $P(x)$ is: $P(x) = (x^2+4)(2x-1)(x+3)$ Equate each factor to zero, then solve each equation to obtain: \begin{array}{ccccc} &x^2+4=0 &\text{or} &2x-1=0 &\text{or} &x+3=0 \\&x^2=-4 &\text{or} &2x=1 &\text{or} &x=-3 \\&x^2=-4 &\text{or} &x=\frac{1}{2} &\text{or} &x=-3 \end{array} The first factor's solutions are the two complex zeros of the function. Thus, the real zeros are $x=\left\{\frac{1}{2}, -3\right\}$. Therefore, the other zero of the given function are: $\color{blue}{\bf 2i, \frac{1}{2}, -3}$