College Algebra (10th Edition)

Published by Pearson

Chapter 5 - Section 5.6 - Complex Zeros; Fundamental Theorem of Algebra - 5.6 Assess Your Understanding - Page 394: 26

Answer

$\color{blue}{\bf -3i, \frac{1}{3}, -2}$

Work Step by Step

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 $3i$, which is ${\bf -3i}$. Two of the four zeros of the polynomial function with real coefficients are: $-3i$ and $3i$ This means that $(x-3i)$ and $(x+3i)$ are both factors of $P(x)$. Multiply these two to obtain: $(x-3i)(x+3i) = x^2-9i^2 = x^2-9(-1) = x^2+9$ Divide $P(x)$ by $x^2+9$ using long division to obtain $3x^2+5x-2$. (refer to the attached image below for the actual long division) Factor $3x^2+5x-2$ to obtain: $3x^2+5x-2=(3x-1)(x+2)$ Thus, the completely factored form of $P(x)$ is: $P(x) = (x^2+9)(3x-1)(x+2)$ Equate each factor to zero, then solve each equation to obtain: \begin{array}{ccccc} &x^2+9=0 &\text{or} &3x-1=0 &\text{or} &x+2=0 \\&x^2=-9 &\text{or} &3x=1 &\text{or} &x=-2 \\&x^2=-9 &\text{or} &x=\frac{1}{3} &\text{or} &x=-2 \end{array} The first factor's solutions are the two complex zeros of the function. Thus, the real zeros are $x=\left\{\frac{1}{3}, -2\right\}$. Therefore, the other zero of the given function are: $\color{blue}{\bf -3i, \frac{1}{3}, -2}$

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