Question 3.2.3: Deciding Whether a Number Is a Zero Decide whether the given...
Deciding Whether a Number Is a Zero
Decide whether the given number k is a zero of ƒ(x).
(a) ƒ(x) = x³ – 4x² + 9x – 6; k = 1
(b) ƒ(x) = x^{4} + x² – 3x + 1; k = -1
(c) ƒ(x) = x^{4} – 2x³ + 4x² + 2x – 5; k = 1 + 2i
(a) To decide whether 1 is a zero of ƒ(x) = x³ – 4x² + 9x – 6, use synthetic division.
\begin{matrix} \text{Proposed zero}→1)&\overline{\begin{matrix} 1 & -4 & 9 & -6 \end{matrix}}&← ƒ(x) = x³ – 4x² + 9x – 6 \\& \begin{matrix}&& 1 &-3 & 6 \end{matrix} \\ &\overline{\begin{matrix}1 &-3& 6 & 0 \end{matrix}} &← \text{Remainder} \end{matrix}
Because the remainder is 0, ƒ(1) = 0, and 1 is a zero of the given polynomial function. An x-intercept of the graph of ƒ(x) = x³ – 4x² + 9x – 6 is the point (1, 0). The graph in Figure 16 supports this.
(b) For ƒ(x)= x^{4} + x² – 3x + 1, remember to use 0 as coefficient for the missing x²-term in the synthetic division.
\begin{matrix} \text{Proposed zero} →-1)&\overline{\begin{matrix} 1 &0 & 1& -3&1\end{matrix}} \\& \begin{matrix}&- 1 &1 & -2&5 \end{matrix} \\ &\overline{\begin{matrix}1 &-1& 2 & -5&6 \end{matrix}} &← \text{Remainder} \end{matrix}
The remainder is not 0, so -1 is not a zero of ƒ(x)= x^{4} + x² – 3x + 1. In fact, ƒ(-1) = 6, indicating that (-1, 6) is on the graph of ƒ(x). The graph in Figure 17 supports this.
(c) Use synthetic division and operations with complex numbers to determine whether 1 + 2i is a zero of ƒ(x) = x^{4} – 2x³+ 4x² + 2x – 5.
\begin{matrix} 1+2i)\overline{\begin{matrix} 1 & -2 && &&4 &&2 && &-5 \end{matrix} } \\ \underline{\begin{matrix}&&&& && 1 & +2i & -5 & -1 & -2i &5 \end{matrix}} & i²=-1 \\ \begin{matrix} &&&1 & -1 & +2i & -1 && 1 & -2i & 0\end{matrix}& ←\text{Remainder} \end{matrix}
The remainder is 0, so 1 + 2i is a zero of the given polynomial function.
Notice that 1 + 2i is not a real number zero. Therefore, it is not associated with an x-intercept on the graph of ƒ(x).
