Question 3.3.3: Using the Rational Zeros Theorem Consider the polynomial fun...
Using the Rational Zeros Theorem
Consider the polynomial function.
ƒ(x) = 6x^{4} + 7x³ – 12x² – 3x + 2
(a) List all possible rational zeros.
(b) Find all rational zeros and factor ƒ(x) into linear factors.
(a) For a rational number \frac{p}{ q} to be a zero, p must be a factor of a_{0} = 2, and q must be a factor of a_{4} = 6. Thus, p can be ±1 or ±2, and q can be ±1, ±2, ±3, or ±6. The possible rational zeros \frac{p}{ q} \text{are} ±1, ±2, ±\frac{1}{2} , ±\frac{1}{3} , ±\frac{1}{6} , \text{and} ±\frac{2}{3} .
(b) Use the remainder theorem to show that 1 is a zero.
\begin{matrix} 1)\overline{\begin{matrix}6 &7 &-12& -3 &2\end{matrix} } \\ \underline{\begin{matrix}&& 6& 13 &1& -2\end{matrix} } \\ \begin{matrix} 6& 13& 1& -2& 0\end{matrix} & ←ƒ(1) = 0\end{matrix}
The 0 remainder shows that 1 is a zero. The quotient is 6x³ + 13x² + x – 2.
ƒ(x) = (x – 1)(6x³ + 13x³ + x – 2) Begin factoring ƒ(x).
Now, use the quotient polynomial and synthetic division to find that -2 is a zero.
\begin{matrix}-2) \overline{\begin{matrix} 6& 13 &1& -2\end{matrix} } \\ \underline{\begin{matrix} && -12& -2 &2\end{matrix} } \\ \begin{matrix} 6 &1 &-1&& 0\end{matrix}& ←ƒ(-2) = 0\end{matrix}
The new quotient polynomial is 6x² + x – 1. Therefore, ƒ(x) can now be completely factored as follows.
ƒ(x) = (x – 1)(x + 2)(6x² + x – 1)
ƒ(x)= (x – 1)(x + 2)(3x – 1)(2x + 1)
Setting 3x – 1 = 0 and 2x + 1 = 0 yields the zeros \frac{1}{3} \text{and} – \frac{1}{2} . In summary, the rational zeros are 1, -2, \frac{1}{3} , \text{and} – \frac{1}{2}. These zeros correspond to the x-intercepts of the graph of ƒ(x) in Figure 18. The linear factorization of ƒ(x) is as follows.
ƒ(x) = 6x^{4} + 7x³ – 12x² – 3x + 2
ƒ(x) = (x – 1)(x + 2)(3x – 1)(2x + 1)
