Question

answer the following questions about the equation below.
6x³ + 37x² - 34x + 7 = 0
(a) list all rational roots that are possible according to the rational zero theorem. choose the correct answer below.
a. ±1,±7
b. ±1,±2,±3,±6,±1/7,±2/7,±3/7,±6/7
c. ±1,±7,±1/2,±7/2,±1/3,±7/3,±1/6,±7/6
d. ±1,±2,±3,±6
(b) use synthetic division to test several possible rational roots in order to identify one actual root.
one rational root of the given equation is 6. (simplify your answer.)

Explanation:

Step1: Recall Rational Zero Theorem

The Rational Zero Theorem states that if a polynomial equation $a_nx^n + a_{n - 1}x^{n-1}+\cdots+a_1x + a_0=0$ has integer coefficients, then the possible rational zeros are of the form $\pm\frac{p}{q}$, where $p$ is a factor of the constant term $a_0$ and $q$ is a factor of the leading - coefficient $a_n$. For the equation $6x^{3}+37x^{2}-34x + 7 = 0$, $a_0 = 7$ and $a_n=6$.

Step2: Find factors of $a_0$ and $a_n$

The factors of $a_0 = 7$ are $\pm1,\pm7$. The factors of $a_n = 6$ are $\pm1,\pm2,\pm3,\pm6$.

Step3: Form possible rational zeros

The possible rational zeros are $\pm\frac{1}{1},\pm\frac{1}{2},\pm\frac{1}{3},\pm\frac{1}{6},\pm\frac{7}{1},\pm\frac{7}{2},\pm\frac{7}{3},\pm\frac{7}{6}$, which simplifies to $\pm1,\pm7,\pm\frac{1}{2},\pm\frac{7}{2},\pm\frac{1}{3},\pm\frac{7}{3},\pm\frac{1}{6},\pm\frac{7}{6}$.

Step4: Test with synthetic division (for part b)

Let's test the possible rational roots using synthetic division. If we test $x = \frac{1}{2}$:
Set up the synthetic - division with the coefficients $6,37,-34,7$.
Bring down the first coefficient $6$:
Multiply $\frac{1}{2}\times6 = 3$, add to the second coefficient: $37 + 3=40$.
Multiply $\frac{1}{2}\times40 = 20$, add to the third coefficient: $-34 + 20=-14$.
Multiply $\frac{1}{2}\times(-14)=-7$, add to the fourth coefficient: $7+( - 7)=0$.
Since the remainder is $0$, $x=\frac{1}{2}$ is a root.

Answer:

(a) C. $\pm1,\pm7,\pm\frac{1}{2},\pm\frac{7}{2},\pm\frac{1}{3},\pm\frac{7}{3},\pm\frac{1}{6},\pm\frac{7}{6}$
(b) $\frac{1}{2}$