Question

answer the following questions about the equation below.
$x^{3}-37x + 6=0$
(a) list all rational roots that are possible according to the rational zero theorem. choose the correct answer below.
a. 1,2,3,6
b. ±1
c. ±1,±2,±3,±6
d. ±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.)
(c) use the root from part (b) to solve the equation.
the solution set is {}.
(simplify your answer. type an exact answer, using radicals as needed. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed.)

Explanation:

Step1: Recall factor - theorem after finding a root

Since \(x = 6\) is a root of \(x^{3}-37x + 6=0\), then \((x - 6)\) is a factor of \(x^{3}-37x + 6\). We use polynomial long - division or synthetic division to divide \(x^{3}-37x + 6\) by \(x - 6\).
Using synthetic division:
The coefficients of the polynomial \(x^{3}-37x + 6\) are \(1,0,-37,6\).
Dividing by \(x-6\) (where the divisor in synthetic division is \(6\)):
Bring down the first coefficient \(1\):
Multiply \(6\times1 = 6\), add to the second coefficient \(0+6 = 6\).
Multiply \(6\times6=36\), add to the third coefficient \(-37 + 36=-1\).
Multiply \(6\times(-1)=-6\), add to the fourth coefficient \(6+( - 6)=0\).
The quotient is \(x^{2}+6x - 1\).
So, \(x^{3}-37x + 6=(x - 6)(x^{2}+6x - 1)\).

Step2: Solve the quadratic equation

To find the remaining roots, we solve the quadratic equation \(x^{2}+6x - 1=0\) using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\).
Here, \(a = 1\), \(b = 6\), \(c=-1\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(6)^{2}-4\times1\times(-1)=36 + 4=40\).
Then \(x=\frac{-6\pm\sqrt{40}}{2}=\frac{-6\pm2\sqrt{10}}{2}=-3\pm\sqrt{10}\).
Combining with the root \(x = 6\) from part (b), the solution set of the equation \(x^{3}-37x + 6=0\) is \(\{6,-3+\sqrt{10},-3 - \sqrt{10}\}\).

Answer:

\(6,-3+\sqrt{10},-3 - \sqrt{10}\)