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Part 1
Find all rational zeros of f. Then​ (if necessary) use the depressed equation to find all roots of the equation ​f(x)equals0.
​f(x)equalsxcubednegative 37xplus 6
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Part 1
Select the correct choice below and fill in the answer box within your choice.
A.
The set of all zeros of the given function is StartSet nothing EndSet
  
.
​(Simplify your answer.  Type an exact​ answer, using radicals as needed.  Use a comma to separate answers as​ needed.) 
B.
There are no real zeros.

We are given the cubic function \( f(x) = x^3 - 37x + 6 \), and we are tasked with finding all rational zeros, then solving for all roots of the equation \( f(x) = 0 \).

### Step 1: Rational Root Theorem
The **Rational Root Theorem** helps to identify possible rational roots of a polynomial. It states that any rational root, written as \( \frac{p}{q} \), must satisfy:
- \( p \) is a factor of the constant term (in this case, 6),
- \( q \) is a factor of the leading coefficient (in this case, 1).

For \( f(x) = x^3 - 37x + 6 \), the constant term is 6, and the leading coefficient is 1. The possible rational roots are the factors of 6 divided by the factors of 1. Thus, the possible rational roots are:
\[
\pm 1, \pm 2, \pm 3, \pm 6
\]

### Step 2: Testing Possible Rational Roots
We now test each of the possible rational roots using synthetic division or substitution into \( f(x) \).

#### Testing \( x = 1 \):
\[
f(1) = 1^3 - 37(1) + 6 = 1 - 37 + 6 = -30 \quad (\text{Not a root})
\]

#### Testing \( x = -1 \):
\[
f(-1) = (-1)^3 - 37(-1) + 6 = -1 + 37 + 6 = 42 \quad (\text{Not a root})
\]

#### Testing \( x = 2 \):
\[
f(2) = 2^3 - 37(2) + 6 = 8 - 74 + 6 = -60 \quad (\text{Not a root})
\]

#### Testing \( x = -2 \):
\[
f(-2) = (-2)^3 - 37(-2) + 6 = -8 + 74 + 6 = 72 \quad (\text{Not a root})
\]

#### Testing \( x = 3 \):
\[
f(3) = 3^3 - 37(3) + 6 = 27 - 111 + 6 = -78 \quad (\text{Not a root})
\]

#### Testing \( x = -3 \):
\[
f(-3) = (-3)^3 - 37(-3) + 6 = -27 + 111 + 6 = 90 \quad (\text{Not a root})
\]

#### Testing \( x = 6 \):
\[
f(6) = 6^3 - 37(6) + 6 = 216 - 222 + 6 = 0 \quad (\text{Root found!})
\]

So, \( x = 6 \) is a root.

### Step 3: Depressed Equation
Now that we know \( x = 6 \) is a root, we can factor \( (x - 6) \) from the polynomial. To do this, we perform synthetic division on \( f(x) = x^3 - 37x + 6 \) by \( (x - 6) \).

Performing synthetic division:

\[
\begin{array}{r|rrrr}
6 & 1 & 0 & -37 & 6 \\
  &   & 6 & 36 & -6 \\
\hline
  & 1 & 6 & -1 & 0 \\
\end{array}
\]

The quotient is \( x^2 + 6x - 1 \). So, we can write:
\[
f(x) = (x - 6)(x^2 + 6x - 1)
\]

### Step 4: Solving the Quadratic Equation
Next, solve \( x^2 + 6x - 1 = 0 \) using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
For \( x^2 + 6x - 1 = 0 \), we have \( a = 1 \), \( b = 6 \), and \( c = -1 \). Substituting into the quadratic formula:
\[
x = \frac{-6 \pm \sqrt{6^2 - 4(1)(-1)}}{2(1)} = \frac{-6 \pm \sqrt{36 + 4}}{2} = \frac{-6 \pm \sqrt{40}}{2} = \frac{-6 \pm 2\sqrt{10}}{2}
\]
\[
x = -3 \pm \sqrt{10}
\]

### Step 5: Final Answer
Thus, the three roots of the equation are:
\[
x = 6, \quad x = -3 + \sqrt{10}, \quad x = -3 - \sqrt{10}
\]

### Final Answer:
The set of all zeros is:
\[
\{6, -3 + \sqrt{10}, -3 - \sqrt{10}\}
\]

Would you like further details or explanations?

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Here are some related questions to deepen understanding:
1. What are the conditions for using the Rational Root Theorem?
2. How does synthetic division help in factoring polynomials?
3. What is the significance of a "depressed equation" in solving polynomials?
4. How can you verify that the solutions to the quadratic are correct?
5. What are the differences between rational and irrational roots?

**Tip:** Always verify potential roots by substituting them back into the original equation to avoid calculation errors.

Find all rational zeros of f. Then, if necessary, use the depressed equation to find all roots of the equation f(x) = 0. f(x) = x^3 - 37x + 6.

Discover how to find rational zeros and solve the cubic polynomial equation f(x) = x^3 - 37x + 6. This problem explores key algebraic concepts such as the Rational Root Theorem, synthetic division, and the quadratic formula. Suitable for students in Grades 10-12.

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