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:

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? --- 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.