Find all zeros of f ( x )

To find the zeros of the cubic function $f(x) = x^3 + 7x^2 - 4x - 78$, we need to solve for $x$ when $f(x) = 0$.

Step 1: Use Rational Root Theorem to find potential rational roots

The Rational Root Theorem suggests that the potential rational roots of the equation are the factors of the constant term divided by the factors of the leading coefficient.

• Constant term: $-78$
• Leading coefficient: $1$

So, the possible rational roots are the factors of $-78$, which are: $\pm 1, \pm 2, \pm 3, \pm 6, \pm 13, \pm 26, \pm 39, \pm 78$

Step 2: Test the potential roots

We will test these possible values using synthetic division or direct substitution.

Testing $x = 2$:

$f(2) = 2^3 + 7(2^2) - 4(2) - 78 = 8 + 28 - 8 - 78 = -50 \quad (\text{Not a root})$

Testing $x = -3$:

$f(-3) = (-3)^3 + 7(-3)^2 - 4(-3) - 78 = -27 + 63 + 12 - 78 = -30 \quad (\text{Not a root})$

Testing $x = 3$:

$f(3) = 3^3 + 7(3^2) - 4(3) - 78 = 27 + 63 - 12 - 78 = 0$ Thus, $x = 3$ is a root.

Step 3: Polynomial division

Now that we know $x = 3$ is a root, we can divide $f(x)$ by $x - 3$ using synthetic division or polynomial long division. This gives:

$f(x) = (x - 3)(x^2 + 10x + 26)$

Step 4: Solve the quadratic equation

We now need to solve $x^2 + 10x + 26 = 0$. Using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 10$, and $c = 26$:

$x = \frac{-10 \pm \sqrt{10^2 - 4(1)(26)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 104}}{2} = \frac{-10 \pm \sqrt{-4}}{2}$

$x = \frac{-10 \pm 2i}{2} = -5 \pm i$

Step 5: Final solution

The zeros of the function are: $x = 3, -5 + i, -5 - i$

Thus, the zeros of $f(x)$ are $3, -5 + i, -5 - i$.

Would you like more details or have any questions? Here are 5 related questions you might explore:

1. How does synthetic division work for polynomials?
2. What is the Rational Root Theorem and when is it used?
3. How do you solve complex roots from a quadratic equation?
4. How can you graph functions with complex roots?
5. What are some examples of functions with purely real roots?

Tip: When dealing with cubic polynomials, identifying one real root simplifies the process of solving the equation.