find zeroes of function and multiplicities 0=x^3+4x^2-36x-144

To find the zeroes and their multiplicities for the function given by the equation

\[
x^3 + 4x^2 - 36x - 144 = 0,
\]

we can follow a systematic approach to find all the roots and determine their multiplicities.

### Step-by-Step Solution:

1. **Identify possible rational roots:**
   According to the Rational Root Theorem, any rational root, if it exists, must be a factor of the constant term (-144) divided by the factor of the leading coefficient (1). The factors of \(-144\) are:

   \[
   \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 8, \pm 9, \pm 12, \pm 16, \pm 18, \pm 24, \pm 36, \pm 48, \pm 72, \pm 144.
   \]

   Since the leading coefficient is \(1\), the possible rational roots are these factors.

2. **Test the possible roots:**
   Let's try testing a few possible roots using substitution or synthetic division. 

   - **Test \(x = 4\):**

     Substitute \(x = 4\) into the polynomial:

     \[
     4^3 + 4(4^2) - 36(4) - 144 = 64 + 64 - 144 - 144 = 0.
     \]

     Since the equation equals zero, \(x = 4\) is a root.

3. **Use synthetic division to factor the polynomial:**
   Now that we know \(x = 4\) is a root, we can use synthetic division to divide the polynomial by \(x - 4\) to find the other factors.

   Performing synthetic division with \(x = 4\):

   \[
   \begin{array}{r|rrrr}
   4 & 1 & 4 & -36 & -144 \\
     &   & 4 & 32  & -16  \\
   \hline
     & 1 & 8 & -4  & 0
   \end{array}
   \]

   The quotient is \(x^2 + 8x - 4\), and the remainder is \(0\). This tells us:

   \[
   x^3 + 4x^2 - 36x - 144 = (x - 4)(x^2 + 8x - 4).
   \]

4. **Find the roots of the quadratic equation:**
   The quadratic equation \(x^2 + 8x - 4 = 0\) can be solved using the quadratic formula:

   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.
   \]

   Here, \(a = 1\), \(b = 8\), and \(c = -4\). Plugging in these values gives:

   \[
   x = \frac{-8 \pm \sqrt{8^2 - 4 \times 1 \times (-4)}}{2 \times 1} = \frac{-8 \pm \sqrt{64 + 16}}{2} = \frac{-8 \pm \sqrt{80}}{2}.
   \]

   Simplify the square root:

   \[
   \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5}.
   \]

   Thus, the roots are:

   \[
   x = \frac{-8 + 4\sqrt{5}}{2} = -4 + 2\sqrt{5}, \quad x = \frac{-8 - 4\sqrt{5}}{2} = -4 - 2\sqrt{5}.
   \]

5. **Determine the multiplicities:**
   - The root \(x = 4\) appeared once during synthetic division, so its multiplicity is 1.
   - The roots \(x = -4 + 2\sqrt{5}\) and \(x = -4 - 2\sqrt{5}\) are each distinct roots of the quadratic equation, and both have multiplicity 1.

### Final Answer:

The zeroes of the function \(x^3 + 4x^2 - 36x - 144\) are:

- \(x = 4\) with multiplicity **1**,
- \(x = -4 + 2\sqrt{5}\) with multiplicity **1**,
- \(x = -4 - 2\sqrt{5}\) with multiplicity **1**.

Would you like to go over any of the steps in more detail, or do you have any questions?

Here are 5 related questions you might find helpful:

1. How can you determine if a polynomial has a root with multiplicity greater than 1?
2. What is the difference between a root and a zero of a function?
3. How would you factor a polynomial of degree 4?
4. Can the Rational Root Theorem be used to find roots for non-integer coefficients?
5. What are the applications of finding roots and their multiplicities in real-world scenarios?

**Tip:** When solving polynomial equations, finding one root can often simplify the problem significantly, allowing you to factor and solve for the remaining roots more easily.

Discover how to find the zeroes and their multiplicities for the polynomial x^3 + 4x^2 - 36x - 144. Follow a step-by-step solution including identifying rational roots, synthetic division, solving quadratic equations, and determining multiplicities. Suitable for students in Grades 10-12.

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