Ind the zeros and their multiplicities. Consider using Descartes' rule of signs and the upper and lower bound theorem to limit your search for rational zerd $$ f(x)=14 x^{3}-55 x^{2}+42 x-9 $$ Part: $$ 0 / 2 $$ Part 1 of 2 Select "None" if applicable. The zeros of $$ f(x) $$ : $$ \square $$

Question

UpStudy AI · Accepted Answer

To find the zeros of the polynomial $$ f(x) = 14x^3 - 55x^2 + 42x - 9 $$ and determine their multiplicities, we'll follow these steps:

1. **Rational Root Theorem:**
   - **Possible Rational Zeros:** The Rational Root Theorem suggests that any rational zero, expressed as $$\frac{p}{q}$$, satisfies:
     - $$ p $$ is a factor of the constant term ($$-9$$): $$ \pm1, \pm3, \pm9 $$.
     - $$ q $$ is a factor of the leading coefficient ($$14$$): $$ \pm1, \pm2, \pm7, \pm14 $$.
   - **Potential Zeros:** Combining these, the possible rational zeros are:
     $$
     \pm1, \pm\frac{1}{2}, \pm3, \pm\frac{3}{2}, \pm9, \pm\frac{9}{2}, \pm\frac{1}{7}, \pm\frac{3}{7}, \pm\frac{9}{7}, \dots
     $$

2. **Testing Possible Zeros:**
   - **Test $$ x = 3 $$:**
     $$
     f(3) = 14(3)^3 - 55(3)^2 + 42(3) - 9 = 378 - 495 + 126 - 9 = 0
     $$
     - **Conclusion:** $$ x = 3 $$ is a zero.
   - **Factor Out $$ (x - 3) $$:**
     - Perform synthetic division or polynomial long division to factor $$ f(x) $$:
       $$
       f(x) = (x - 3)(14x^2 - 13x + 3)
       $$
   
3. **Finding Remaining Zeros:**
   - **Quadratic Factor:** Solve $$ 14x^2 - 13x + 3 = 0 $$ using the quadratic formula:
     $$
     x = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 14 \cdot 3}}{2 \cdot 14} = \frac{13 \pm \sqrt{169 - 168}}{28} = \frac{13 \pm 1}{28}
     $$
     - **Solutions:**
       $$
       x = \frac{14}{28} = \frac{1}{2} \quad \text{and} \quad x = \frac{12}{28} = \frac{3}{7}
       $$
   - **Conclusion:** The remaining zeros are $$ x = \frac{1}{2} $$ and $$ x = \frac{3}{7} $$.

4. **Multiplicity of Zeros:**
   - Each zero $$ x = 3 $$, $$ x = \frac{1}{2} $$, and $$ x = \frac{3}{7} $$ appears only once in the factorization.
   - **Multiplicity:** All zeros have **multiplicity 1**.

5. **Summary of Zeros:**
   $$
   \begin{align*}
   x &= 3 \quad \text{(Multiplicity 1)} \
   x &= \frac{1}{2} \quad \text{(Multiplicity 1)} \
   x &= \frac{3}{7} \quad \text{(Multiplicity 1)}
   \end{align*}
   $$

**Final Answer:**

The zeros of $$ f(x) $$ are:
- $$ x = 3 $$ with multiplicity 1,
- $$ x = \frac{1}{2} $$ with multiplicity 1, and
- $$ x = \frac{3}{7} $$ with multiplicity 1.
The zeros of $$ f(x) $$ are: - $$ x = 3 $$ (multiplicity 1) - $$ x = \frac{1}{2} $$ (multiplicity 1) - $$ x = \frac{3}{7} $$ (multiplicity 1)

Ind the zeros and their multiplicities. Consider using Descartes' rule of signs and the upper and lower bound theorem to limit your search for rational zerd \( f(x)=14 x^{3}-55 x^{2}+42 x-9 \) Part: \( 0 / 2 \) Part 1 of 2 Select "None" if applicable. The zeros of \( f(x) \) : \( \square \)

Real Tutor Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Algebra Questions