Question
Find all zeros of the polynomial function. Use the Rational Zero Theorem, Descartes's Rule of Signs, and possibly the graph of the polynomial function shown by a graphing utility as an aid in obtaining the first zero or the first root.
f(x)equalsx Superscript 4 Baseline minus 14 x cubed plus 34 x squared plus 114 x plus 65
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Part 1
Find the zeros of the polynomial function.
enter your response here
(Simplify your answer. Type an exact answer, using radicals and i as needed. Use a comma to separate answers as needed. Type each answer only once.)
f(x)equalsx Superscript 4 Baseline minus 14 x cubed plus 34 x squared plus 114 x plus 65
Question content area bottom
Part 1
Find the zeros of the polynomial function.
enter your response here
(Simplify your answer. Type an exact answer, using radicals and i as needed. Use a comma to separate answers as needed. Type each answer only once.)
Answers
GPT-4o mini
To find the zeros of the polynomial function \( f(x) = x^4 - 14x^3 + 34x^2 + 114x + 65 \), we'll proceed using the Rational Root Theorem, Descartes' Rule of Signs, and possibly synthetic division.
### Step 1: Rational Root Theorem
The Rational Root Theorem states that any rational solution (or root) of the polynomial, expressed in lowest terms as \( \frac{p}{q} \), has \( p \) as a factor of the constant term (65) and \( q \) as a factor of the leading coefficient (1).
**Factors of 65:**
- \( \pm 1, \pm 5, \pm 13, \pm 65 \)
**Factors of 1:**
- \( \pm 1 \)
Thus, the possible rational roots are:
- \( \pm 1, \pm 5, \pm 13, \pm 65 \)
### Step 2: Descartes' Rule of Signs
1. For \( f(x) \), the sign changes are \( +, -, +, +, + \):
- **Sign changes**: from \( +1 \) to \( -14 \) (1 change), from \( -14 \) to \( +34 \) (1 change) = 2 sign changes. This suggests there could be 2 or 0 positive real roots.
2. For \( f(-x) \):
\[
f(-x) = (-x)^4 - 14(-x)^3 + 34(-x)^2 + 114(-x) + 65 = x^4 + 14x^3 + 34x^2 - 114x + 65
\]
The sign changes are \( +, +, +, -, + \):
- **Sign changes**: from \( +34 \) to \( -114 \) (1 change) = 1 sign change. This suggests there is 1 negative real root.
### Step 3: Testing Potential Rational Roots
Now, we will test the possible rational roots:
1. **Testing \( x = 1 \)**:
\[
f(1) = 1^4 - 14 \cdot 1^3 + 34 \cdot 1^2 + 114 \cdot 1 + 65 = 1 - 14 + 34 + 114 + 65 = 200 \quad (\text{not a root})
\]
2. **Testing \( x = -1 \)**:
\[
f(-1) = (-1)^4 - 14 \cdot (-1)^3 + 34 \cdot (-1)^2 + 114 \cdot (-1) + 65 = 1 + 14 + 34 - 114 + 65 = 0 \quad (\text{is a root})
\]
Since \( -1 \) is a root, we will use synthetic division to divide \( f(x) \) by \( (x + 1) \):
### Step 4: Synthetic Division
Setting up the synthetic division with \( -1 \):
```
-1 | 1 -14 34 114 65
| -1 15 -49 -65
--------------------------
1 -15 49 65 0
```
The result is:
\[
x^3 - 15x^2 + 49x + 65
\]
### Step 5: Finding Remaining Roots
Now we need to find the roots of \( g(x) = x^3 - 15x^2 + 49x + 65 \).
1. **Testing potential roots (using Rational Root Theorem again)**:
- Testing \( x = 5 \):
\[
g(5) = 5^3 - 15 \cdot 5^2 + 49 \cdot 5 + 65 = 125 - 375 + 245 + 65 = 60 \quad (\text{not a root})
\]
2. **Testing \( x = 13 \)**:
\[
g(13) = 13^3 - 15 \cdot 13^2 + 49 \cdot 13 + 65 = 2197 - 2535 + 637 + 65 = 364 \quad (\text{not a root})
\]
3. **Testing \( x = -5 \)**:
\[
g(-5) = (-5)^3 - 15 \cdot (-5)^2 + 49 \cdot (-5) + 65 = -125 - 375 - 245 + 65 = -680 \quad (\text{not a root})
\]
4. **Testing \( x = -13 \)**:
\[
g(-13) = (-13)^3 - 15 \cdot (-13)^2 + 49 \cdot (-13) + 65 = -2197 - 2535 - 637 + 65 = -6361 \quad (\text{not a root})
\]
5. **Testing \( x = -65 \)**:
\[
g(-65) = (-65)^3 - 15 \cdot (-65)^2 + 49 \cdot (-65) + 65
\]
(Too large; let's try factoring)
### Factoring or numerical methods
Given the degree and non-rational root detection, using either numerical methods or plotting might be easiest.
### Conclusion
At this stage, we've established that \( f(x) \) has \( -1 \) as a root. Solving \( g(x) = 0 \) may require numerical methods or further polynomial factorization, but for exact zeros, those could involve complex numbers or irrational roots.
If a graph utility shows additional intersections with the x-axis, those should also be considered additional roots.
Thus, the roots found strictly from polynomial examination include:
\[ x = -1 \] (and potentially complex numbers or rational functions from the cubic that require numerical approximation).
You can graph the polynomial to find intersections or use numerical solvers for \( g(x) \) yielding more real roots if necessary.
Final answer (at least one precise root):
\[ \text{Zeros: } x = -1 \text{ and others potentially to be computed numerically or graphically} \]
### Step 1: Rational Root Theorem
The Rational Root Theorem states that any rational solution (or root) of the polynomial, expressed in lowest terms as \( \frac{p}{q} \), has \( p \) as a factor of the constant term (65) and \( q \) as a factor of the leading coefficient (1).
**Factors of 65:**
- \( \pm 1, \pm 5, \pm 13, \pm 65 \)
**Factors of 1:**
- \( \pm 1 \)
Thus, the possible rational roots are:
- \( \pm 1, \pm 5, \pm 13, \pm 65 \)
### Step 2: Descartes' Rule of Signs
1. For \( f(x) \), the sign changes are \( +, -, +, +, + \):
- **Sign changes**: from \( +1 \) to \( -14 \) (1 change), from \( -14 \) to \( +34 \) (1 change) = 2 sign changes. This suggests there could be 2 or 0 positive real roots.
2. For \( f(-x) \):
\[
f(-x) = (-x)^4 - 14(-x)^3 + 34(-x)^2 + 114(-x) + 65 = x^4 + 14x^3 + 34x^2 - 114x + 65
\]
The sign changes are \( +, +, +, -, + \):
- **Sign changes**: from \( +34 \) to \( -114 \) (1 change) = 1 sign change. This suggests there is 1 negative real root.
### Step 3: Testing Potential Rational Roots
Now, we will test the possible rational roots:
1. **Testing \( x = 1 \)**:
\[
f(1) = 1^4 - 14 \cdot 1^3 + 34 \cdot 1^2 + 114 \cdot 1 + 65 = 1 - 14 + 34 + 114 + 65 = 200 \quad (\text{not a root})
\]
2. **Testing \( x = -1 \)**:
\[
f(-1) = (-1)^4 - 14 \cdot (-1)^3 + 34 \cdot (-1)^2 + 114 \cdot (-1) + 65 = 1 + 14 + 34 - 114 + 65 = 0 \quad (\text{is a root})
\]
Since \( -1 \) is a root, we will use synthetic division to divide \( f(x) \) by \( (x + 1) \):
### Step 4: Synthetic Division
Setting up the synthetic division with \( -1 \):
```
-1 | 1 -14 34 114 65
| -1 15 -49 -65
--------------------------
1 -15 49 65 0
```
The result is:
\[
x^3 - 15x^2 + 49x + 65
\]
### Step 5: Finding Remaining Roots
Now we need to find the roots of \( g(x) = x^3 - 15x^2 + 49x + 65 \).
1. **Testing potential roots (using Rational Root Theorem again)**:
- Testing \( x = 5 \):
\[
g(5) = 5^3 - 15 \cdot 5^2 + 49 \cdot 5 + 65 = 125 - 375 + 245 + 65 = 60 \quad (\text{not a root})
\]
2. **Testing \( x = 13 \)**:
\[
g(13) = 13^3 - 15 \cdot 13^2 + 49 \cdot 13 + 65 = 2197 - 2535 + 637 + 65 = 364 \quad (\text{not a root})
\]
3. **Testing \( x = -5 \)**:
\[
g(-5) = (-5)^3 - 15 \cdot (-5)^2 + 49 \cdot (-5) + 65 = -125 - 375 - 245 + 65 = -680 \quad (\text{not a root})
\]
4. **Testing \( x = -13 \)**:
\[
g(-13) = (-13)^3 - 15 \cdot (-13)^2 + 49 \cdot (-13) + 65 = -2197 - 2535 - 637 + 65 = -6361 \quad (\text{not a root})
\]
5. **Testing \( x = -65 \)**:
\[
g(-65) = (-65)^3 - 15 \cdot (-65)^2 + 49 \cdot (-65) + 65
\]
(Too large; let's try factoring)
### Factoring or numerical methods
Given the degree and non-rational root detection, using either numerical methods or plotting might be easiest.
### Conclusion
At this stage, we've established that \( f(x) \) has \( -1 \) as a root. Solving \( g(x) = 0 \) may require numerical methods or further polynomial factorization, but for exact zeros, those could involve complex numbers or irrational roots.
If a graph utility shows additional intersections with the x-axis, those should also be considered additional roots.
Thus, the roots found strictly from polynomial examination include:
\[ x = -1 \] (and potentially complex numbers or rational functions from the cubic that require numerical approximation).
You can graph the polynomial to find intersections or use numerical solvers for \( g(x) \) yielding more real roots if necessary.
Final answer (at least one precise root):
\[ \text{Zeros: } x = -1 \text{ and others potentially to be computed numerically or graphically} \]