Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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![**Problem Statement:**
Find the value of \( k \) that makes the divisor \( x - 11 \) a factor of the polynomial \( x^3 - 13x^2 + 40x + k \).
**Solution Explanation:**
To determine the value of \( k \) for which \( x - 11 \) is a factor of the polynomial, we use the Factor Theorem. This theorem states that if \( x - c \) is a factor of the polynomial \( f(x) \), then \( f(c) = 0 \).
Here, \( c = 11 \). Thus, we need to evaluate the polynomial at \( x = 11 \) and set it equal to zero:
\[ f(x) = x^3 - 13x^2 + 40x + k \]
\[ f(11) = (11)^3 - 13(11)^2 + 40(11) + k = 0 \]
Calculate:
\[ (11)^3 = 1331 \]
\[ 13(11)^2 = 1573 \]
\[ 40(11) = 440 \]
Substitute back:
\[ 1331 - 1573 + 440 + k = 0 \]
Simplify:
\[ 198 + k = 0 \]
Solve for \( k \):
\[ k = -198 \]
Thus, the value of \( k \) that makes \( x - 11 \) a factor of the polynomial is \( k = -198 \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd80d3a39-76f0-4817-b68e-1bece675673e%2Fee583fb9-cc63-49dc-9d93-7f2c37586862%2Fsei0zvi_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Find the value of \( k \) that makes the divisor \( x - 11 \) a factor of the polynomial \( x^3 - 13x^2 + 40x + k \).
**Solution Explanation:**
To determine the value of \( k \) for which \( x - 11 \) is a factor of the polynomial, we use the Factor Theorem. This theorem states that if \( x - c \) is a factor of the polynomial \( f(x) \), then \( f(c) = 0 \).
Here, \( c = 11 \). Thus, we need to evaluate the polynomial at \( x = 11 \) and set it equal to zero:
\[ f(x) = x^3 - 13x^2 + 40x + k \]
\[ f(11) = (11)^3 - 13(11)^2 + 40(11) + k = 0 \]
Calculate:
\[ (11)^3 = 1331 \]
\[ 13(11)^2 = 1573 \]
\[ 40(11) = 440 \]
Substitute back:
\[ 1331 - 1573 + 440 + k = 0 \]
Simplify:
\[ 198 + k = 0 \]
Solve for \( k \):
\[ k = -198 \]
Thus, the value of \( k \) that makes \( x - 11 \) a factor of the polynomial is \( k = -198 \).
![**Problem Statement:**
Given the polynomial function \( f(x) = 8x^3 - 5x + k \), determine the value of \( k \) so that \( x + 5 \) is a factor of the polynomial \( f \).
**Solution:**
1. **Synthetic Division Setup:**
- The goal is to find \( k \) such that \( x + 5 \) (which corresponds to \( x = -5 \)) is a factor. This means the remainder should be zero when dividing the polynomial by \( x + 5 \).
2. **Synthetic Division Process:**
- Coefficients of the polynomial: \( 8, 0, -5, k \).
- Use \( -5 \) for synthetic division.
\[
\begin{array}{r|rrrr}
-5 & 8 & 0 & -5 & k \\
& & -40 & 200 & -975 \\
\hline
& 8 & -40 & 195 & 0 \\
\end{array}
\]
3. **Explanation of Steps:**
- Write the leading coefficient (8) below the line.
- Multiply -5 (the value used for division) by 8 (the number just written below the line). Write the result, -40, above the line under the next coefficient.
- Add the result to the next coefficient: 0 + (-40) = -40.
- Repeat this process:
- Multiply -5 by -40 to get 200.
- Add to get 195: \( -5 + 200 = 195 \).
- Multiply -5 by 195 to get -975 and add to \( k \).
4. **Result:**
- For the remainder to be zero, \( k \) must be 975.
- Thus, \( k = 975 \) ensures that \( x + 5 \) is a factor of the polynomial.
**Conclusion:**
The required value for \( k \) is 975, ensuring \( x + 5 \) is a factor of \( f(x) \).
**Reference:**
Video explanation for this solution is available [here](https://www.loom.com/share/fc798de56eed4c99905fa6797d62f6e6).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd80d3a39-76f0-4817-b68e-1bece675673e%2Fee583fb9-cc63-49dc-9d93-7f2c37586862%2Fbpxwfv_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Given the polynomial function \( f(x) = 8x^3 - 5x + k \), determine the value of \( k \) so that \( x + 5 \) is a factor of the polynomial \( f \).
**Solution:**
1. **Synthetic Division Setup:**
- The goal is to find \( k \) such that \( x + 5 \) (which corresponds to \( x = -5 \)) is a factor. This means the remainder should be zero when dividing the polynomial by \( x + 5 \).
2. **Synthetic Division Process:**
- Coefficients of the polynomial: \( 8, 0, -5, k \).
- Use \( -5 \) for synthetic division.
\[
\begin{array}{r|rrrr}
-5 & 8 & 0 & -5 & k \\
& & -40 & 200 & -975 \\
\hline
& 8 & -40 & 195 & 0 \\
\end{array}
\]
3. **Explanation of Steps:**
- Write the leading coefficient (8) below the line.
- Multiply -5 (the value used for division) by 8 (the number just written below the line). Write the result, -40, above the line under the next coefficient.
- Add the result to the next coefficient: 0 + (-40) = -40.
- Repeat this process:
- Multiply -5 by -40 to get 200.
- Add to get 195: \( -5 + 200 = 195 \).
- Multiply -5 by 195 to get -975 and add to \( k \).
4. **Result:**
- For the remainder to be zero, \( k \) must be 975.
- Thus, \( k = 975 \) ensures that \( x + 5 \) is a factor of the polynomial.
**Conclusion:**
The required value for \( k \) is 975, ensuring \( x + 5 \) is a factor of \( f(x) \).
**Reference:**
Video explanation for this solution is available [here](https://www.loom.com/share/fc798de56eed4c99905fa6797d62f6e6).
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