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Find the value of k for the polynomial g(x)= 2x^3+4x^2-3x+k so that x+3 is a factor of the polynomial g. Example of how to solve attached.

Find the value of k for the polynomial g(x)= 2x^3+4x^2-3x+k so that x+3 is a factor of the polynomial g. Example of how to solve attached.

Algebra and Trigonometry (6th Edition)
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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Find the value of k for the polynomial g(x)= 2x^3+4x^2-3x+k so that x+3 is a factor of the polynomial g. Example of how to solve attached.

Given \( f(x) = 8x^3 - 5x + k \). Determine the value of \( k \) so that \( x+5 \) is a factor of the polynomial \( f \). ### Explanation To find when \( x+5 \) is a factor, perform synthetic division on \( f(x) \) using \( -5 \). 1. **Setup for synthetic division:** - Coefficients: \( 8, 0, -5, k \) - Divisor: \( -5 \) 2. **Synthetic Division Process:** - **Bring down the first coefficient:** - \( 8 \) - **Multiply and add down:** - \( -5 \times 8 = -40 \) - Add: \( 0 + (-40) = -40 \) - Repeat the process: - \( -5 \times (-40) = 200 \) - Add: \( -5 + 200 = 195 \) - Final step: - \( -5 \times 195 = -975 \) - Add: \( k + (-975) = 0 \) 3. **Conclusion:** - For \( x+5 \) to be a factor, the remainder must be 0. - Thus, \( k = 975 \). Therefore, \( k = 975 \) for \( x+5 \) to be a factor of the polynomial. ### Visual Notation: - The divisor \(-5\) is used on the left. - Each row performs multiplication of the divisor and addition with the numbers below the line. - The final sum is ensured to be zero to verify the factor condition. The process shows that \( k \) must equal 975 to achieve a remainder of 0.
Transcribed Image Text:Given \( f(x) = 8x^3 - 5x + k \). Determine the value of \( k \) so that \( x+5 \) is a factor of the polynomial \( f \). ### Explanation To find when \( x+5 \) is a factor, perform synthetic division on \( f(x) \) using \( -5 \). 1. **Setup for synthetic division:** - Coefficients: \( 8, 0, -5, k \) - Divisor: \( -5 \) 2. **Synthetic Division Process:** - **Bring down the first coefficient:** - \( 8 \) - **Multiply and add down:** - \( -5 \times 8 = -40 \) - Add: \( 0 + (-40) = -40 \) - Repeat the process: - \( -5 \times (-40) = 200 \) - Add: \( -5 + 200 = 195 \) - Final step: - \( -5 \times 195 = -975 \) - Add: \( k + (-975) = 0 \) 3. **Conclusion:** - For \( x+5 \) to be a factor, the remainder must be 0. - Thus, \( k = 975 \). Therefore, \( k = 975 \) for \( x+5 \) to be a factor of the polynomial. ### Visual Notation: - The divisor \(-5\) is used on the left. - Each row performs multiplication of the divisor and addition with the numbers below the line. - The final sum is ensured to be zero to verify the factor condition. The process shows that \( k \) must equal 975 to achieve a remainder of 0.
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