In Exercises 11-18, show that the given value of x is a zero of the polynomial. Use the zero to completely factor the polynomial. 11. p(x)=x³5x² + 8x - 4; x = 2 13. p(x) = -x x³ + 18x² + 16x32; x = 1 14. p(x) = 2x³ 15. p(x) = 3x³ 2x² + 3x - 2; x 17. p(x) = 3x³ + x² + 24x + 8; x = 2/3 12. p(x)= x³7x+6; x = 2 1 3 11x² + 17x6; x = 16. p(x) = 2x³x² + 6x = 3; x = 18. p(x) = 2x5 + x² - 2x - 1; x = 17 N 1|2

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,...
Question

number 15

In Exercises 11–18, show that the given value of \( x \) is a zero of the polynomial. Use the zero to completely factor the polynomial.

11. \( p(x) = x^3 - 5x^2 + 8x - 4; \, x = 2 \)

12. \( p(x) = x^3 - 7x + 6; \, x = 2 \)

13. \( p(x) = -x^4 - x^3 + 18x^2 + 16x - 32; \, x = 1 \)

14. \( p(x) = 2x^3 - 11x^2 + 17x - 6; \, x = \frac{1}{2} \)

15. \( p(x) = 3x^3 - 2x^2 + 3x - 2; \, x = \frac{2}{3} \)

16. \( p(x) = 2x^3 - x^2 + 6x - 3; \, x = \frac{1}{2} \)

17. \( p(x) = 3x^3 + x^2 + 24x + 8; \, x = -\frac{1}{3} \)

18. \( p(x) = 2x^5 + x^4 - 2x - 1; \, x = -\frac{1}{2} \)
Transcribed Image Text:In Exercises 11–18, show that the given value of \( x \) is a zero of the polynomial. Use the zero to completely factor the polynomial. 11. \( p(x) = x^3 - 5x^2 + 8x - 4; \, x = 2 \) 12. \( p(x) = x^3 - 7x + 6; \, x = 2 \) 13. \( p(x) = -x^4 - x^3 + 18x^2 + 16x - 32; \, x = 1 \) 14. \( p(x) = 2x^3 - 11x^2 + 17x - 6; \, x = \frac{1}{2} \) 15. \( p(x) = 3x^3 - 2x^2 + 3x - 2; \, x = \frac{2}{3} \) 16. \( p(x) = 2x^3 - x^2 + 6x - 3; \, x = \frac{1}{2} \) 17. \( p(x) = 3x^3 + x^2 + 24x + 8; \, x = -\frac{1}{3} \) 18. \( p(x) = 2x^5 + x^4 - 2x - 1; \, x = -\frac{1}{2} \)
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