Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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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. \( p(x) = 2x^5 + x^4 - 2x - 1; \, x = -\frac{1}{2} \)
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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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