Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Appendix C - Solving Polynomial Equations - Exercise Set C - Page A33: 15

Answer

$$p\left( x \right) = {\left( {x + 1} \right)^2}\left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right)$$

Work Step by Step

$$\eqalign{ & p\left( x \right) = {x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18 \cr & {\text{The constant term is }} - 18{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18} \right) \cr & {\text{Evaluating the divisors}} \cr & p\left( { - 1} \right) = {\left( { - 1} \right)^5} + 4{\left( { - 1} \right)^4} - 4{\left( { - 1} \right)^3} - 34{\left( { - 1} \right)^2} - 45\left( { - 1} \right) - 18 \cr & p\left( { - 1} \right) = - 1 + 4 + 4 - 34 + 45 - 18 \cr & p\left( { - 1} \right) = 0 \cr & \cr & p\left( x \right) = 0{\text{ for }}x = - 1,{\text{ then }}\left( {x + 1} \right){\text{ is a factor of }}p\left( x \right) \cr & p\left( x \right) = {x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18 = \left( {x + 1} \right)q\left( x \right) \cr & {\text{By the long division}}\,\,{\text{ }}x + 1\left){\vphantom{1{{x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^5} + 4{x^4} - 4{x^3} - 34{x^2} - 45x - 18}}}{\text{ we obtain}} \cr & p\left( x \right) = \left( {x + 1} \right)\left( {{x^4} + 3{x^3} - 7{x^2} - 27x - 18} \right) \cr & {\text{with }}q\left( x \right) = {x^4} + 3{x^3} - 7{x^2} - 27x - 18 \cr & \cr & {\text{The constant term is }} - 18{\text{ }}\left( {{\text{which has divisors }} \pm 1, \pm 2, \pm 3, \pm 6, \pm 9, \pm 18} \right) \cr & {\text{Evaluating the divisors}} \cr & q\left( { - 1} \right) = {\left( { - 1} \right)^4} + 3{\left( { - 1} \right)^3} - 7{\left( { - 1} \right)^2} - 27\left( { - 1} \right) - 18 \cr & q\left( { - 1} \right) = 0 \cr & \cr & q\left( x \right) = 0{\text{ for }}x = - 1,{\text{ then }}\left( {x + 1} \right){\text{ is a factor of }}q\left( x \right) \cr & q\left( x \right) = {x^4} + 3{x^3} - 7{x^2} - 27x - 18 = \left( {x + 1} \right)t\left( x \right) \cr & {\text{By the long division}}\,\,{\text{ }}x + 1\left){\vphantom{1{{x^4} + 3{x^3} - 7{x^2} - 27x - 18}}}\right. \!\!\!\!\overline{\,\,\,\vphantom 1{{{x^4} + 3{x^3} - 7{x^2} - 27x - 18}}}{\text{ we obtain}} \cr & q\left( x \right) = \left( {x + 1} \right)\left( {{x^3} + 2{x^2} - 9x - 18} \right) \cr & {\text{with }}t\left( x \right) = {x^3} + 2{x^2} - 9x - 18 \cr & \cr & {\text{Thus}}{\text{,}} \cr & p\left( x \right) = \left( {x + 1} \right)\left( {x + 1} \right)\left( {{x^3} + 2{x^2} - 9x - 18} \right) \cr & {\text{Factoring }}{x^3} + 2{x^2} - 9x - 18{\text{ by grouping terms}} \cr & {x^3} + 2{x^2} - 9x - 18 = {x^2}\left( {x + 2} \right) - 9\left( {x + 2} \right) \cr & {x^3} + 2{x^2} - 9x - 18 = \left( {x + 2} \right)\left( {{x^2} - 9} \right) \cr & {x^3} + 2{x^2} - 9x - 18 = \left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right) \cr & \cr & {\text{Thus}}{\text{,}} \cr & p\left( x \right) = \left( {x + 1} \right)\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right) \cr & p\left( x \right) = {\left( {x + 1} \right)^2}\left( {x + 2} \right)\left( {x + 3} \right)\left( {x - 3} \right) \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.