Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 1 - Systems of Linear Equations - 1.3 Applications of Systems of Linear Equations - 1.3 Exercises - Page 32: 7

Answer

$$ p(x)=-6-3x+x^2- x^3+x^4. $$
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Work Step by Step

Suppose that $$ p(x)=a_{0}+a_{1} x+a_{2} x^{2}+a_3x^3+a_4x^4. $$ Now, we have $$ \begin{array}{l} {p(-2)=a_{0}+a_{1}(-2)+a_{2}(-2)^{2}+a_3(-2)^3+a_4(-2)^4=a_{0}-2a_1+4a_2-8a_3+16a_4=28} \\ {p(-1)=a_{0}+a_{1}(-1)+a_{2}(-1)^{2}+a_3(-1)^3+a_4(-)^4=a_{0}-a_1+a_2-a_3 +a_4=0} \\ {p(0)=a_{0}+a_{1}(0)+a_{2}(0)^{2}+a_3(0)^3+a_4(0)^4=a_{0} =-6}\\ {p(1)=a_{0}+a_{1}(1)+a_{2}(1)^{2}+a_3(1)^3+a_4(1)^4=a_{0}+ a_{1}+ a_{2}+ a_3+a_4=-8}\\ {p(2)=a_{0}+a_{1}(2)+a_{2}(2)^{2}+a_3(2)^3+a_4(2)^4=a_{0}+2 a_{1}+4 a_{2}+8a_3+16a_4=0}\\ \end{array} $$ The above system has the solution $$a_{0}=-6, \quad a_{1}=-3, \quad a_{2}=1, \quad a_{3}=-1, \quad a_{4}=1.$$ Hence, $$ p(x)=-6-3x+x^2- x^3+x^4. $$
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