College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter R - Section R.4 - Factoring Polynomials - R.4 Exercises - Page 39: 78

Answer

$\text{The factored form of the given expression is }$$\displaystyle \left( x^{4} +x^{2} +1\right)\left( x^{4} -x^{2} +1\right).$

Work Step by Step

$\displaystyle \begin{array}{{>{\displaystyle}l}} \begin{array}{ c l l } x^{8} +x^{4} +1 & =x^{8} +2x^{4} +1-x^{4} & \mathrm{Add} \ x^{4} \text{ to the second term and subtract } x^4. \\ & & \\ & =\left( x^{8} +2x^{4} +1\right) -x^{4} & \begin{array}{{>{\displaystyle}l}} \mathrm{Apply\ associative\ property}\\ ( a+b) +c=a+( b+c) \end{array}\\ & & \\ & =\left( x^{4} +1\right)^{2} -\left( x^{2}\right)^{2} & \begin{array}{{>{\displaystyle}l}} \mathrm{Apply\ special\ products\ property}\\ \left( a^{2} +2ab+b^{2}\right) =( a+b)^{2} \end{array}\\ & & \\ & =\left( x^{4} +1+x^{2}\right)\left( x^{4} +1-x^{2}\right) & \begin{array}{{>{\displaystyle}l}} \mathrm{Apply\ the\ difference\ of\ squares\ property}\\ a^{2} +b^{2} =( a+b)( a-b) \end{array}\\ & & \\ & =\left( x^{4} +x^{2} +1\right)\left( x^{4} -x^{2} +1\right) & \begin{array}{{>{\displaystyle}l}} \mathrm{Apply\ the\ commuative\ property}\\ a+b=b+a \end{array} \end{array} \end{array}$
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