Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter R - Elementary Algebra Review - R.5 Polynomials and Factoring - R.5 Exercise Set: 24

Answer

$(t+p)(t-p)(t^2-tp+p^2)(t+tp+p^2)$

Work Step by Step

The expressions $ t^6 $ and $ p^6 $ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $ t^6-p^6 ,$ is a difference of $2$ squares. Using the factoring of the difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the expression above is equivalent to \begin{array}{l}\require{cancel} (t^3)^2-(p^3)^2 \\\\= (t^3+p^3)(t^3-p^3) .\end{array} The expressions $ t^3 $ and $ p^3 $ are both perfect cubes (the cube root is exact). Hence, $ t^3+p^3 $ is a $\text{ sum }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (t^3+p^3)(t^3-p^3) \\\\= [(t)^3+(p)^3](t^3-p^3) \\\\= (t+p)[(t)^2-t(p)+(p)^2](t^3-p^3) \\\\= (t+p)(t^2-tp+p^2)(t^3-p^3) .\end{array} The expressions $ t^3 $ and $ p^3 $ are both perfect cubes (the cube root is exact). Hence, $ t^3-p^3 $ is a $\text{ difference }$ of $2$ cubes. Using the factoring of the sum or difference of $2$ cubes which is given by $a^3+b^3=(a+b)(a^2-ab+b^2)$ or by $a^3-b^3=(a-b)(a^2+ab+b^2)$ the expression above is equivalent to \begin{array}{l}\require{cancel} (t+p)(t^2-tp+p^2)(t^3-p^3) \\\\= (t+p)(t^2-tp+p^2)[(t)^3-(p)^3] \\\\= (t+p)(t^2-tp+p^2)(t-p)[(t)^2+t(p)+(p)^2] \\\\= (t+p)(t^2-tp+p^2)(t-p)(t+tp+p^2) \\\\= (t+p)(t-p)(t^2-tp+p^2)(t+tp+p^2) .\end{array}
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