## Intermediate Algebra (12th Edition)

$(1+x^{8})(1+x^{4})(1+x^{2})(1+x)(1-x)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $1-x^{16} ,$ use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ The expressions $1$ and $x^{16}$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $1-x^{16} ,$ 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} (1)^2-(x^{8})^2 \\\\= (1+x^{8})(1-x^{8}) .\end{array} The expressions $1$ and $x^{8}$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $1-x^{8} ,$ 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} (1+x^{8})[(1)^2-(x^{4})^2] \\\\= (1+x^{8})[(1+x^{4})(1-x^{4})] \\\\= (1+x^{8})(1+x^{4})(1-x^{4}) .\end{array} The expressions $1$ and $x^{4}$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $1-x^{4} ,$ 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} (1+x^{8})(1+x^{4})[(1)^2-(x^{2})^2] \\\\ (1+x^{8})(1+x^{4})[(1+x^{2})(1-x^{2})] \\\\ (1+x^{8})(1+x^{4})(1+x^{2})(1-x^{2}) .\end{array} The expressions $1$ and $x^{2}$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $1-x^{2} ,$ 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} (1+x^{8})(1+x^{4})(1+x^{2})[(1)^2-(x)^{2}] \\\\= (1+x^{8})(1+x^{4})(1+x^{2})[(1+x)(1-x)] \\\\= (1+x^{8})(1+x^{4})(1+x^{2})(1+x)(1-x) .\end{array}