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

$10(x^2+1)(x+1)(x-1)$
Factoring the $GCF= 10 ,$ the given expression is equivalent to \begin{array}{l}\require{cancel} 10x^4-10 \\\\= 10(x^4-1) .\end{array} The expressions $x^4$ and $1$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^4-1 ,$ 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} 10(x^4-1) \\\\= 10[(x^2)^2-(1)^2] \\\\= 10(x^2+1)(x^2-1) .\end{array} The expressions $x^2$ and $1$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $x^2-1 ,$ 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} 10(x^2+1)(x^2-1) \\\\= 10(x^2+1)[(x)^2-(1)^2] \\\\= 10(x^2+1)[(x+1)(x-1)] \\\\= 10(x^2+1)(x+1)(x-1) .\end{array}