# Chapter 5 - Section 5.4 - A General Approach to Factoring - 5.4 Exercises: 69

$(x+3)(x^2+1)(x+1)(x-1)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ Group the terms of the given expression, $x^5+3x^4-x-3 ,$ such that the factored form of the groupings will result to a factor that is common to the entire expression. Then, factor the $GCF$ in each group. and factor the $GCF$ of the entire expression. Finally, factor the resulting difference of $2$ squares. $\bf{\text{Solution Details:}}$ Grouping the first and second terms and the third and fourth terms, the given expression is equivalent to \begin{array}{l}\require{cancel} (x^5+3x^4)-(x+3) .\end{array} Factoring the $GCF$ in each group results to \begin{array}{l}\require{cancel} x^4(x+3)-(x+3) .\end{array} Factoring the $GCF= (x+3)$ of the entire expression above results to \begin{array}{l}\require{cancel} (x+3)(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} (x+3)[(x^2)^2-(1)^2] \\\\= (x+3)[(x^2+1)(x^2-1)] \\\\= (x+3)(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} (x+3)(x^2+1)[(x)^2-(1)^2] \\\\= (x+3)(x^2+1)[(x+1)(x-1)] \\\\= (x+3)(x^2+1)(x+1)(x-1) .\end{array}

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