## Intermediate Algebra: Connecting Concepts through Application

$(x+1)(x-1)(x^2-5)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $x^4-6x^2+5 ,$ find two numbers, $m_1$ and $m_2,$ whose product is $c$ and whose sum is $b$ in the quadratic expression $x^2+bx+c.$ Express the factored form as $(x+m_1)(x+m_2).$ Then use the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ In the trinomial expression above, the value of $c$ is $5$ and the value of $b$ is $-6 .$ The possible pairs of integers whose product is $c$ are \begin{array}{l}\require{cancel} \{ 1,5 \}, \{ -1,-5 \} .\end{array} Among these pairs, the one that gives a sum of $b$ is $\{ -1,-5 \}.$ Hence, the factored form of the expression above is \begin{array}{l}\require{cancel} (x^2-1)(x^2-5) .\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)^2-(1)^2](x^2-5) \\\\= (x+1)(x-1)(x^2-5) .\end{array}