#### Answer

$(x+1)(x-1)(x^2-5)$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$
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}