## Intermediate Algebra (12th Edition)

$(z+2)(z-2)(z^2-5)$
$\bf{\text{Solution Outline:}}$ To factor the given expression, $z^4-9z^2+20 ,$ use the factoring of trinomials in the form $x^2+bx+c.$ Then factor further the resulting expression using the factoring of the difference of $2$ squares. $\bf{\text{Solution Details:}}$ In the trinomial expression above, $b= -9 ,\text{ and } c= 20 .$ Using the factoring of trinomials in the form $x^2+bx+c,$ the two numbers whose product is $c$ and whose sum is $b$ are $\left\{ -4,-5 \right\}.$ Using these two numbers, the factored form of the expression above is \begin{array}{l}\require{cancel} (z^2-4)(z^2-5) .\end{array} The expressions $z^2$ and $4$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $z^2-4 ,$ 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} [(z)^2-(2)^2](z^2-5) \\\\= [(z+2)(z-2)](z^2-5) \\\\= (z+2)(z-2)(z^2-5) .\end{array}