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

$(t+10+2y)(t+10-2y)$
Grouping the first $3$ terms the given expression is equivalent to \begin{array}{l}\require{cancel} 100+20t+t^2-4y^2 \\\\= (100+20t+t^2)-4y^2 \\\\= (t^2+20t+100)-4y^2 .\end{array} Using the factoring of trinomials in the form $x^2+bx+c,$ the $\text{ expression }$ \begin{array}{l}\require{cancel} (t^2+20t+100) \end{array} has $c= 100$ and $b= 20 .$ The two numbers with a product of $c$ and a sum of $b$ are $\left\{ 10,10 \right\}.$ Using these two numbers, the $\text{ expression }$ above is equivalent to \begin{array}{l}\require{cancel} (t+10)(t+10)-4y^2 \\\\= (t+10)^2-4y^2 .\end{array} The expressions $(t+10)^2$ and $4y^2$ are both perfect squares (the square root is exact) and are separated by a minus sign. Hence, $(t+10)^2-4y^2 ,$ 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} (t+10)^2-(2y)^2 \\\\= (t+10+2y)(t+10-2y) .\end{array}