## Algebra 1

$y^{2}$-16y+64 Using the rule of $a^{2}$-2ab+$b^{2}$= $(a-b)^{2}$ In this case, the a= y and b= 8 as substituting these gives us the original polynomial $a^{2}$-2ab+$b^{2}$ $y^{2}$-2(y)(8)+$8^{2}$ $y^{2}$-16y+64 Thus: $y^{2}$-16y+64 = $(y-8)^{2}$ = (y-8)(y-8)