# Chapter 8 - Polynomials and Factoring - 8-7 Factoring Special Cases - Practice and Problem-Solving Exercises - Page 515: 41

$8(s-4)^{2}$

#### Work Step by Step

Given the polynomial $8s^{2}$ - 64s + 128 We see that all three terms have a common factor of 8, so we factor out the 8. $8(s^{2} - 8s + 16)$ We see that the polynomial has the first and last term squared and the middle term is -2 times the first and last term. Thus it follows the rule of $a^{2}$ - 2ab + $b^{2}$ = $(a-b)^{2}$ 8($(s)^{2}$ - 8s + $4^{2}$) In this polynomial a= s and b=4 8($(s)^{2}$ - 8s + $4^{2}$) = $8(s-4)^{2}$

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