How do you factor: #y= s^2-8s+7 #?

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Answer 1 · 2018-06-22T15:01:55

#y=s^2-8x+7hArrcolor(blue)(y=(s-7)(s-1))#

Since #(s+a)(s+b)=s^2+(a+b)s+ab#
we are looking for constants #a# and #b#
such that #a xx b=+7#
and #(a+b)=-8#

Answer 2 · 2018-06-22T15:02:35

The same processes as when you use #x#

#y=(s-1)(s-8)#

Note that #1xx7=7 and 1+7=8#

but we need negative 8 so we use

#(-1)xx(-7)=+7 and (-1)+(-7)=-8# as required

Thus #y=(s-1)(s-8)#