Question 221295
Usually  when  we  see  4  different  terms  we  try  "  grouping"
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st +5b -bs -5t,,,rearrange terms
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st - bs +5b-5t
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s(t-b) +5(b-t),,,,change  signs  2nd  term
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s(t-b) -5(t-b),,,,,,,,,,,,,,now  comes  hard  part,,,,,treat  this  as  s(x) -5(x),,where 
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 x = (t-b),,,,then  x(s-5) is  factored  form,,,subst  back  in  for x
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  ,,,,,,(t-b)(s-5)
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(t-b)(s-5),,,,,answer
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