How do you factor the trinomial $c^2 - 8 = - 7 c$ ?

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Answer 1 · 2018-06-16T08:32:24

$(c+8)(c-1)$

$"rearrange in standard form"$

$c^2+7c-8$

$"the factors of - 8 which sum to + 7 are + 8 and - 1"$

$c^2+7c-8=(c+8)(c-1)$

How do you factor the trinomial c^2 - 8 = - 7 cc^2 - 8 = - 7 c?