SOLUTION: 6. Factor completely: x^2-3x-28
(x-7)(x+4)
(x-7)(x-4)
(x+7)(x-4
(x+7)(x+4)
Algebra.Com
Question 956977: 6. Factor completely: x^2-3x-28
(x-7)(x+4)
(x-7)(x-4)
(x+7)(x-4
(x+7)(x+4)
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Factor completely: x^2-3x-28
----
Find a pair of factors of 28 that differ by 3.
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