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Mathematics 10 Online
OpenStudy (anonymous):

factor n^2-19n+48 please help!

OpenStudy (anonymous):

n^2-19n+48 = (n - ?)(n - ?) The factors of 48 are 1,48 2,24 3,16 4,12 The answer is (n-3)(n-16),, which can be see through use of the FOIL method. n*n + -3n + -16n + -3*-16 = n^2 -19n + 48

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