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factor n^2-19n+48
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n^2-19n+48 n^2-16n-3n+48 (n^2-16n)+(-3n+48) n(n-16)-3(n-16) (n-3)(n-16) This tells us that n^2-19n+48 completely factors to (n-3)(n-16)
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1. List factors of 48 List: 48, 1 24, 2 12, 4 6, 8 16, 3 48, -1 24, - 2 12, -4 6, -8 -3, 16 And so on... 2. Which of these factors adds up to -19? Well the pair is -3 and -16 3. Since n is to the second power you'll get this as your answer... (n - 3) (n - 16) If I was not clear at some point, please do tell me and I'll elaborate.
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