Kj91
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Report Thread starter 7 years ago
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Can anyone help me with this question please. (n n-3) = 84 when I cancel this down I was left with n^3-3n^2+2n-252=0. I can't get this to factor by substituting in factors of 252 into the equation to equal zero. How do you factor this? Thanks.
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Kj91
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(Original post by TenOfThem)
n(n-1)(n-2) = 3 * 84

Not sure this is solvable since 3*84 = 6*6*7 = 4*7*9 = other possibilities

252 is not a product of 3 consecutive numbers
Could this be a mistake in the book? Since it has an answer of n=9. I cannot get 9 to work on this though.
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TenOfThem
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(Original post by Kj91)
Could this be a mistake in the book? Since it has an answer of n=9. I cannot get 9 to work on this though.
Sorry

I used 3 instead of 3!

I am dim


So

n(n-1)(n-2) = 6*84 = 9*8*7

so n=9
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Kj91
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(Original post by TenOfThem)
Sorry

I used 3 instead of 3!

I am dim


So

n(n-1)(n-2) = 6*84 = 9*8*7

so n=9
Thanks for you help. I forgot the 3! as well.
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