SOLUTION: Usually I know how to factor up to four terms (by grouping method), but this particular problem caught me because it is five terms. Here is the problem: Factor the expression compl

Question 137047This question is from textbook McDougal Littell's Algebra 1
: Usually I know how to factor up to four terms (by grouping method), but this particular problem caught me because it is five terms. Here is the problem: Factor the expression completely: 7n^5+7n^4-3n^2-6n-3. Please explain it, and help! Thank you so much. :) This question is from textbook McDougal Littell's Algebra 1

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
f(n) = 7n^5+7n^4-3n^2-6n-3
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= 7n^4(n+1) -3(n^2+2n+1)
= 7n^4(n+1) -3(n+1)^2
= (n+1)[7n^4 -3(n+1)]
= (n+1)[7n^4-3n-3]
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Comment: 7n^4 -3n -3 is not factorable in the Real Number system.
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Cheers,
Stan H.

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