How do you factor $3 n^{3} - 4 n^{2} - 6 n + 8$ $3n^3-4n^2-6n+8$ ?

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Answer 1 · 2015-05-16T01:53:48

Group first term and third term:

$f (x) = 3 n (n^{2} - 2) - 4 (n^{2} - 2) = (n^{2} - 2) (3 n - 4)$ $f(x) = 3n(n^2 - 2) - 4(n^2 - 2) = (n^2 - 2)(3n - 4)$

How do you factor 3n3−4n2−6n+83n^3-4n^2-6n+8?