How do you factor $9 t^{3} + 18 t - t^{2} - 2$ $9t^3+18t-t^2-2$ ?

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Answer 1 · 2015-06-03T19:40:52

$9 t^{3} + 18 t - t^{2} - 2 = (9 t^{3} + 18 t) - (t^{2} + 2)$ $9t^3+18t-t^2-2 = (9t^3+18t)-(t^2+2)$

$= 9 t (t^{2} + 2) - (t^{2} + 2)$ $=9t(t^2+2)-(t^2+2)$

$= (9 t - 1) (t^{2} + 2)$ $=(9t-1)(t^2+2)$

Since $t^{2} + 2 > 0$ $t^2+2 > 0$ for all $t in RR$ , it has no linear factors with real coefficients. So our factorization is complete.

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