How do you factor $0 = x^{3} + 4 x^{2} - 3 x - 18$ $0 = x^3 + 4x^2 -3x - 18$ ?

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Answer 1 · 2015-05-02T21:54:57

First, this polynomial doesn't have (x - 1) and (x + 1) as factor.

Try: x = 2 -> f(x) = 8 + 16 - 6 - 18 = 0, then (x - 2) is a factor.

Next, divide or guess:

$f (x) = (x - 2) (x^{2} + 6 x + 9) = (x - 2) . {(x + 3)}^{2}$ $f(x) = (x - 2)(x^2 + 6x + 9) = (x -2).(x + 3)^2$

How do you factor 0 = x^3 + 4x^2 -3x - 180=x3+4x2−3x−180 = x^3 + 4x^2 -3x - 18?