How do you factor completely $x^{3} - x^{2} - x + 1$ $x^3-x^2-x+1$ ?

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Answer 1 · 2015-07-09T03:38:04

Factor $f (x) = x^{3} - x^{2} - x + 1$ $f(x) = x^3 - x^2 - x + 1$

Since (a + b + c + d = 0), one factor is: (x - 1)

$y = (x - 1) (x^{2} - 1) = {(x - 1)}^{2} (x + 1)$ $y = (x - 1)(x^2 - 1) = (x - 1)^2(x + 1)$

ChecK: $f (x) = x^{3} - x - x^{2} + 1$ $f(x) = x^3 - x - x^2 + 1$ . OK

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