How do you find the zeroes of $p(x)= x^3+x^2-x-1$ ?

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Answer 1 · 2015-05-11T23:19:53

$p(x) = x^3 + x^2 - x - 1$

$= (x^3 + x^2) - (x + 1)$

$= x^2*(x + 1) - 1*(x + 1)$

$= (x^2 - 1)(x + 1)$

$= (x - 1)(x + 1)(x + 1)$

This is zero for $x = -1$ (twice) and $x = 1$ .

The process I used here is called factoring by grouping.

How do you find the zeroes of p(x)= x^3+x^2-x-1p(x)= x^3+x^2-x-1?