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

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Answer 1 · 2016-05-07T11:39:16

There are 4 terms - group them in pairs.
= $(x + 2) (x^{2} - 1)$ $(x + 2)(x^2 - 1)$ if $p = 2$ $p=2$

$x^{3} + p x^{2} - x - 2$ $x^3 + px^2 - x - 2$ = $(x^{3} + p x^{2}) + (- x - 2)$ $(x^3 + px^2) + ( - x - 2)$

= $x^{2} (x + p) - (x + 2)$ $x^2(x + p) - (x + 2) " "$ This will only factor if p = 2

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

I suspect that the question should have read something like,
"find the value of p by factorising."

How do you factor completely x3+px2−x−2 x^3 + px^2 - x - 2?