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

May 7, 2016

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

Explanation:

${x}^{3} + p {x}^{2} - x - 2$ = $\left({x}^{3} + p {x}^{2}\right) + \left(- x - 2\right)$

= ${x}^{2} \left(x + p\right) - \left(x + 2\right) \text{ }$ This will only factor if p = 2

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

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