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Factorise : $x^{3}-3 x^{2}-9 x-5$
Solution
$x^{3}-3 x^{2}-9 x-5$
We have $p(x)=x^{3}-3 x^{2}-9 x-5$
By trial, let us find : $p (1)=(1)^{3}-3(1)^{2}-9(1)-5=3-3-9-5$
$=-14 \neq 0 $
Now $p(-1)=(-1)^{3}-3(-1)^{2}-9(-1)-5=-1-3(1)+9-5$
$=-1-3+9-5=0$
$\therefore$ By factor theorem, $[ x -(-1)]$ is a factor of $p ( x )$.
Now, $\frac{x^{3}-3 x^{2}-9 x-5}{x-(-1)}=x^{2}-4 x-5$
$\therefore x^{2}-3 x^{2}-9 x-5 =(x+1)\left(x^{2}-4 x-5\right) $
$=(x+1)\left[x^{2}-5 x+x-5\right]$
[Splitting $-4$ into $-5$ and $+1$]
$=(x+1)[x(x-5)+1(x-5)]$
$=(x+1)[(x-5)(x+1)]$
$=(x+1)(x-5)(x+1)$