2. Polynomials
hard

Factorise : $x^{3}-3 x^{2}-9 x-5$

Option A
Option B
Option C
Option D

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 $+1$]

                  $=(x+1)[x(x-5)+1(x-5)]$

                  $=(x+1)[(x-5)(x+1)]$

                  $=(x+1)(x-5)(x+1)$

Standard 9
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.