Q

Factorise : (ii) x^3 - 3x ^2 - 9x - 5

5.(ii) Factorise :

(ii)    $x^3 - 3x^2 -9x -5$

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Given polynomial is  $x^3 - 3x^2 -9x -5$

Now, by hit and trial method we observed that  $(x+1)$  is one of the factore of given polynomial.

By long division method, we will get

We know that  Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 - 3x^2 -9x -5=(x+1)(x^2-4x-5)$

$= (x+1)(x^2-5x+x-5)$

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

Therefore, on factorization of  $x^3 - 3x^2 -9x -5$ we will get  $(x+1)(x-5)(x+1)$

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