Question

Factorise - x³- 3x²- 9x- 5

Answer 1

here is the answer

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Answer 2

let f(x)= x3_3x2_9x_5

> f(_1) = (_1)3_3(_1)2_9(_1)_5

> f(_1)=0

so, according to factor theorem x+1 is a factor of f(x)

On dividing f(x) by x+1 we get

x2_4x_5, which on factorisation gives two more factors which are x_5 and x+1.

So all the factors of f(x) are (x+1), ( x+1) and ( x_5)