Question

Factorise x³ – 3x² – 9x – 5

Answer 1

Let f(x) = x³ – 3x² – 9x – 5 By trial, 

f(- 1) = (- 1)³ – 3(- 1)² – 9(- 1) – 5 

=-1 – 3 + 9 – 5 =0 

∴ (x + 1) is a factor of f(x). 

[ ∵ by factor theorem] 

Now dividing f(x) by (x + 1).

f(x)=(x + 1)(x² – 4x – 5) 

But x² – 4x – 5 = x² – 5x + x – 5 

= x (x – 5) + 1 (x – 5)

=(x – 5)(x + 1) 

∴ f(x)=(x + 1)(x + 1)(x – 5)

Answer 2

substitute 5 for x in x³ - 3x²-9x -5

so 5 is a zero of the expression X3 -3x²   _{-9x - 5} 

_{this means (x-5) is a factor of x}³ -3x -9x -5

divide the equation by x-5 

factories then we found the equation x²+2x+1

this means x³-3x²-9x-5.  =(x-5) X2+ 2x +1

=(x-5) (x+1)^²