Question

factories :x³-2x²-5x-6 by factor theorem

Answer 1

  for a cubic equaiton you need to take guesses at first root so try 1 for x and we get 

1^3 - 2(1^2) - 5(1) + 6 = 

1 - 2 - 5 + 6 = 

- 1 + 1 = 0 so 1 is a root and can be written as x - 1 now do long division 

................x^2 - x - 6 
................x^3 - 2x^2 - 5x + 6 
x - 1.........x^3 - x^2 
.......................- x^2 - 5x 
......................- x^2 + x 
................................- 6x + 6 
................................- 6x + 6 

so we have (x - 1)(x^2 - x - 6) as factors the second term can factored further and we get 

(x - 1)(x - 3)(x + 2)

Hope This Helps :)

Answer 2

x^3+2x^2−5x−6  

=x^3+x^2+x^2−5x−6

=x^2(x+1)+x^2+x−6x−6

=x^2(x+1)+x(x+1)−6(x+1)

=(x+1)(x^2+x−6)

=(x+1)(x^2+3x−2x−6)

=(x+1)[x(x+3)−2(x+3)]

=(x+1)(x+3)(x−2)