Question

4) Factorise: x^3 – 2x^2 – 5x - 6​

Answer 1

Answer:

First observe that the sum of the coefficients is  

0 , so  x = 1  is a root. Then divide by  ( x − 1 )  to get a quadratic that is easier to factor and thus solve to find two other roots  x = 3  and  x  =− 2 .  

Explanation:

Let  f ( x ) = x 3 − 2 x 2 − 5 x + 6

First note that the sum of the coefficients is  0 , so  x = 1  is a zero of  f ( x ) ...  

f ( 1 ) = 1 − 2 − 5 + 6 = 0  

Divide  

f ( x )  by  ( x − 1 )  to find:  

f ( x ) = x 3 − 2 x 2 -5 x + 6 = ( x − 1 ) ( x 2 − x − 6 )  

Then  x 2 − x − 6 = ( x − 3 ) ( x + 2 )

So the other two roots of  

f ( x ) = 0  are  x = 3  and  x = − 2

Alternatively, use the rational roots theorem to know that any rational roots of  f ( x ) = 0  must be of the form  p q where  p ,  q  are integers,  q ≠ 0 ,  p  a factor of the constant term  6  and  q  a factor of the coefficient  1  of the term ( x 3) of highest degree. So the only possible rational roots are:

± 1 ,  ± 2 ,  ± 3  and  ± 6  

Try each to find  f ( 1 ) = 0 ,  f ( − 2 ) = 0  and  f ( 3 ) = 0 .