To find a factor of f(x) we assume different values of x and substitute it in f(x)

First we consider the factors of the constant term −6 i.e., ±1,±2,±3,±6
Now we substitute the values of x in f(x)
If f(x)=0 for some value a
Then (x−a) is a factor of f(x)
[Note : This is a hit and trial method, So you have to iterate the process until you get the required x]

Substituting x=−1 in f(x) , We get, f(−1)=(−1)3+2(−1)2−5(−1)−6 =−1+2(1)+5−6 =−1+2+5−6 =−7+7 =0 ∴ By factor theorem, (x+1) is a factor of f(x)

Now dividing f(x) by (x+1), we get x3+2x2−5x−6=(x+1)(x2+3x−2x−6)=(x+1){x(x+3)−2(x+3)}=(x+1)(x−2)(x+3)

To find a factor of f(x) we assume different values of x and substitute it in f(x)

First we consider the factors of the constant term −6 i.e., ±1,±2,±3,±6

Now we substitute the values of x in f(x)

If f(x)=0 for some value a

Then (x−a) is a factor of f(x)

[Note : This is a hit and trial method, So you have to iterate the process until you get the required x]

Substituting x=−1 in f(x) , We get,

f(−1)=(−1)3+2(−1)2−5(−1)−6

=−1+2(1)+5−6

=−1+2+5−6

=−7+7

=0

∴ By factor theorem,

(x+1) is a factor of f(x)

Now dividing f(x) by (x+1), we get

x3+2x2−5x−6=(x+1)(x2+3x−2x−6)=(x+1){x(x+3)−2(x+3)}=(x+1)(x−2)(x+3)