To start off, everything that has a factor of x in common so . . factor out the x -- this means divide each term by the x.

x(x^2 - x - 6)

Now you will go through the some what laborious process of factor the remaining trinomial . . .

make a list of factors of 6 (I'll ignore the sign for now)

1 * 6

2 * 3

No using these factors and assigning signs as you might need them to make it work . . .can you add or subract to find -1, the coeffiecient of the middle term?

1+6 = 7, -1 + -6 = -7, -1 + 6 = 5 . . .this combo just won't work

2 + -3 = -1 . . .so use 2 and -3 when we go back to rewrite the problem . . .

x(x^2 + **2**x + **-3**x - 6) next group and factor the smaller groups

x[ (x^2 + 2x) + (-3x - 6)]

looking at (x^2 + 2x) they have a x in common so

x(x + 2)

looking at (-3x - 6), they have a -3 in common so . .

-3(x + 2) -> remember -6/-3 = +2

so now we have . . .

x[x(x + 2) + -3(x + 2)]

Notice that there is a matching (x + 2) for each grouping . . if you pull this out front .. . you will use the rest as the final factor . . . .

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

This is your factored form!

As it appears to me the answer posted above is not sticking to the main question, that is to find factors of given expression. This can be done in a much simpler way as follows.

x^3 - x^2 - 6x = x*[x^2 - x - 6] = x*[x^2 - 3x + 2x - 6]

= x*[x*(x - 3) + 2*(x - 3)] = x*(x - 3) *(x + 2)

Therefore factors of the given expression are x, (x - 3), and (x + 2)

The factored form of x^3-x^2-6x is required.

First factor out the common term x which is present in all the terms of the expression.

x^3-x^2-6x

= x(x^2 - x - 6)

Now write -1 as a sum of two numbers such that their product is equal to -6. This can be done as -1 = 3 - 2

= x(x^2 - 3x + 2x - 6)

= x(x(x - 3) + 2(x - 3))

= x(x - 3)(x + 2)

The factored form of x^3-x^2-6x = x(x - 3)(x + 2)

x^3-x^2-6

=x(x^2-6x-6)=x(x^2+2x-3x-6)=x{x(x+2)-3(x+2)}=x{(x+2)(x3)}=x(x+2)(x-3).

Alternative procedure:

The given expression has obviously a factor x. So, x^3-x^2-6=**x **(x^2-x-6)

Now, consider the quadratic factor x^2-x-6 :

Let x^2-x-6 = (x+A)(x+B) . Let us determine A and B.

x^2-x-6=x^2+(A+B)x+AB

Comparing the coefficients of x^2, x and constant terms on both sides we get:

1/1=-1/(A+B) = -6/(AB) . From this we get two independent equations by which we can determine A and B as follows:

AB=-6 (1) and

A+B=-1. (2)

A-B=sqrt[(A+B)^2-4AB)^(1/2)]= [1-4(-6)]^(1/2)=5

A-B =5 (3)

From (2) and (3), by adding 2A=-1+5=4 or A = 4/2=2

And (3)-(2) : -2B=5-(-1)=6 or or B= 6/(-2)= -3.

Therefore, x^2-x-6=(x+2)(x-3).

Therefore,x^3-x-6x =x(x^2-x-6)= x(x+2)(x-3).

Hope this helps.