# What is the factor form for x^3-x^2-6x?i just cant seem 2 get it

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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)

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!

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.