What is the factor form for x^3-x^2-6x?
i just cant seem 2 get it
3 Answers | Add Yours
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 + 2x + -3x - 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 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.
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 =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^3-x-6x =x(x^2-x-6)= x(x+2)(x-3).
Hope this helps.
Join to answer this question
Join a community of thousands of dedicated teachers and students.Join eNotes