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 + 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 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(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)
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.
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