# factoring question

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#### Guest

##### Guest
I had a problem on a test

factor 3x^3-6x^2-24x

the teacher said that there were three different answers that could be true

i got one of them: (3x-12)(x+2), but how do you get any other answers?

#### Eva Barragan

##### New member
you have to factor out 3x first to find the binomial
there is only one awnser

#### Denis

##### Senior Member
brendaursula said:
I had a problem on a test
factor 3x^3-6x^2-24x
the teacher said that there were three different answers that could be true
i got one of them: (3x-12)(x+2), but how do you get any other answers?
3x^3 - 6x^2 - 24x
= 3x(x^2 - 2x - 8)
= 3x(x - 4)(x + 2) : now fully factored

#### soroban

##### Elite Member
Hello, brendaursula!

Factor: $$\displaystyle \:3x^3\,-\,6x^2\,-\.24x$$

the teacher said that there were three different answers that could be true. . . . no!

i got one of them: $$\displaystyle (3x\,-\,12)(x\,+\,2)$$, . . . incorrect
$$\displaystyle \;\;$$but how do you get any other answers?
The instruction 'Factor" always means "Factor completely."

First, take out any common factors: $$\displaystyle \,3x(x^2\,-\,2x\,-\,8)$$

Then factor the trinomials: $$\displaystyle \,3x(x\,+\,2)(x\,-\,4)$$

Now it's factored completely.
$$\displaystyle \;\;$$(The order of the factors makes no difference.)

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I hope you teacher realizes that it was a silly thing to say to a math class.

By his/her mind-set, there are four ways to factor 24:

$$\displaystyle \;\;\;24\;=\;1\,\times\,24,\;\;2\,\times\,12,\;\;3\,\time\,8,\;\;4\,\times\,6$$

But there is only one way: $$\displaystyle \,24\;=\;2\,\times\,2\,\time\,2\times\,3 \;=\;2^3\cdot3$$

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Even if your teacher is correct ... and we don't have to factor completely ...
$$\displaystyle \;\;$$there are eight "right answers":

$$\displaystyle 3x^3\,-\,6x^2\,-\,24x\;=$$

$$\displaystyle \;\;3x(x\,+\,2)(x\,-\,4)$$
$$\displaystyle \;\;3(x^2\,+\,2x)(x\,-\,4)$$
$$\displaystyle \;\;3(x\,+\,2)(x^2\,-\,4x)$$
$$\displaystyle \;\;x(3x\,+\,6)(x\,-\,4)$$
$$\displaystyle \;\;x(x\,+\,2)(3x\,-\,12)$$
$$\displaystyle \;\;(3x^2\,+6x)(x\,-\,4)$$
$$\displaystyle \;\;(x\,+\,2)(3x^2\,-\,12x)$$

$$\displaystyle \;\;$$and, of course: $$\displaystyle \,1\,\times\,(3x^3\,-\,6x^2\,-\,24x)$$

It should be obvious why we should factor completely.