factoring question

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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?
 
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
 
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: (3x12)(x+2)\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: 3x(x22x8)\displaystyle \,3x(x^2\,-\,2x\,-\,8)

Then factor the trinomials: 3x(x+2)(x4)\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":

3x36x224x  =\displaystyle 3x^3\,-\,6x^2\,-\,24x\;=

    3x(x+2)(x4)\displaystyle \;\;3x(x\,+\,2)(x\,-\,4)
    3(x2+2x)(x4)\displaystyle \;\;3(x^2\,+\,2x)(x\,-\,4)
    3(x+2)(x24x)\displaystyle \;\;3(x\,+\,2)(x^2\,-\,4x)
    x(3x+6)(x4)\displaystyle \;\;x(3x\,+\,6)(x\,-\,4)
    x(x+2)(3x12)\displaystyle \;\;x(x\,+\,2)(3x\,-\,12)
    (3x2+6x)(x4)\displaystyle \;\;(3x^2\,+6x)(x\,-\,4)
    (x+2)(3x212x)\displaystyle \;\;(x\,+\,2)(3x^2\,-\,12x)

    \displaystyle \;\;and, of course: 1×(3x36x224x)\displaystyle \,1\,\times\,(3x^3\,-\,6x^2\,-\,24x)

It should be obvious why we should factor completely.
 
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