factoring question

[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]

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

 

[Denis]

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]

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.