factoring question

G

Guest

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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?
 

Eva Barragan

New member
Joined
Feb 8, 2006
Messages
44
you have to factor out 3x first to find the binomial
there is only one awnser
 

Denis

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Feb 17, 2004
Messages
1,461
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
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Jan 28, 2005
Messages
5,588
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.
 
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