math help?

Question

how do we factor this are their any factors?
3x^4-4x^3-12x^2+2

Answer 1

3x^4-4x^3-12x^2+2 =
x³(3x - 4) - 2(6x² - 1) = (x³ - 2)(3x - 4)(6x² - 1)
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Answer 2

The only rational possible zeros are (plus or minus) 2, 2/3, 1, 1/3. None of these work. Therefore the 4 zeros (which do exist) are irrational. That makes it really tough to factor!

The only way I know is trial and error. You know the factors must be trinomials. The first in each trinomial must be 3x^2 and x^2; the last two must be -2 and -1

(3x^2 + ____ - 2)(x^2 + _____ -1)

This gives -3x^2 and -2 x^2 and you need a total of -12 x^2 so the missing terms must multiply to be -7x^2. And since one will

(3x^2 - 7x - 2)(x^2 + x - 1)

Check it out and see if it works; if not switch the -2 and -1 and try again! Oh I give up, sorry

Answer 3

....3...-4... -12... ..2
1.........3 .....-1....-13
... 3... -1... -13... -11

There is a root between 0 and 1. So, there must be at least another one.

....3....-4... -12.......2
2.........6 ......4 ....-12
... 3... .2... ..-6.....-10.

....3...-4... -12... ..2
3.........9 ...15.......9
... 3... .5... ..3......11

And the other one is between 2 and 3.

So, try the possible rational roots in these intervals. If there are no rational roots, then, they are irrational, just calculate them in an approximated way

Perhaps you have up to 2 more roots.

I suggest you to take derivatives and to see how the functions growths


Ana

Answer 4

3x^4-4x^3-12x^2+2

3x^4-4x^3-12x^2=-2
x^2(3x^2-4x-12)=-2
The ones in the parentheses do not have a square.

Answer 5

Yes, find the possible zeros: +/- 1, +/- 2, +/- 2/3
now use synthetic division and test each zeros. There are no equation on how to factor a fourth root.
Hope this help, I didn't have time to work out the whole problem.