help w. factoring plz explain steps that you did?

2008-11-09 1:26 pm
12x^4-5x^2y^2-2y^2
更新1:

im sry yest you guys are correct the question is 12x^4-5x^2y^2-2y^4

回答 (5)

2008-11-09 1:40 pm
✔ 最佳答案
First, I'm going to assume the last term is -2y^4 because if it's really -2y^2 then the expression can't be factored.

Next, I take the coefficient of the first term (12) times the coefficient of the last term (-2) and multiply them together (-24). List the factors of -24 and find the ones that add up to the middle coefficient (-5)
-1,24=23
1,-24=-23
-2,12=10
2,-12=-10
-3,8=5
3,-8=-5 ****** Ding ding ding ding. These are the factors
Set up your parentheses:
(??x^2 + ?? y^2)(??x^2 + ??y^2)
You fill in the ?? with factors of 2 and 12 that multiply out to be 3 and -8 when you FOIL.
(3x^2-2y^2)(4x^2+y^2)

_/
2016-11-09 7:39 pm
constantly start up with the start, if it starts off with ^4, the factors will the two be ^a million and ^3 or ^2 and ^2. for the reason that there are no ^1s or ^3s on your equation, you know to pass with ^2's (n^2 )(n^2 ) using an identical technique on the w^4, you finally end up with (n^2 w^2)(n^2 w^2) Now you verify the sign of the w's and word its effective, so the indications would desire to be the two effective, or the two destructive. because of fact the sign of the n^2w^2 is effective, you know the two factors would be effective, leaving: (n^2+w^2)(n^2+w^2) because of fact the factors.
2008-11-09 2:32 pm
12x^4 - 5x^2y^2 - 2y^4
= 12x^4 + 3x^2y^2 - 8x^2y^2 - 2y^4
= (12x^4 + 3x^2y^2) - (8x^2y^2 + 2y^4)
= 3x^2(4x^2 + y^2) - 2y^2(4x^2 + y^2)
= (4x^2 + y^2)(3x^2 - 2y^2)
2008-11-09 1:38 pm
can't factor it. Did you write it correctly? I'd expect either the last term to be -2y^4, or the middle term to have y's not y^2's
2008-11-09 1:36 pm
=(3x^2 -2y)(4x^2+y)


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