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is it 4y^3... ??
how did you get that >?
i looked at your probelm and then decided to ask you if 4y^3 was what you meanrt to type....then waited for you to answer :)
if there are no typos in your problem than I would begin by factoring out a "2" tha they all have in common
yes its 4y^3
is the equation of the curve "4y^3 +2y -6" ? or is that the 1st derivative of it?
thats the first derivative..the eqauation of the porabola is x-y^2=0
..... if all you are looking to do is factor "4y^3+2y-6" then I can help.... but as far as it fitting in with all the information that you are leaving out, thatll be up to you :)
yes i can figure the rest out .. its the factoring that for some reason is hurting me ...
factor a 2: 2(2y^3 +y -3) = 0 2y^3 +y -3 = 0 is easier to play with
lets find a "pool" of options to narrow our trail and error with ok" factors of the last # -------------- factors of the first # 1,3 --- now we can get dirty with it :) 1,2
we could use long division to try it, but synthetic division is more compact....
lets try y =1 1 | 2 0 1 -3 0 2 2 3 ------------- 2 2 3 0 <- remainder is zero, this is a factor (y-1)
(y-1) (2y^2 +2y +3)
the quad facotrs easily now right?
or not at all ;) sqrt(2^2 - 4(3)(2)) = sqrt(4 - 24) = sqrt(-20) doesnt factor... your only root is y = 1
which means, there there is only one critical point; and it is at y=1
any of that make sense?
makes perfect sense.. thanks you much ...
youre welcome :)