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anonymous
 5 years ago
How do you solve the cubic function x^314x^2+16x+12=0 without using a calculator?
anonymous
 5 years ago
How do you solve the cubic function x^314x^2+16x+12=0 without using a calculator?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0can do trial and error :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first try factor by grouping

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lets make a pool of useable numbers that will narrow our solutions: factors of last# + 1,12,2,6,3,4 :  factors of first# + 1

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0descartes has a sign theorum; the number of sign changes tells us the number of + solutiuons available: possible 2 (+) solutions in this case

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the next best thing I can do with this is to start trying and erroring :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x=1; lets try that.. 1 1 14 16 12 0 1 13 3  1 13 3 15 < not zero in the end; not a solution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x=2; lets try that.. 2 1 14 16 12 0 2 24 16  1 12 8 28 < not a zero, not a solution...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I feel like this could take days to actually figure out haha...what if the answer is not a whole number?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x = 6 maye?? 6 1 14 16 12 0 6 48 ?? < not make it zero; not a solution  1 8 36

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol..there has to be at least one solution. All degrees higher than a 2 can be factored to at least a linear and a irreducible quadratic

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0and one of our pool options has got to work :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x=3; lets try that.. 3 1 14 16 12 0 3 33 ........ not a 12 that we need here  1 1127 ..........not the zero we need here

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01? x=2; lets try that.. 1 1 14 16 12 0 1 15 21  1 15 21 ..... < not a zero, not a solution

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0copy paste edit and forgot to clean that x=2 outta there lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x=2; lets try that.. 2 1 14 16 12 0 2 32 56  1 16 48 ..... < not a zero, not a solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No doubt there are solutions for x but is it possible for all 3 solutions for x to be a fraction? If that's the case, then guessing whole numbers will lead to infinite frustration ha.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04? 4 1 14 16 12 0 4 40 96  1 10 24 28 < not a zero, not a solution well, since our first# is a 1..that makes our pool limited to integers only; becasue they all have a 1 on the bottom :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.04? 4 1 14 16 12 0 4 72 342  1 18 88 ....... < not a zero, not a solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0try 800189074059/1562500000000

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are no integer factors

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I must have made a mistake in coming up with this cubic function because I don't see how my textbook would expect me to come up with that as an option ha

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0HOW YOU FIND THAT WITHOUT A CALCULATor...caps got locked .....too lazy to erase and re type lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you know calculus, newtons method works pretty well at approximating zeroes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no i cheated and used calculator :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I cheated and used a calculator as well haha

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i was using one after the numbers started to all look the same ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0why do the normal rules not apply to this equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The original problem was solve the system: y1'=4y_1+2y_2+2y_3 y2'=2y_1+4y_2+2y_3 y3'=2y_1+2y_2+4y_3

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0lol..did you give me a broken quesition :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and I used the characteristic eqution lambda I  A=0 and by using that I cam up with that cubic function and I didn't know how to solve it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I didn't purposely give you a broken problem if I did haha...I'm just as confused as you. I just came up with that function after taking the determinate of: [(x4), 2, 2; 2, (x4), 2; 2, 2, (x4)] and equalling it to zero.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0maaa!!! lina broke the math again ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i wonder if its possible to decompose a cubic ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm sure there is somehow...ya, I think I must up the algebra of that problem. I resolved and came up with x^312x^2+36x32=0 and I think 2 works for this one...sorry about the crazy cubic. ha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm i think i figured it out x = 2 and 8 if you dont expand everything when doing the determinant and leave it as (x4) it makes it easier say u = x4 then u^312u16 =0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yeah, 2 and 8 are roots to that one :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ya, that's the correct answer. I just did what amistre suggested to solve the "broken one" I gave him ha. I guessed 2 and it worked. Then I had to do long division to figure out that (x^210x+16) was the other factor.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0<html> <body> <input type=button value="start" onclick="qwe()"> </body> <script language=javascript> var y,x,j,m,n function qwe() { for (x=0;x<33;x=x+1) {y = (x*x*x)+(12*x*x)+(36*x)32 if (y==0){alert(x+" is a root")} } alert("no root") } </script> </html>

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i should put an input box to evaluate any function entered lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well yes thats the smart way to do it

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0its hard to make these in the dark :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0amistre , thats wrong , a polynomial of degree greater than 2 doesnt have to factor into a linear and irreducible quadratic. you can have say x^4 + 1 = 0 . no linear factor
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