How do you solve the cubic function x^3-14x^2+16x+12=0 without using a calculator?

- anonymous

How do you solve the cubic function x^3-14x^2+16x+12=0 without using a calculator?

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- chestercat

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- amistre64

can do trial and error :)

- dumbcow

first try factor by grouping

- amistre64

lets make a pool of useable numbers that will narrow our solutions:
factors of last# +- 1,12,2,6,3,4
-------------: -------------
factors of first# +- 1

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## More answers

- amistre64

descartes has a sign theorum; the number of sign changes tells us the number of + solutiuons available: possible 2 (+) solutions in this case

- amistre64

the next best thing I can do with this is to start trying and erroring :)

- amistre64

x=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

x=2; lets try that..
2| 1 -14 16 12
0 2 -24 16
---------------
1 -12 8 28 <-- not a zero, not a solution...

- anonymous

I feel like this could take days to actually figure out haha...what if the answer is not a whole number?

- amistre64

x = 6 maye??
6| 1 -14 16 12
0 6 -48 ?? <-- not make it zero; not a solution
---------------
1 -8 -36

- amistre64

lol..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

and one of our pool options has got to work :)

- amistre64

x=3; lets try that..
3| 1 -14 16 12
0 3 -33 ........ not a -12 that we need here
---------------
1 -11-27 ..........not the zero we need here

- amistre64

-1?
x=2; lets try that..
-1| 1 -14 16 12
0 -1 15 -21
---------------
1 -15 21 ..... <-- not a zero, not a solution

- amistre64

copy paste edit and forgot to clean that x=2 outta there lol

- amistre64

x=-2; lets try that..
-2| 1 -14 16 12
0 -2 32 -56
---------------
1 -16 48 ..... <-- not a zero, not a solution

- anonymous

No 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

4?
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

-4?
-4| 1 -14 16 12
0 -4 72 -342
---------------
1 -18 88 ....... <-- not a zero, not a solution

- anonymous

try 800189074059/1562500000000

- dumbcow

there are no integer factors

- anonymous

I 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

HOW YOU FIND THAT WITHOUT A CALCULATor...caps got locked .....too lazy to erase and re type lol

- dumbcow

if you know calculus, newtons method works pretty well at approximating zeroes

- dumbcow

no i cheated and used calculator :)

- anonymous

I cheated and used a calculator as well haha

- amistre64

i was using one after the numbers started to all look the same ;)

- amistre64

why do the normal rules not apply to this equation?

- anonymous

The 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

lol..did you give me a broken quesition :)

- anonymous

and 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

I 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:
[(x-4), -2, -2; -2, (x-4), -2; -2, -2, (x-4)] and equalling it to zero.

- amistre64

maaa!!! lina broke the math again ;)

- amistre64

i wonder if its possible to decompose a cubic ...

- anonymous

I'm sure there is somehow...ya, I think I must up the algebra of that problem. I resolved and came up with x^3-12x^2+36x-32=0 and I think 2 works for this one...sorry about the crazy cubic. ha

- dumbcow

hmm i think i figured it out
x = 2 and 8
if you dont expand everything when doing the determinant and leave it as (x-4) it makes it easier
say u = x-4 then u^3-12u-16 =0

- amistre64

yeah, 2 and 8 are roots to that one :)

- anonymous

Ya, 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^2-10x+16) was the other factor.

- amistre64

- amistre64

##### 1 Attachment

- amistre64

i should put an input box to evaluate any function entered lol

- dumbcow

well yes thats the smart way to do it

- amistre64

its hard to make these in the dark :)

- anonymous

amistre , 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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