Question

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

Answer 1

can do trial and error :)

Answer 2

first try factor by grouping

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

x=2; lets try that..

2| 1 -14 16 12
0 2 -24 16
---------------
1 -12 8 28 <-- not a zero, not a solution...

Answer 8

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

Answer 9

x = 6 maye??

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

-1?

x=2; lets try that..

-1| 1 -14 16 12
0 -1 15 -21
---------------
1 -15 21 ..... <-- not a zero, not a solution

Answer 14

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

Answer 15

x=-2; lets try that..

-2| 1 -14 16 12
0 -2 32 -56
---------------
1 -16 48 ..... <-- not a zero, not a solution

Answer 16

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.

Answer 17

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 :)

Answer 18

-4?

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

Answer 19

try 800189074059/1562500000000

Answer 20

there are no integer factors

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

no i cheated and used calculator :)

Answer 25

I cheated and used a calculator as well haha

Answer 26

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

Answer 27

why do the normal rules not apply to this equation?

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

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.

Answer 32

maaa!!! lina broke the math again ;)

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

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

Answer 37

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.

Answer 38

<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>

Answer 39

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

Answer 40

well yes thats the smart way to do it

Answer 41

its hard to make these in the dark :)

Answer 42

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