quartic Polynomial (2 Viewers)

[M@ster P]

need help with this

Solve 6x^4 + 5x^3 - 24x^2 - 15x +18 = 0 if the sum of two of its roots is zero.



appreciated

 

[M@ster P]

wat do u mean testing, test what values?

 

[bored of sc]

i meant subbing in some values for x to see if it satisfies the equation and therefore is a solution.

Is trial and error the only way to work this out? Is there a formula for the roots of a quartic polynomial?

 

[M@ster P]

lol sorry are u still there i still dont get it!!!!!!!!!!!!!!!

 

[3unitz]

Is trial and error the only way to work this out? Is there a formula for the roots of a quartic polynomial?

you can use the formulas i gave before and solve for 3 unknowns:

Sum alpha = -b/a
Sum alpha.beta = c/a
Sum alpha.beta.gamma = -d/a
alpha.beta.gamma.delta = e/a

might be a little bit more messy, but would still would work.

 

[M@ster P]

i meant subbing in some values for x to see if it satisfies the equation and therefore is a solution.

eg. say you have x^3 + 8x^2 - 7 = 0
sub in -1 for x, we see this satisfies the equation and is a solution.
this example is a monic polynomial (the highest power's coefficient is 1).

for non-monic polynomials like the polynomial in your question, theres a few more options we can consider as possible solutions.

eg. consider the non-monic polynomial P(x) = 4x^3 + 2x^2 - 14x + 3
if we look at the coefficient of the highest power and the last term (4 and 3 respectively) its also possible that we can try the values:
1/2, 3/2, 1/4, 3/4 and their negatives, as well as the obvious integer possibilities 1, 3.

with your question, 6 and 18 where the two coefficients i looked at, and tried similar combinations
eg. 1/6, - 1/18, 6/18, 3/18, 6/9

6/9 turned out to satisfy the equation so 6/9 = 2/3 was one of the solutions

man i dnt get it how the values for my question which you subbed in are 1/6, - 1/18, 6/18, 3/18, 6/9. Are these the factors of 6 and 18? im lost at this part

 

[M@ster P]

you can use the formulas i gave before and solve for 3 unknowns:

Sum alpha = -b/a
Sum alpha.beta = c/a
Sum alpha.beta.gamma = -d/a
alpha.beta.gamma.delta = e/a

might be a little bit more messy, but would still would work.

yeah this is the method i would rather prefer, can someone use this method where you use the 4 formulas to work this question out. Sorry guys but i am just not getting it.

 

[3unitz]

man i dnt get it how the values for my question which you subbed in are 1/6, - 1/18, 6/18, 3/18, 6/9. Are these the factors of 6 and 18? im lost at this part

consider P(x) = (x - a)(x - b), our solutions are a and b, if we expand this we get:

P(x) = x^2 - xb - ax + ab
P(x) = x^2 - x(b + a) + ab

you can see the independant term has factors a and b.

lets say we introduce a "c" (a non zero integer) to the coefficient of x^2 to make the polynomial non-monic, we will get:

P(x) = cx^2 - x(b + a) + ab

we can factor out the c to get:

P(x) = c [x^2 - x[(b+a)/c] + ab/c]

now considering the monic polynomial on the right, ab/c will have factors of the roots.

another way of seeing this, is say i gave you the equation

P(x) = x^2 - x[(b+a)/c] + ab/c

and told you to find the roots, you might use these identities:

alpha + beta = - B/A
alpha.beta = C/A

here our alpha.beta = ab/c, hence ab/c has factors of the roots.

because ab is our independent term, and c is the coefficient of the highest power, any polynomial:
Ax^n + Bx^n-1 + ... + Yx + Z
will have n roots, and the factors of Z/A are possible roots.

notice that the answers i gave to your question were: sqrt 3, - sqrt 3, -3/2, 2/3. multiply them all together and you get 3, which just so happens to be equal to 18/6, the independent term divided by the coefficient of the highest power.

 

[3unitz]

yeah this is the method i would rather prefer, can someone use this method where you use the 4 formulas to work this question out. Sorry guys but i am just not getting it.

let solutions be: x, -x, y, z

x + -x + y + z = -b/a
y + z = -5/6 -----------(1)

x.-x + x.y + x.z + -x.y + -x.z + y.z = c/a
-x^2 + yz = -24/6 -----------(2)

x.-x.y.z = e/a
-x^2.y.z = 18/6 -----------(3)

solve these 3 for the 3 unknowns:

firstly, looking at (2) if we multiply both sides by -x^2 we get:
-x^2(-x^2 + yz) = -x^2 (-24/6)
x^4 - x^2.y.z = 4x^2
notice now we can sub in (3) to get:
x^4 + 18/6 = 4x^2
x^4 - 4x^2 + 3 = 0
let x^2 = u:
u^2 - 4u + 3 = 0
(u - 1)(u - 3) = 0
u = 1, u = 3
x = +/- 1, x = +/- sqrt 3

check these 2 values (1 and sqrt3) by subbing them into the original polynomial to see which one it is. here we will find x = sqrt 3

subbing sqrt 3 back into (3):
-3.y.z = 3
yz = -1 -----------(4)

and from (1):
y + z = -5/6
y = -5/6 - z

subbing into (4):
(-5/6 - z)z = -1
solve the quadratic for z, and test the 2 values you get back into the polynomial to see which one satisfies, then use this value to get y.

as you can see its a lot more work.

 

[M@ster P]

man i really appreciate your help, i like this method better as its the only one i understand thoroughly.

 

[u-borat]

Master P, sometimes you'll need both the trial/error method and the sum of roots method together to solve a question.

 

[lolokay]

if the equation 6x^4 + 5x^3 - 24x^2 - 15x +18 = 0
were put in the form 6(x+a)(x-a)(x+c)(x+d)
=6x^4 + 6(c+d)x^3 + 6(-a^2+cd)x^2 + 6(-a^2(c+d))x + -6a^2cd = 0 (not hard to expand since you can see which expanded terms will cancel out)

you now get;
c + d = 5/6
-a^2(c + d) = -15/6
a^2 = 3
a = +/- sqrt3
so 2 roots are -/+sqrt3

-a^2cd = 18/6
cd = -1
d = -1/c
c - 1/c = 5/6
6c^2 - 5c - 6 = 0

c = (5 +- sqrt(25 + 144))/12
=3/2, -2/3
so the other 2 roots are -3/2 and 2/3

 

[M@ster P]

i meant subbing in some values for x to see if it satisfies the equation and therefore is a solution.

eg. say you have x^3 + 8x^2 - 7 = 0
sub in -1 for x, we see this satisfies the equation and is a solution.
this example is a monic polynomial (the highest power's coefficient is 1).

for non-monic polynomials like the polynomial in your question, theres a few more options we can consider as possible solutions.

eg. consider the non-monic polynomial P(x) = 4x^3 + 2x^2 - 14x + 3
if we look at the coefficient of the highest power and the last term (4 and 3 respectively) its also possible that we can try the values:
1/2, 3/2, 1/4, 3/4 and their negatives, as well as the obvious integer possibilities 1, 3.

with your question, 6 and 18 where the two coefficients i looked at, and tried similar combinations
eg. 1/6, - 1/18, 6/18, 3/18, 6/9

6/9 turned out to satisfy the equation so 6/9 = 2/3 was one of the solutions

ok in this post you mentioned you looked at the coefficient with the highest power which is 6 and the last term which is 18.
ok i get that part and then you mentioned trying combinations such as 1/6, - 1/18, 6/18, 3/18, 6/9.

ok now were these combinations derived from the factors of 6 and 18?

the factors of 6 are 1,2,3,6 and the factors of 18 are 1,2,3,6,9.18. So were the combinations based on this?
And if they were, there were many more combinations that could have been listed meaning that it would have taken a long time to get the root.