10 Golden Points, anyone???

Question

Hi..I just needed help with a math question below. If ur a type of a person who is going to answer me by saying, "Do ur homework urself", then excuse me bcuz I know I can do this question myself. I just have my own reasons for asking ur help. Please help me by showing me the steps and explainin them so that i can understand better.

a) Prove that if x^3 + mx + n is divisible by (x - k)^2, then (m / 3)^3 + (n / 2)^2 = 0.

b) Prove that if x^3 +mx + n and 3x^2 + m have a common factor (x - k) then 4m^3 + 27n^2 = 0.


In addition, Im extremely sorry if i sounded rude above. But i promise I will appreciate ur help by giving 10 points surely to anyone who answers my question in such a way that I understand the solution properly (although there is no solution as it involves proving). Thanks in advance, guys!!

Answer 1

b) f1(x) = x^3 +mx + n
f2(x) = 3x^2 + m
according to remainder theorem if (x-k) is common factor, then
f1(k) = 0, or k^3 + mk+ n = 0......(i)
f2(k) = 0, or 3k^2 + m = 0......(ii)

we now need to eleminate k,
3k^3 + 3mk+ 3n = 0 [multiply (i) by 3]
and 3k^3 + mk = 0 [multiply (ii) by k]
or, 2mk + 3n = 0
or, k = -(3n/2m) ......(ii)

mk^3 + m^2k+ mn = 0 [multiply (i) by m]
and 3nk^2 + mn = 0 [multiply (ii) by n]

or mk^3 + m^2k - 3nk^2 = 0

or, mk^2 - 3nk + m^2 = 0
or, m (3n/2m)^2 - 3n(-(3n/2m)) + m^2 = 0
or, (9/4) n^2 /m+ (9/2) * n^2/m + m^2 =0
or, 4m^3 + 27n^2 = 0

Part a:
if f(x) = x^3 + mx + n is divisible by (x - k)^2,
then, f(k) = 0,
or, k^3 + mk + n = 0
or n = -(k^3 + mk)
and, m = -(k^2 + n/k)
then (m / 3)^3 + (n / 2)^2
= - (k^2 + n/k)^3 / 27 + (k^3 + mk)^2 / 4
= -(k^6 + 3k^4n/k + 3k^2(n/k)^2 + (n/k)^3) / 27 + (k^6 + 2mk^4 + m^2k^2) / 4
= -(k^6 + 3nk^3 + 3n^2 + (n/k)^3) / 27 + (k^6 + 2mk^4 + m^2 k^2) / 4

= a(k^3 + mk + n)^2 [some a can be determined through calculation]
= 0

Answer 2

if (x^3+mx+n) is divisible by (x-k)^2 then

(x^3+mx+n) = (x-k)^2(x-a) for some a
= (x^2-2kx + k^2) (x-a)
= x^3- (2k+a)x^2 +x(2ka + k^2) - ak^2

comparing coefficients as it is an identity

a = - 2k
m = 2ka +k^2 (putting a in
= - 4k^2 + k^2 = - 3k^2
n = ak^2 = 2k^3
now (m/3) ^3 + (n/2)^2 = (-k^2)^3 + (k^3)^2 = 0

hence proved

b)
this follows from

we see that x^3+mx + n has differential 3x^2+m
as (x-k) is a common factor (x-k)^2 is a factor of x^3+mx +n

so from the above proof

(m/3)^3+(n/2)^2 =0
or m^3/27+n^2/4 =0 (expand)
or 4m^3 + 27 n^2 =0 (multiply by 108)
QED

Answer 3

ok...im surprised that i understood the question and perhaps worked out some way to prove it.

a) let x^3 + mx + n = f(x)
so f(x) = (x - k)^2 *(ax+b)
= (x^2 - 2kx + k^2)(ax+b) // you have to solve for ax+b now.

** since coefficient of x^3in f(x) is 1, so a = 1
since coefficient of x^2 in f(x) = 0, b has to be 2k.
Cos only 2k * x^2 - 2kx * x = 0

Since you worked out the factors, expand and you will get
x^3 -(3k^2)x + (2k^3) = f(x)
Compare coefficients, m = -3k^2
n = 2k^3
Compute them into (m / 3)^3 + (n / 2)^2 = 0 and u will get ur zero.
[PROVEN]

b) same method
(i worked out earlier that 4m^3 + 27n^2 = 0 is the same as (m / 3)^3 + (n / 2)^2 = 0.)

3x^2 + m = (x-k)(ax+b)
solve for ax+b with the same method and u get 3x+3.
so m is now -3k^2 like just now.
COmpute into f(x), and u get
x^3 - (3k^2)x + n = (x-k)(ax^2+bx+c)
Solve it out the same way and u get x^2 + kx - 2k^2.
n is now 2k^3 as before.

COmpute them into the equation 4m^3 + 27n^2 = 0. and you will get your zero once again.
[PROVEN]

Answer 4

Kool Gurl
Our first concern is to help you and others who genuinely need help. The 10 points we earn if our answer is chosen as the best is of little importance.

Answer 5

The problem with your question is the divisors you have chosen has terms which are not in the given equation meaning those cannot be the divisors. You divide the equation given by any one of the divisors given you will know, it gives reminders, if they vere perfect divisors then there won't be any reminder.

Answer 6

i'm sorry!