Polynomial Degree 4 ... NEED SOLUTION

[lrtcoolman]

QUESTION: The polynomial p(x) = x⁴ - 4x³ + hx² - 6x + 2 has a factor in the form (x-m)², where m ∈ N. Find the values of m and h.

To those who professional in Math, please solve this ... Thanks.

 

[daon2]

QUESTION: The polynomial p(x) = x⁴ - 4x³ + hx² - 6x + 2 has a factor in the form (x-m)², where m ∈ N. Find the values of m and h.

To those who professional in Math, please solve this ... Thanks.

Did you know the constant term of (x-m)^2 must divide the constant term of p(x)?

 

[Deleted member 4993]

Did you know the constant term of (x-m)^2 must divide the constant term of p(x)?

Daon,

Are you assuming "m" is a rational number?

 

[Mrspi]

Daon,

Are you assuming "m" is a rational number?

I'm not Daon, but I don't think we need to ASSUME that "m" is rational; the conditions of the problem specify that "m is an element of N" and N is generally used to indicate the set of natural numbers.

 

[daon2]

I assumed m is a positive integer, but it is true in any commutative ring as long as not every constant is 0:

(a_nx^n+...+a_1x+a_0)(b_mx^m+...+b_1x+b_0) = (higher-degree terms) + a_0b_0

 

[Deleted member 4993]

I'm not Daon, but I don't think we need to ASSUME that "m" is rational; the conditions of the problem specify that "m is an element of N" and N is generally used to indicate the set of natural numbers.

Ooops .... I did not see that....

 

[lrtcoolman]

Reply

Did you know the constant term of (x-m)^2 must divide the constant term of p(x)?

Yes. (x-m)^2 is the factor of p(x).
As p(x) is a 4 degree equation. After expand the factor = (x^2 -2mx + m^2) ... need another quadratic factor = (x^2 + mx +c)

 

[lrtcoolman]

Reply

Daon,

Are you assuming "m" is a rational number?

m is a natural number which is a non-negative integers {0, 1, 2, 3, ...}

 

[lrtcoolman]

Reply

If x⁴ - 4x³ + hx² - 6x + 2 = 0 and x = 1,
then what does h equal?

h = 7...
GREAT Thanks!

 

[lrtcoolman]

Reply

If x⁴ - 4x³ + hx² - 6x + 2 = 0 and x = 1,
then what does h equal?

How you get the m = 1? comparing?

 

[mmm4444bot]

m is a natural number which is a non-negative integers {0, 1, 2, 3, ...}

In this thread, zero is not a Natural number. The lowest-possible value for m is 1.

 

[mmm4444bot]

How you get the m = 1? comparing?

You can get m = 1 by realizing that the constant term of (x-m)^2 must evenly divide the constant term of p.

You told Daon that you understand this fact, but did you actually write the ratio of constant terms and contemplate it?

If not, try -- and see if you can reason out that m must be 1.

 

[Deleted member 4993]

How you get the m = 1? comparing?

No by logic...

The constant term in p(x) is 2. The constant term in (x-m)² is m²

so m² must be a factor of 2.

m is a natural number - so - m = ????