Can x3 be factored into common quadratic factor for P(x) and Q(x)?

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The discussion focuses on finding integer values for m and n such that the polynomials P(x) and Q(x) share a common quadratic factor. It is established that setting m = n = 0 results in both polynomials having a common factor of x². However, participants suggest that there may be other solutions, such as m = 5 and n = 3, which also yield a common quadratic factor. The approach involves expressing both polynomials in terms of their factors and matching coefficients to solve for m and n. The conversation emphasizes the need for further exploration of potential solutions beyond the initial findings.
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Homework Statement


Find the value of m and n, where m and n are integer, so that P(x) = x3 + mx2 – nx - 3m and
Q(x) = x3 + (m – 2) x2 –nx – 3n have common quadratic factor.


Homework Equations





The Attempt at a Solution


Is m = n = 0 one of the solution?

If m = n = 0,then :
P(x) = x3 = x2 (x)

Q(x) = x3 - 2x2 = x2 (x-2)

Can I say that they have common quadratic factor, which is x2 ?

The point is : Am I right to say x3 can be be factorized to x2 x or even (x) (x) (x) ?

Thanks
 
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That works, but there might be another solution with both functions having a quadratic factor of x2 + bx + c. I've filled up a couple of pieces of paper without finding it, though.
 
m=5 and n=3 works.
 
If p(x) and q(x) have a common quadratic factor, they can be written

p(x) = (x+a)(x^2+cx+d)
q(x) = (x+b)(x^2+cx+d)

so that

p(x)-q(x) = (a-b)(x^2+cx+d).

Try plugging the given polynomials into the LHS and match coefficients to determine a, b, c, d, m, and n.
 
Hi Mark and vela

Thanksssss !
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...

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