Polynomials and real numbers

Question

If two polynomial equations have real solutions, then will the equation that is the result of adding, subtracting, or multiplying the two polynomial equations also have real solutions? Will the resulting equation have complex solutions if one or both of the original equations have complex solutions?

Answer 1

FALSE!!!

x^2 + x - 6 = ( x + 3)(x - 2) has in fact two INTEGER solutions, namely x=-3 and x=2

-x + 10 has in fact one integer solution, namely x=10.

But x^2 + x - 6 + -x + 10 = x^2 + 4 which has only two IMAGINARY (not real) solutions.

This counterexample can be altered for subtraction, namely by subtracting (x-10) which

has the same integer solution 10

Multiplication is proven by contradiction. That is, Suppose the polynomials P(x) and Q(x)

have real solutions (not necessarily rational) while P(x)Q(x) have imaginary/complex

solutions.

P(x) = (x- m1)(x-m2)(x-m3).... (x-xm) where m1,m2,m3,... xm are the m solutions of P

Q(x) = (x-n1)x-n2)(x-n3)... (x- xn) where n1,n2,n3,....xn are the n solutions of Q

Their product creates a polynomial of m+n real solutions which contradicts the assumption.

For the converse, that is the polynomials having complex/imaginary solutions:

2x^2+8 and x^2 + 12 have imaginary/complex solutions, but their difference

x^2 - 4 has two integer solutions, namely +or- 2. Addition can be disproven again by changing the sign.

Multiplication on the other hand can be disproven by multiplying by conjugates.

Polynomials (x-3i )(x+3i) and (x-2i)(x+2i)(x-1) for example have complex/imaginary solutions

while their product has one integer solution x=1