P(x) is a polynomial of even degree. Also, all the coefficients of P(x) are odd numbers.
Is it possible for P(x) to have a rational root?
• If so, provide an example.
• If not, prove that it is NOT possible for P(x) to have a rational root.
Start with a quadratic with leading coefficient a=1 and roots {p,q}:
x^2+bx+c = (x-p)(x-q)
If the roots are rational they must be integers, such that b=-(p+q) and c=pq
For c to be odd, p and q are both odd, but then p+q is even.
For higher even degree, 2n, we get the same contradiction:
The constant term is the product of pq... from which they are all odd, but their sum (since there are an even number of them) is even, but b (from bx^(2n-1), the second term) must also be odd.
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Posted by Jer
on 2023-10-30 09:15:06 |
Idea for a=1
