Prove that if p, q, r and s are odd integers, then the equation
x^10 + p x^9 - q x^7 + r x^4 - s = 0
has no integer roots.
To prove that the equation has no integer roots, we can use the Rational Root Theorem in combination with parity analysis.
The Rational Root Theorem states that if a polynomial equation with integer coefficients has a rational root, it must be of the form p/q, where p is a factor of the constant term and q is a factor of the leading coefficient.
Let's analyze the given equation: x^10 + px^9 - qx^7 + rx^4 - s = 0.
1. Parity analysis:
Since p, q, r, and s are odd integers, let's consider the parity (even or odd) of each term in the equation:
- The first term, x^10, is an even power of x, so it is always positive regardless of the value of x.
- The second term, px^9, is an odd power of x multiplied by an odd integer p. Therefore, this term can take both positive and negative values depending on the value of x.
- The third term, qx^7, is an odd power of x multiplied by an odd integer q. Thus, this term can also be positive or negative depending on the value of x.
- The fourth term, rx^4, is an even power of x multiplied by an odd integer r. Hence, this term can take both positive and negative values depending on the value of x.
- The constant term, -s, is a negative odd integer.
2. Possible rational roots:
By applying the Rational Root Theorem, the possible rational roots of the equation are all the factors of -s (the constant term) divided by all the factors of 1 (the leading coefficient):
Possible rational roots: ±[factors of -s] / [factors of 1]
However, since all the factors of 1 are ±1 and all the factors of -s are ±1 and ±s, we can conclude that the only possible rational roots are ±1.
3. Evaluating for x = ±1:
Let's substitute x = ±1 into the given equation and analyze the result:
- For x = 1: 1^10 + p(1)^9 - q(1)^7 + r(1)^4 - s = 1 + p - q + r - s
- For x = -1: (-1)^10 + p(-1)^9 - q(-1)^7 + r(-1)^4 - s = 1 - p + q - r - s
From the parity analysis, we know that the second, third, and fourth terms can take both positive and negative values, while the first and the constant terms are always positive and negative, respectively.
Since the sum of these terms for x = 1 and x = -1 is not equal to zero, the equation x^10 + px^9 - qx^7 + rx^4 - s = 0 has no integer roots.
Therefore, we have proven that if p, q, r, and s are odd integers, the equation x^10 + px^9 - qx^7 + rx^4 - s = 0 has no integer roots.