Problem 82 - Algebra and Trigonometry

Question

Show that the polynomial does not have any rational zeros. $$ P(x)=x^{50}-5 x^{25}+x^{2}-1 $$

Step-by-Step Solution

Answer

The polynomial has no rational zeros as neither \(x = 1\) nor \(x = -1\) is a root.

Step 1: Identify the Rational Root Theorem

The Rational Root Theorem suggests that any rational root of a polynomial equation with integer coefficients, \(P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_0\), is of the form \(\frac{p}{q}\), where \(p\) is a divisor of the constant term \(a_0\) and \(q\) is a divisor of the leading coefficient \(a_n\).

Step 2: Apply the Rational Root Theorem

In the given polynomial \(P(x) = x^{50} - 5x^{25} + x^2 - 1\), the constant term \(a_0\) is \(-1\) and the leading coefficient \(a_n\) is \(1\). The possible rational roots are thus the divisors of \(-1\), which are \(\pm 1\).

Step 3: Test Possible Roots

Test \(x = 1\) by substituting it into the polynomial: \[ P(1) = 1^{50} - 5 \times 1^{25} + 1^2 - 1 = 1 - 5 + 1 - 1 = -4. \] Thus, \(x = 1\) is not a root. Now, test \(x = -1\): \[ P(-1) = (-1)^{50} - 5 \times (-1)^{25} + (-1)^2 - 1 = 1 + 5 + 1 - 1 = 6. \] Thus, \(x = -1\) is also not a root.

Step 4: Conclusion

Since neither \(x = 1\) nor \(x = -1\) is a zero of \(P(x)\), and these are the only possible rational roots allowed by the Rational Root Theorem, the polynomial \(P(x) = x^{50} - 5x^{25} + x^2 - 1\) does not have any rational zeros.

Key Concepts

Polynomial Equations

Polynomial equations are mathematical expressions involving a sum of powers of one or more variables multiplied by coefficients. The variable is often denoted by \( x \). In our context, we are focusing on a single-variable polynomial equation.
The general form of a polynomial equation can be written as:\[P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0\]where,

• \( a_n, a_{n-1}, \ldots, a_1, a_0 \) are coefficients, which are numbers like integers or real numbers.
• \( n \) represents the degree of the polynomial, which is the highest power of the variable in the equation.

Polynomial equations can describe a wide range of phenomena and are widely used in mathematics and science. For example, the equation given in the exercise is:\[P(x) = x^{50} - 5 x^{25} + x^2 - 1\]Here, the highest degree is 50, indicating this polynomial is of degree 50.

Integer Coefficients

Integer coefficients in a polynomial mean that all the numbers multiplying the powers of the variable \( x \) are integers. Integer means whole numbers, which can be positive, negative, or zero.
Why does this matter? It’s critical because integer coefficients allow us to apply certain mathematical theorems, like the Rational Root Theorem, to find solutions more easily.

• In the example polynomial equation \( x^{50} - 5x^{25} + x^2 - 1 \), the coefficients are 1, -5, 1, and -1, all of which are integers.
• This polynomial, therefore, qualifies for techniques focusing on integer calculations.

In scenarios where the coefficients are non-integers, additional rules or numerical methods may be required to solve or analyze the equation.

Rational Roots

Rational roots are potential solutions to polynomial equations that can be expressed as a fraction \( \frac{p}{q} \), where both \( p \) and \( q \) are integers, and \( q eq 0 \). The Rational Root Theorem is a useful tool in determining the possible rational roots of a polynomial equation.
This theorem stipulates that if a polynomial with integer coefficients has any rational root \( \frac{p}{q} \), then:

• \( p \) must be a divisor of the constant term \( a_0 \).
• \( q \) must be a divisor of the leading coefficient \( a_n \).

To apply this to the polynomial \( x^{50} - 5x^{25} + x^2 - 1 \):

• The constant term \( a_0 \) is \(-1\), and the divisors of \(-1\) are \( \pm 1 \).
• The leading coefficient \( a_n \) is 1, whose divisors are also \( \pm 1 \).
• Thus, the only possible rational roots are \( x = 1 \) and \( x = -1 \).

Upon testing these potential roots, neither \( x = 1 \) nor \( x = -1 \) gives zero when substituted into the polynomial, confirming there are no rational roots for the given polynomial.