Problem 30 - Algebra and Trigonometry with Analytic Geometry

Question

If a polynomial of the form $$ x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}, $$ where each \(a_{k}\) is an integer, has a rational root \(r\), show that \(r\) is an integer and is a factor of \(a_{0}\).

Step-by-Step Solution

Answer

A rational root of the given polynomial is an integer and is a factor of \( a_0 \).

Step 1: Understanding Rational Roots

If a polynomial has a rational root \( r = \frac{p}{q} \), we use the Rational Root Theorem, which states that \( p \) is a factor of the constant term \( a_0 \) and \( q \) is a factor of the leading coefficient (which is 1 for our polynomial).

Step 2: Applying the Rational Root Theorem

Since the leading coefficient of the polynomial is 1, \( q \) must be a factor of 1. Therefore, \( q = \pm 1 \), and the root \( r = \frac{p}{q} = p \), implying that \( r \) must be an integer.

Step 3: Establishing r as a Factor of a_0

Given that \( r = p \) is an integer, and \( p \) must be a factor of \( a_0 \) as per the Rational Root Theorem, \( r \) is both an integer and a factor of the constant term \( a_0 \).

Key Concepts

Rational Root Theorem

The Rational Root Theorem is a powerful tool in algebra for determining the possible rational roots of a polynomial. When working with a polynomial, such as \[ x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0 \]where all coefficients \(a_k\) are integers, the theorem can greatly simplify the search for rational solutions. It states that if \(r = \frac{p}{q}\) is a rational root of the polynomial, then \(p\) (the numerator) must be a factor of the constant term \(a_0\), while \(q\) (the denominator) must be a factor of the leading coefficient, which is the coefficient of \(x^n\).

• Useful for identifying potential rational roots.
• Saves time by reducing the possibilities to only integer factor candidates of \(a_0\) and the leading coefficient.

When the leading coefficient is 1, as in our example, the denominator \(q\) must be \(\pm 1\). This directly influences our next topic: integer roots.

Integer Root

An integer root is a specific type of root which is a whole number, making calculations and predictions much simpler. For the polynomial problem at hand, because of the Rational Root Theorem and the condition that the leading coefficient is 1, any rational root \(r\) must be an integer. This is because \(q\), the factor of the leading coefficient, is limited to \(\pm 1\).

• Since \(r = \frac{p}{1} = p\), \(r\) must be an integer.
• Integer roots are easy to verify and calculate.

Knowing that any rational root must first be an integer simplifies finding the roots, as only whole numbers need to be tested. It also ensures straightforward divisibility properties with the constant term, which we will explore further.

Factor of Constant Term

The factor of the constant term, often referred to as \(a_0\), plays a crucial role in identifying a polynomial's rational roots. By the Rational Root Theorem, any integer root \(r\) must be a factor of this constant term. This means if \(r\) is an integer solution to the polynomial equation, it should divide \(a_0\) without leaving a remainder.

• Finding factors of \(a_0\) helps determine possible integer roots efficiently.
• Only integer roots that are factors of \(a_0\) need to be considered.

Thus, when analyzing polynomials for roots, first list the integer factors of \(a_0\). Each of these factors is a candidate for an integer root. This approach streamlines root-finding by focusing only on feasible solutions, aligning with the theorem's criteria.