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Chapter 1: Problem 59

Given any \(p(x) \in \mathbb{R}[x]\) and \(\alpha \in \mathbb{R}\), show that there is a unique polynomial \(q(x) \in \mathbb{R}[x]\) such that \(p(x)=(x-\alpha) q(x)+p(\alpha)\). Deduce that \(\alpha\) is a root of \(p(x)\) if and only if the polynomial \((x-\alpha)\) divides \(p(x)\).

Short Answer

Expert verified
Given any polynomial function \(p(x)\) and a constant \(\alpha\), we can perform polynomial long division by dividing \(p(x)\) by \((x-\alpha)\). Let the quotient be \(q(x)\) and the remainder be \(r(x)\). Notice that the degree of \(r(x)\) must be less than the degree of \((x-\alpha)\), so \(r(x)\) can only be a constant. On substituting \(x = \alpha\), we get \(p(\alpha) = r(\alpha)\), which proves the existence and uniqueness of the polynomial \(q(x)\). To deduce that \(\alpha\) is a root of \(p(x)\) if and only if the polynomial \((x-\alpha)\) divides \(p(x)\), we consider two cases: (i) If \(\alpha\) is a root of \(p(x)\), then \(p(\alpha) = 0\) and thus, \((x-\alpha)\) divides \(p(x)\). (ii) If \((x-\alpha)\) divides \(p(x)\), then \(p(伪) = (伪-伪)q(伪) = 0\), so \(\alpha\) is a root of \(p(x)\).

Step by step solution

01

(Step 1) Polynomial Long Division

Given any polynomial function p(x), perform polynomial long division by dividing p(x) by (x-伪). Let the quotient be q(x) and the remainder be r(x).
02

(Step 2) Analyzing Division Result

From the polynomial long division, we have: \[p(x) = (x-\alpha)q(x) + r(x)\] Now, let's focus on the degree of r(x). Notice that the degree of r(x) must be less than the degree of (x-伪), which is 1. Thus, r(x) can only be a constant.
03

(Step 3) Proof of Uniqueness for q(x)

On substituting x = 伪, we get: \[p(\alpha) = (伪-\alpha)q(\alpha) + r(\alpha)\] Since (伪-伪) = 0, this equation simplifies to: \[p(\alpha) = r(\alpha)\] So now we have: \[p(x) = (x-\alpha)q(x) + p(\alpha)\] This proves that there indeed exists a unique polynomial q(x) satisfying the given equation.
04

(Step 4) Deduce Conditions for 伪 to be Root of p(x)

Now, we need to deduce that 伪 is a root of p(x) if and only if the polynomial (x-伪) divides p(x). (i) If 伪 is a root of p(x), then p(伪) = 0. This means that the remainder, r(伪) = p(伪). So r(x) = 0, and thus, p(x) = (x-伪)q(x). This implies that (x-伪) divides p(x). (ii) If (x-伪) divides p(x), then p(x) = (x-伪)q(x) and r(x) = 0. In this case, p(伪) = (伪-伪)q(伪) = 0. So 伪 is a root of p(x). This deduces that 伪 is a root of p(x) if and only if the polynomial (x-伪) divides p(x).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polynomial Root
A polynomial root is a value that makes the polynomial equal zero. For a polynomial \(p(x)\), if \(p(\alpha) = 0\), then \(\alpha\) is called a root of the polynomial. This value implies that when substituting \(x\) with \(\alpha\), the entire expression of \(p(x)\) results in zero. Identifying the roots of a polynomial helps simplify the polynomial or even solve polynomial equations effectively.

Understandably, polynomial roots are key for:
  • Determining where a graph of the polynomial intersects the x-axis.
  • Factoring the polynomial into simpler components.
Finding a root means finding an \(\alpha\) such that substituting it into the polynomial gives us zero, which plays a crucial role in polynomial division and many other mathematical procedures.
Polynomial Remainder Theorem
The Polynomial Remainder Theorem provides a handy shortcut in polynomial division. It states that if a polynomial \(p(x)\) is divided by \(x-\alpha\), the remainder is \(p(\alpha)\). In simpler terms, when you substitute \(\alpha\) into your polynomial, the result will be the remainder of the division.

Here's how it works in practice:
  • If you divide \(p(x)\) by \(x-\alpha\), your expression becomes: \(p(x) = (x-\alpha)q(x) + r(x)\).
  • By substituting \(x = \alpha\), the left part \((x-\alpha)q(x)\) becomes zero, leaving \(r(x) = p(\alpha)\).
This theorem is extremely useful because it saves time. Rather than calculating the remainder through long division step-by-step, you can just plug in the value. Plus, if \(p(\alpha) = 0\), you effortlessly find that \(\alpha\) is indeed a root.
Root-Factor Theorem
The Root-Factor Theorem builds on the concepts of polynomial roots and factors. If \(\alpha\) is a root of a polynomial \(p(x)\), then \(p(x)\) can be expressed as \((x-\alpha)q(x)\), showing that \(x-\alpha\) is a factor of \(p(x)\). This theorem tells us how roots connect to the factors of the polynomial.

This has several implications:
  • If \(\alpha\) is a root, then dividing \(p(x)\) by \(x-\alpha\) will yield no remainder (remainder = 0).
  • This means the polynomial can be decomposed into linear factors, making it simpler to analyze or solve.
Understanding the Root-Factor Theorem is crucial for verifying if a number is a root and consequently simplifying or solving polynomials efficiently.

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Most popular questions from this chapter

Prove that the function \(f:[0, \infty) \rightarrow \mathbb{R}\) defined by the following is an algebraic function: (i) \(f(x)=1+\sqrt[3]{x}\) (ii) \(f(x)=\sqrt{x}+\sqrt{2 x}\), (iii) \(f(x)=\sqrt{x}+\sqrt[3]{x}\).

Show that \(n ! \leq 2^{-n}(n+1)^{n}\) for every \(n \in \mathbb{N}\), and that equality holds if and only if \(n=1\).

Give an example of a nonconstant function \(f:(-1,1) \rightarrow \mathbb{R}\) such that \(f\) has a local extremum at 0, and 0 is a point of inflection for \(f\).

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Let \(p(x), q(x) \in \mathbb{R}[x]\) be nonzero polynomials of degrees \(m\) and \(n\) respectively. Let \(e \in \mathbb{N}\) be such that \(m
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