Problem 49

Question

Suppose that $p$ is a polynomial. We say that $\alpha$ is a root of $p$ of multiplicity $m$ if there is a polynomial $q$ with $q(\alpha) \neq 0$ and $p(x)=(x-\alpha)^{m} q(x)$. Prove that if $\alpha$ is a root of $p$ of multiplicity $m,$ then $\alpha$ is a root of $p^{\prime}$ of multiplicity $m-1$.

Step-by-Step Solution

Verified

Answer

If $ \alpha $ is a root of $ p $ with multiplicity $ m $, then $ \alpha $ is a root of $ p' $ with multiplicity $ m-1 $ due to factor structure and differentiation.

1Step 1: Understand the Definition of Multiplicity

We begin by reviewing the definition: A root $ \alpha $ of a polynomial $ p $ is said to have multiplicity $ m $ if there exists a polynomial $ q(x) $, where $ q(\alpha) eq 0 $, such that $ p(x) = (x-\alpha)^m q(x) $. This means that the root $ \alpha $ appears in the factorization of $ p(x) $ exactly $ m $ times.

2Step 2: Apply Product Rule of Differentiation

To find the derivative $ p'(x) $, we use the product rule of differentiation: If $ f(x) = u(x) v(x) $, then $ f'(x) = u'(x)v(x) + u(x)v'(x) $. Set $ u(x) = (x-\alpha)^m $ and $ v(x) = q(x) $ in $ p(x) = (x-\alpha)^m q(x) $. Hence, \[ p'(x) = m(x-\alpha)^{m-1}q(x) +(x-\alpha)^m q'(x). \]

3Step 3: Determine Multiplicity of $ \alpha $ in $ p'(x) $

The expression $ m(x-\alpha)^{m-1}q(x) +(x-\alpha)^m q'(x) $ shows that 1. $ (x-\alpha)^{m-1} $ is a factor in the first term, and 2. $ (x-\alpha)^m $ is a factor in the second term.The multiplicity of $ \alpha $ in $ p'(x) $ must be the smallest power of $ (x-\alpha) $ among the terms, which is $ m-1 $, ensuring $ \alpha $ is a root of multiplicity $ m-1 $.

4Step 4: Verification of Condition $ q(\alpha) \neq 0 $

Since $ q(\alpha) eq 0 $ originally by definition, $ m (x-\alpha)^{m-1} q(x) $ does not cancel at $ \alpha $ unless $ m-1 $ factors of $ (x-\alpha) $ cancel itself completely. Thus, the non-zero $ q(\alpha) $ keeps $ \alpha $ as a root of $ p'(x) $ in multiplicity $ m-1 $.Mathematically this implies the existence of $ q' $ where terms align as required as $ m $ decreases to $ m-1 $.

Key Concepts

Polynomial DifferentiationProduct Rule of DifferentiationRoots of Polynomial

Polynomial Differentiation

Polynomial differentiation is a process where we find the derivative of a polynomial function. A polynomial is simply an expression that contains variables raised to whole number powers and coefficients. Differentiation helps us understand how a polynomial changes or evolves at any particular point in its domain.
During differentiation, the power of each term's variable is reduced by one, as reflected in the derivative formula. Let's say you have a polynomial function like this: \[f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\]Then, the derivative, denoted as $f'(x)$, is given by:
\[f'(x) = n a_nx^{n-1} + (n-1) a_{n-1}x^{n-2} + ... + a_1\]The derivative effectively reduces the degree of each variable term by one and multiplies by the original power, allowing us to analyze the behavior of the polynomial at different values of $x$. This process is fundamental for finding slopes of tangents, identifying local maxima or minima, and examining a function's continuity.
Differentiation gives us valuable insights into polynomials beyond their roots, helping highlight aspects such as rate of change and concavity.

Product Rule of Differentiation

The product rule of differentiation is a key tool when differentiating functions that are products of two or more other functions. This rule states that if you have a function $f(x)$ defined as a product of two functions $u(x)$ and $v(x)$, then the derivative $f'(x)$ is given as:
\[f'(x) = u'(x)v(x) + u(x)v'(x)\]This rule is crucial because it allows us to differentiate complex expressions that result from multiplying simpler functions.

Applying the Product Rule

Let's imagine you have a polynomial $p(x) = (x-\alpha)^m q(x)$. To find $p'(x)$, you would set $u(x) = (x-\alpha)^m$ and $v(x) = q(x)$. Applying the product rule yields:
\[p'(x) = m(x-\alpha)^{m-1}q(x) + (x-\alpha)^m q'(x)\]This result shows how the expression's differentiation factors directly relate to the polynomial's terms and structure. Using the product rule, we can effectively handle cases, like the exercise, where functions within polynomials are multiplied, leading to accurate differentiation.

Roots of Polynomial

The roots of a polynomial are the values of $x$ for which the polynomial evaluates to zero. If you have a polynomial $p(x) = a_nx^n + ... + a_0$, finding its roots means solving $p(x) = 0$.

Multiplicity of Roots

Multiplicity refers to the number of times a root appears in the polynomial. If a polynomial $p(x)$ can be expressed as $(x-\alpha)^m q(x)$, then $\alpha$ is a root with multiplicity $m$. This means $\alpha$ will satisfy $p(x) = 0$ exactly $m$ times.
In our exercise, we see that if $\alpha$ is a root of $p(x)$ with multiplicity $m$, then the derivative $p'(x)$ reflects this in having a root $\alpha$ with multiplicity $m-1$, showcasing how differentiation reduces the repeating nature of a root by one.

The multiplicity impacts the shape of the graph of the polynomial at the root.
Higher multiplicity roots indicate that the polynomial "touches" or "flattens" at the x-axis.

Understanding roots and their multiplicity is essential for graphing polynomials and studying their characteristics.

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