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If polynomials \(f(x)\) and \(g(x)\) satisfy \(f(a)=g(a),\) show that \(f(x)-g(x)=(x-a) h(x)\) for some polynomial \(h(x)\).

Short Answer

Expert verified
If \(f(a) = g(a)\), then \(f(x) - g(x) = (x-a) h(x)\) by the factor theorem.

Step by step solution

01

Expand and Set Up Given Information

Given that the polynomials \(f(x)\) and \(g(x)\) satisfy \(f(a) = g(a)\), we can rewrite this by subtracting \(g(a)\) from both sides, yielding \(f(a) - g(a) = 0\). Thus, \(f(x) - g(x)\) must be zero at \(x = a\).
02

Factor the Difference

Since \(f(x) - g(x)\) equals zero when \(x = a\), this suggests that \(x - a\) is a factor of \(f(x) - g(x)\). According to the factor theorem, \(f(x) - g(x) = (x-a) h(x)\) for some polynomial \(h(x)\).
03

Determine the Existence of Polynomial \(h(x)\)

Because \(x - a\) is a factor of \(f(x) - g(x)\), we can express \(f(x) - g(x) = (x-a) h(x)\). By dividing \(f(x) - g(x)\) by \(x-a\), we get \(h(x)\) as a polynomial quotient. Polynomial division ensures that \(h(x)\) exists unless \(f(x) - g(x)\) is a zero polynomial, in which case \(h(x)\) is still valid as the zero polynomial.

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Key Concepts

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

Factor Theorem
The Factor Theorem is a vital concept when dealing with polynomial factorization. It helps us determine when a polynomial is divisible by a linear factor. Specifically, it states that a polynomial \( P(x) \) has \( x-a \) as a factor if and only if \( P(a) = 0 \). This concept is incredibly useful because it provides a straightforward way to identify factors through evaluation at specific points.
In our problem, since \( f(a) = g(a) \), it implies that \( f(x) - g(x) \) is zero when \( x = a \). By the Factor Theorem, \( x-a \) is a factor of the polynomial \( f(x) - g(x) \).
This understanding lays the groundwork for expressing the difference \( f(x) - g(x) \) in a factorized form, utilizing the linear factor \( x-a \). Once we see that \( f(x) - g(x) \) equals zero at \( x = a \), we conclude that a polynomial \( h(x) \) exists such that \( f(x) - g(x) = (x-a)h(x) \). This expression showcases the core application of the Factor Theorem in simplifying polynomial expressions.
Polynomial Division
Polynomial division is similar to numerical long division, except it deals with variables instead of numbers. It's a process used to divide one polynomial by another, potentially simpler polynomial. When we encounter a statement like \( f(x) - g(x) = (x-a)h(x) \), polynomial division helps us formally find the polynomial \( h(x) \).
In the context of our exercise, once \( x-a \) is identified as a factor of \( f(x) - g(x) \), dividing \( f(x) - g(x) \) by \( x-a \) allows us to determine the polynomial \( h(x) \). This step validates the existence of \( h(x) \) and helps in breaking down complex expressions into more manageable parts.
Understanding polynomial division is crucial because it not only provides proofs, like showing the existence of \( h(x) \), but also simplifies solving polynomial expressions. If during division the remainder is zero, \( x-a \) is indeed a factor, which confirms that the division was successful and fits into our equation neatly.
Roots of Polynomials
Finding the roots of a polynomial is about identifying the values of \( x \) that make the polynomial equal to zero. The roots are directly connected to the Factor Theorem and Polynomial Division, as they are the points where the polynomial intersects the x-axis.
In terms of our given exercise, identifying \( f(a) - g(a) = 0 \) implies that \( a \) is a root of the polynomial \( f(x) - g(x) \). This discovery is significant, as each root corresponds to a factor of \( x-a \), contributing to the factorization of the polynomial.
Understanding the roots helps in breaking down the polynomial expressions, making it easier to factorize them into linear components like \( (x-a) \). Once the roots are found, one can use rules like the Factor Theorem to confirm and validate the existence of polynomial expressions that can simplify complex mathematical problems.

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Most popular questions from this chapter

\(\mathbf\). Let \(B=\left\\{\mathbf{v}_{1}, \ldots, \mathbf{v}_{n}\right\\}\) be a minimal spanning set for a vector space \(V\). That is, \(V\) cannot be spanned by fewer than \(n\) vectors. Show that \(B\) is a basis of \(V\)

Consider the differential equation \(f^{\prime}+a f^{\prime}+b f=g,\) where \(g\) is some fixed function. Assume that \(f_{0}\) is one solution of this equation. a. Show that the general solution is \(c f_{1}+d f_{2}+f_{0}\), where \(c\) and \(d\) are constants and \(\left\\{f_{1}, f_{2}\right\\}\) is any basis for the solutions to \(f^{\prime \prime}+a f^{\prime}+b f=0\). b. Find a solution to \(f^{\prime \prime}+f^{\prime}-6 f=2 x^{3}-x^{2}-2 x\).

A polynomial \(p(x)\) is called even if \(p(-x)=p(x)\) and odd if \(p(-x)=-p(x)\). Let \(E_{n}\) and \(O_{n}\) denote the sets of even and odd polynomials in \(\mathbf{P}_{n}\). a. Show that \(E_{n}\) is a subspace of \(\mathbf{P}_{n}\) and find \(\operatorname{dim} E_{n}\). b. Show that \(O_{n}\) is a subspace of \(\mathbf{P}_{n}\) and find \(\operatorname{dim} O_{n}\).

Let \(D_{n}\) denote the set of all functions \(f\) from the set \(\\{1,2, \ldots, n\\}\) to \(\mathbb{R}\). a. Show that \(\mathbf{D}_{n}\) is a vector space with pointwise addition and scalar multiplication. b. Show that \(\left\\{S_{1}, S_{2}, \ldots, S_{n}\right\\}\) is a basis of \(\mathbf{D}_{n}\) where, for each \(k=1,2, \ldots, n,\) the function \(S_{k}\) is defined by \(S_{k}(k)=1,\) whereas \(S_{k}(j)=0\) if \(j \neq k\).

Let \(D, I,\) and \(X\) denote finite, nonempty sets of vectors in a vector space \(V\). Assume that \(D\) is dependent and \(I\) is independent. In each case answer yes or no, and defend your answer. a. If \(X \supseteq D,\) must \(X\) be dependent? b. If \(X \subseteq D,\) must \(X\) be dependent? c. If \(X \supseteq I,\) must \(X\) be independent? d. If \(X \subseteq I,\) must \(X\) be independent?

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