Problem 36 - Precalculus : Building Concepts and Connections

Question

Determine together \(q(x)\) is a factor of \(p(x)\) Here, \(p(x)\) is the first polynomial and \(q(x)\) is the second polynomial. justify your answer. $$2 x^{5}-1 ; x-2$$

Step-by-Step Solution

Answer

No, \(q(x) = x - 2\) is not a factor of \(p(x) = 2x^5 - 1\).

Step 1: Identify the Root of q(x)

The root of \(q(x)\) is the value of \(x\) that makes \(q(x) = 0\). For our problem, this root is \(x = 2\), which is obtained by solving the quadratic equation \(x - 2 = 0\).

Step 2: Substitute the Root into p(x)

The next step is to substitute the root of \(q(x)\), which is \(2\), into \(p(x)\). Thus, we will find \(p(2)\), which is \(2*(2)^5 - 1\). Calculating this gives us \(63\).

Step 3: Analyze the Result

We see that \(p(2) = 63\), which is not equal to zero. According to the Factor Theorem, for \(q(x) = x - 2\) to be a factor of \(p(x) = 2x^5 - 1\), we would need \(p(2)\) to equal zero. Since \(p(2) \neq 0\), we can conclude that \(q(x) = x - 2\) is not a factor of \(p(x) = 2x^5 - 1\).

Key Concepts

Factor Theorem

The Factor Theorem is an essential concept in algebra that establishes a direct relationship between the roots of a polynomial and its factors. In simple terms, the theorem states that if a value of \(x = r\) is a root of the polynomial \(p(x)\), then \(x - r\) is a factor of \(p(x)\). Conversely, if \(x - r\) is a factor of \(p(x)\), then \(r\) will be a root of \(p(x)\).

Let's apply this theorem to an example: suppose we are given a polynomial \(p(x)\) and we wish to determine if \(q(x) = x - r\) is a factor of \(p(x)\).

• We must first identify the potential root of \(p(x)\), which is \(r\), by setting \(q(x) = 0\) and solving for \(x\).
• Next, we substitute this root back into \(p(x)\) to calculate \(p(r)\).
• If \(p(r) = 0\), then according to the Factor Theorem, \(q(x)\) is a factor of \(p(x)\). Otherwise, \(q(x)\) is not a factor.

In the exercise provided, we have \(q(x) = x - 2\) and \(p(x) = 2x^5 - 1\). By the theorem, the root \(2\) must satisfy \(p(2) = 0\) for \(x - 2\) to be a factor of \(p(x)\). However, we calculated \(p(2)\) and obtained \(63\), not \(0\). Thus, we conclude that \(x - 2\) is not a factor of \(p(x)\), illustrating the practical use of the Factor Theorem in polynomial factorization.

Polynomial Division

Polynomial division operates on the same basic principle as numerical division, but instead of numbers, it involves dividing polynomials. The goal is to determine whether one polynomial (the divisor) can be evenly divided into another (the dividend), without leaving a remainder. The result of such a division is a polynomial called the quotient, and the process can also help identify the remainder.

The division operation can be performed using either long division or synthetic division, depending on the complexity of the polynomials. These methods enable us to break down polynomials into simpler components, much like factoring. If the remainder is zero, it indicates that the divisor is indeed a factor of the dividend. If there is a non-zero remainder, the divisor is not a factor.

In the context of the exercise, to check if \(q(x) = x - 2\) is a factor of \(p(x) = 2x^5 - 1\), one would perform polynomial division. However, as per the Factor Theorem, since substituting the root \(x = 2\) into \(p(x)\) yields a non-zero remainder, there's no need for actual division — we already know that \(q(x)\) is not a factor.

Roots of Polynomial

The roots of a polynomial \(p(x)\) are the values of \(x\) that satisfy the equation \(p(x) = 0\). Finding these roots is a crucial aspect of understanding polynomial behavior, as they provide significant insights into the function defined by the polynomial.

Roots can be real or complex numbers, and the number of roots a polynomial has is related to its degree; a polynomial of degree \(n\) can have up to \(n\) roots, including multiplicity. The Fundamental Theorem of Algebra assures us that every non-constant polynomial has at least one complex root.

• A simple root is one that corresponds to a factor that crosses the x-axis only once.
• A repeated root corresponds to a factor that touches the x-axis and turns around, indicating a multiplicity greater than one.

For the exercise we are considering, we sought to determine if there was a root at \(x = 2\), which would have corresponded to a zero of \(p(x)\). The calculation of \(p(2)\) and the subsequent non-zero result informed us that \(2\) is not a root of the polynomial \(2x^5 - 1\), underlining the connection between factors and roots of a polynomial: a lack of a root at a certain point guarantees the absence of a corresponding linear factor.