Problem 34 - 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. $$-2 x^{4}-7 x^{3}+5 ; x+2$$

Step-by-Step Solution

Answer

After performing the Synthetic Division, the final result is non-zero. Therefore, \(x + 2\) is not a factor of the polynomial \(-2x^{4}-7x^{3}+5\).

Step 1: Setup Synthetic Division

Start by setting up synthetic division. Write down the coefficients of \(p(x)\) in decreasing order of their terms. In this case, those would be -2, -7, 0, 0, and 5. The zeros fill in for any missing terms in the polynomial. The value against which you'll divide these is the solution for \(q(x) = 0\), which is -2.

Step 2: Perform Synthetic Division

Begin Synthetic Division by bringing down the first coefficient (-2), then multiply by -2 (-2 x -2 = 4). Add this result to the second coefficient to get a new second coefficient (-7 + 4 = -3). Repeat this process for the remaining coefficients until you have gone through all coefficients of \(p(x)\).

Step 3: Analyze the Remainder

If the final result from the synthetic division is zero, it means that \(x + 2\) is a factor of the given polynomial. If the final result is non-zero, then \(x + 2\) is not a factor of the given polynomial.

Key Concepts

Polynomial Factorization

Factorizing polynomials is a crucial skill in precalculus that helps students to simplify, solve, and graph polynomial functions more easily. The process of polynomial factorization involves breaking down a complex polynomial into simpler, irreducible polynomials called factors that, when multiplied together, give back the original polynomial. It's like finding out what ingredients went into a cake so that you can understand how to make it yourself.

When we talk about a polynomial such as \(q(x)\) being a factor of another polynomial like \(p(x)\), we mean that if you divide \(p(x)\) by \(q(x)\), there will be no remainder; \(q(x)\) fits perfectly into \(p(x)\) a certain number of times. For instance, if \(x + 2\) were a factor of \( -2x^4 - 7x^3 + 5\), dividing the latter by the former would result in no remainders. Synthetic division is a handy tool for testing this relationship, as it allows us to divide polynomials in a more streamlined way than traditional long division.

Polynomials in Precalculus

Polynomials are an essential topic in precalculus, serving as building blocks for more complex mathematical concepts. A polynomial is an expression consisting of variables, coefficients, and exponents where the exponents are all whole numbers, and the coefficients are real numbers. They can look intimidating, but they are just a way of putting multiple varying quantities together to see how they add up. Precalculus introduces students to the different types of polynomials, like linear, quadratic, cubic, etc., based on their degree, which is the highest exponent in the polynomial.

Understanding the structure of a polynomial is key to mastering synthetic division. As seen in the exercise, the polynomial \(p(x) = -2x^4 - 7x^3 + 5\) might be missing some terms (like those of \(x^2\) and \(x\)) which we represent with zero coefficients for the synthetic division process. This thoroughness ensures that we consider every potential term up to the highest degree when applying synthetic division, providing clarity and accuracy in determining factors.

Remainder Theorem

The Remainder Theorem is a powerful tool that connects the dots between polynomials and their factors. It states that if a polynomial \(f(x)\) is divided by a linear divisor of the form \(x - c\), the remainder is the value \(f(c)\). In the context of synthetic division, applying the Remainder Theorem allows you to quickly check whether a given linear polynomial is a factor of another polynomial without doing long division.

In our example, we divide the polynomial \( -2x^4 - 7x^3 + 5\) by \(x + 2\) using synthetic division. According to the Remainder Theorem, if \(x + 2\) is indeed a factor of \(p(x)\), then \(p(-2)\) should equal zero. The synthetic division process follows this theorem step by step to reveal the remainder after division. If the remainder is anything other than zero, we can conclude that \(x + 2\) is not a factor of \(p(x)\), directly applying the insights provided by the Remainder Theorem.

Analyze the Remainder

Upon completing the synthetic division routine in the exercise, we look at the final value in the synthetic division array. This value is synonymous with the remainder obtained by substituting \(c\) into \(p(x)\) — it's our shortcut to uncovering whether \(q(x)\) is a factor of \(p(x)\), as prescribed by the Remainder Theorem.