Problem 43 - College Algebra

Question

For the following exercises, use synthetic division to determine whether the first expression is a factor of the second. If it is, indicate the factorization. $$ x+\frac{1}{3}, 3 x^{4}+x^{3}-3 x+1 $$

Step-by-Step Solution

Answer

\(x + \frac{1}{3}\) is not a factor; remainder is 2, not zero.

Step 1: Setup Synthetic Division

First, we'll set up the synthetic division process for \(x + \frac{1}{3}\). Normally, with \(x - c\), we plug \(c\) into our process. But here, since \(x + \frac{1}{3}\) is written in the form of \(x - (-\frac{1}{3})\), we will use \(-\frac{1}{3}\). We'll list the coefficients of the polynomial \(3x^4 + x^3 + 0x^2 - 3x + 1\), which are \([3, 1, 0, -3, 1]\).

Step 2: Perform Synthetic Division

Bring down the leading coefficient (3). Multiply it by \(-\frac{1}{3}\) and add it to the next coefficient: \[3 \cdot \left( -\frac{1}{3} \right) = -1 \quad \text{then add to 1 to get 0.} \]Continue this process step by step:- Multiply \(0 \times -\frac{1}{3} = 0\), add to 0 (next coefficient) to get 0.- Multiply \(0 \times -\frac{1}{3} = 0\), add to -3 to get -3.- Multiply \(-3 \times -\frac{1}{3} = 1\), add to 1 to get 2.

Step 3: Determine if Remainder is Zero

The last number in the synthetic division row is the remainder. Our final row from the synthetic division is \([3, 0, 0, -3, 2]\). Since the remainder is 2, \(x + \frac{1}{3}\) is not a factor of the polynomial, as a factor would result in a remainder of 0.

Step 4: Conclusion

Since the remainder from the synthetic division is not zero, \(x + \frac{1}{3}\) is not a factor, and hence no factorization of \(3 x^{4}+x^{3}-3 x+1\) involving \(x + \frac{1}{3}\) can be written.

Key Concepts

Polynomial Division

Polynomial division is an essential technique in algebra, similar to long division with numbers. It's used to divide one polynomial by another, usually leading to results expressed as a quotient and a remainder. The main aim here is to determine how many times the divisor polynomial fits into the dividend polynomial.

There are two main methods of polynomial division: long division and synthetic division. While long division is straightforward, synthetic division offers a more simplified and quicker approach when dividing polynomials by linear expressions of the form \(x - c\). The method focuses on coefficients rather than entire terms, making calculations faster and less complicated.

Remainder Theorem

The Remainder Theorem is a powerful tool in algebra, relating the remainder of a polynomial division to the value of the polynomial at a specific point. If a polynomial \(f(x)\) is divided by \(x - c\), according to the theorem, the remainder is equal to \(f(c)\).

This theorem is particularly useful for quickly determining whether a linear expression is a factor of a polynomial. If the substitution of \(c\) into the polynomial results in zero, \((x - c)\) is a factor of the polynomial. This saves time by eliminating the need to perform complex division when trying to identify factors.

Factorization

Factorization involves expressing a polynomial as a product of its factors, which are polynomials of lesser degree. Understanding factorization is critical in solving polynomial equations and can simplify expressions for easier computation and analysis.

For example, to factor a polynomial means to write it as the multiplication of two or more polynomials. If a polynomial can be divided by a linear factor \((x - c)\) with no remainder, then \((x - c)\) is a factor of that polynomial. In turn, this information is used to reduce the polynomial into simpler, solvable components.

Algebraic Expressions

Algebraic expressions are mathematical phrases that can include numbers, variables, and operators. They form the foundation of algebraic reasoning and problem solving. A polynomial is a specific type of algebraic expression characterized by the sum of terms, each consisting of a variable raised to a non-negative integer power, multiplied by a coefficient.

Understanding algebraic expressions is key to manipulating and simplifying expressions, solving equations, and interpreting mathematical relationships. Concepts like like terms, distributive property, and operations with polynomials are all applied when working with algebraic expressions to make complex problems tractable.