Problem 15 - Algebra and Trigonometry with Analytic Geometry

Question

Use the factor theorem to show that \(x-c\) is a factor of \(f(x)\). f(x)=x^{4}-6 x^{2}+4 x-8 ; \quad c=-3$

Step-by-Step Solution

Answer

\(x + 3\) is not a factor of \(f(x)\) because \(f(-3) \neq 0\).

Step 1: Identify the Given Polynomial and Value of c

We are given the polynomial \( f(x) = x^4 - 6x^2 + 4x - 8 \) and need to show that \( x + 3 \) is a factor (since \( c = -3 \) makes \( x - c = x + 3 \)).

Step 2: Understand the Factor Theorem

The factor theorem states that \( x - c \) is a factor of \( f(x) \) if and only if \( f(c) = 0 \). We need to substitute \( c = -3 \) into the polynomial and see if it equals zero.

Step 3: Substitute c into the Polynomial

Substitute \( x = -3 \) into \( f(x) \):\[ f(-3) = (-3)^4 - 6(-3)^2 + 4(-3) - 8 \].

Step 4: Calculate Each Term

1. Calculate \((-3)^4 = 81\) 2. Calculate \(-6(-3)^2 = -6 imes 9 = -54\) 3. Calculate \(4(-3) = -12\) 4. The constant term \(-8\) remains as \(-8\).

Step 5: Sum the Calculated Terms

Sum all terms: \[ 81 - 54 - 12 - 8 = 7 \].

Step 6: Conclusion from the Calculations

Since \( f(-3) = 7 \), which is not zero, this means \( x + 3 \) is not a factor of \( f(x) \).

Key Concepts

Polynomial Division

Polynomial division is similar to the long division method you might remember from arithmetic, but it deals with polynomials instead of numbers. The core idea is to divide a polynomial by another polynomial to get a quotient and a remainder. Here's how it applies:

• Dividend: The given polynomial you're working with, in this case, \( f(x) = x^4 - 6x^2 + 4x - 8 \).
• Divisor: This would be the polynomial you're dividing by, often in the form \( x-c \), where \( c \) is a constant. For our example, it's \( x+3 \) because \( c = -3 \).
• Quotient: The result of the division before considering the remainder.
• Remainder: What's left over after division, which should be zero if \( x-c \) is truly a factor.

The division process involves taking parts of the dividend step-by-step, starting with the term of the highest degree and finding how many times the leading term of the divisor fits into it. This mechanism can help simplify polynomials and find factors manually. Back to our exercise, you could attempt polynomial division to verify that \( x+3 \) isn't a factor by showing a non-zero remainder.

Remainder Theorem

The remainder theorem is a nifty shortcut in polynomial math. It states that if you divide a polynomial \( f(x) \) by a linear divisor \( x-c \), the remainder you'll get is \( f(c) \). You can see why this is so useful:

• No need for long tedious division if you're just after the remainder.
• Helps quickly check if \( x-c \) is a factor, as you simply evaluate \( f(c) \). If it's zero, then \( x-c \) is indeed a factor.

To apply it to our exercise, we calculated \( f(-3) \) by substituting \( -3 \) into the polynomial \( f(x) = x^4 - 6x^2 + 4x - 8 \). This gave us a remainder of 7, confirming that \( x+3 \) is not a factor, since the remainder is not zero. The theorem's power lies in quickly determining factor presence without full division.

Polynomial Functions

Polynomial functions form the backbone of algebra, characterized by the presence of variables raised to whole number powers. Each term in a polynomial function has a coefficient and a power, like \( ax^n \). Some key highlights about polynomial functions include:

• Terms are ordered by descending powers.
• They can be of different degrees, determined by the highest power of \( x \).
• Polynomials are incredibly versatile and appear in equations, graphs, and real-world applications.

In the context of the exercise, we dealt with a fourth-degree polynomial \( f(x) = x^4 - 6x^2 + 4x - 8 \), which means the degree is 4 since the highest power of \( x \) is 4. Polynomial functions can describe complex relationships, and operations like factorization can simplify them for easier handling and interpretation. Understanding the structure and properties of polynomial functions opens up a vast realm of mathematical exploration and application.