Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

Use the factor theorem to show that \(x-c\) is a factor of \(f(x)\). \(12 f(x)=x^{4}+3 x^{2}-12\) \(c=-2\)

Step-by-Step Solution

Answer

\(x + 2\) is not a factor of \(f(x)\) because \(f(-2) \neq 0\).

Step 1: Understand the Factor Theorem

The factor theorem states that \( x-c \) is a factor of \( f(x) \) if and only if \( f(c) = 0 \). This means we need to plug in \( c \) into the polynomial and check if the result is zero.

Step 2: Simplify the Given Function

The given function is \( 12f(x) = x^4 + 3x^2 - 12 \). To work with this, divide both sides by 12 to get \( f(x) = \frac{x^4 + 3x^2 - 12}{12} \).

Step 3: Substitute \( c = -2 \) into \( f(x) \)

Substitute \( x = -2 \) in the function \( f(x) \), which gives us \( f(-2) = \frac{(-2)^4 + 3(-2)^2 - 12}{12} \).

Step 4: Compute \( f(-2) \)

Calculate each term:1. \((-2)^4 = 16\)2. \(3(-2)^2 = 3 imes 4 = 12\)3. Substitute these into the equation: \ \(f(-2) = \frac{16 + 12 - 12}{12} = \frac{16}{12}\)4. Simplify \( \frac{16}{12} = \frac{4}{3} \) which is not equal to zero.

Step 5: Conclude the Factor Verification

Since \( f(-2) \) is not zero \( \left( \frac{4}{3} eq 0 \right) \), by the factor theorem, \( x + 2 \) is not a factor of \( f(x) \). This is because substituting \( c = -2 \) did not result in zero.

Key Concepts

Polynomial Division

Polynomial division is a technique similar to long division but applied to polynomials. It allows you to divide one polynomial by another and find the quotient and remainder. This method is especially practical when simplifying expressions or when factoring polynomials.
For instance, the polynomial division was indirectly used when we divided the polynomial by 12 to simplify the function:

• Original polynomial: \(12f(x) = x^4 + 3x^2 - 12\)
• To make the function manageable, divide both sides by 12: \(f(x) = \frac{x^4 + 3x^2 - 12}{12}\)

Here, the purpose was not to find a remainder but to express \(f(x)\) in a simplified form. This step was crucial before applying the substitution method in our factor theorem test.

Substitution Method

The substitution method involves plugging a specific value into a formula to calculate a result. In the context of the factor theorem exercise, you substitute the value of \(c\) into the polynomial, which helps determine whether \(x-c\) is a factor.

• Substitute \(c = -2\) into the simplified polynomial function \(f(x)\)
• Calculate \(f(-2) = \frac{(-2)^4 + 3(-2)^2 - 12}{12}\)

The substitution method allowed for direct evaluation of the polynomial at a specific point, in this case, \(x = -2\). The primary goal was to see if this substitution results in zero, which would indicate that \(x + 2\) is a factor of \(f(x)\). However, the outcome \(\frac{4}{3}\) showed that \(x + 2\) is not a factor, as the value was not zero.

Roots of Polynomials

Roots of a polynomial are values of \(x\) for which the polynomial equals zero. They are essential for understanding the behavior and factorization of polynomials. In connection with the factor theorem, identifying roots helps establish the possible linear factors of a polynomial.
When using the factor theorem:

• A root for \(f(x)\) at \(x = c\) means \(f(c) = 0\)
• If \(f(c) = 0\), then \(x-c\) is a factor

In this specific exercise, searching for roots revolves around substituting \(c = -2\) and checking if the polynomial becomes zero. Since \(f(-2)\) yielded \(\frac{4}{3}\) instead of zero, \(-2\) is not a root of the polynomial, affirming that \(x + 2\) is not a factor according to the factor theorem.