Problem 38 - Precalculus: Functions and Graphs

Question

Use synthetic division to show that \(c\) is a zero of \(f(x)\). $$f(x)=27 x^{4}-9 x^{3}+3 x^{2}+6 x+1 ; \quad c=-\frac{1}{3}$$

Step-by-Step Solution

Answer

\(c = -\frac{1}{3}\) is not a zero of \(f(x)\) as the remainder is \(-\frac{2}{3}\).

Step 1: Write down the coefficients

List the coefficients of the polynomial function \(f(x) = 27x^4 - 9x^3 + 3x^2 + 6x + 1\). They are \(27, -9, 3, 6, 1\).

Step 2: Set up synthetic division

Place \(c = -\frac{1}{3}\) on the left and the coefficients \(27, -9, 3, 6, 1\) on the right in the synthetic division tableau.

Step 3: Bring down the leading coefficient

Bring down the leading coefficient \(27\) to start the division process.

Step 4: Perform synthetic division

Multiply \(-\frac{1}{3}\) by the number brought down, add this result to the next coefficient, and continue this process:- Multiply: \(-\frac{1}{3} imes 27 = -9\) - Add \(-9\) to \(-9\) to get \(0\).- Multiply: \(-\frac{1}{3} imes 0 = 0\)- Add \(0\) to \(3\) to get \(3\).- Multiply: \(-\frac{1}{3} imes 3 = -1\)- Add \(-1\) to \(6\) to get \(5\).- Multiply: \(-\frac{1}{3} imes 5 = -\frac{5}{3}\)- Add: \(-\frac{5}{3}\) to \(1\) results in \(-\frac{2}{3}\).

Step 5: Analyze the remainder

The last number in the row is the remainder of the division. Here, the remainder is \(-\frac{2}{3}\). Since it is not zero, \(c = -\frac{1}{3}\) is not a zero of \(f(x)\).

Key Concepts

Polynomial Function

A polynomial function is an expression involving variables and coefficients, consisting of terms in the form of ax^n, where 'a' is the coefficient and 'n' is a non-negative integer. For example, in the polynomial function given in the exercise, \[ f(x) = 27x^4 - 9x^3 + 3x^2 + 6x + 1 \] this polynomial is of degree 4 because the highest power of the variable 'x' is 4. The coefficients are 27, -9, 3, 6, and 1 corresponding to the terms in the polynomial. Polynomial functions are used to model a wide range of real-world scenarios including economic models, physics equations, and in engineering. One crucial aspect of working with polynomials is to determine their zeros, which is essential in solving equations and understanding graph behaviors. To solve for or validate the zeros, techniques such as synthetic division are often utilized, offering a quick and reliable method for dividing polynomials by a binomial of the form (x - c).

Zero of the Function

A zero of a function is any value of 'x' that makes the function equal to zero. Hence, for a given polynomial function like \( f(x) \), a zero will satisfy the equation \( f(c) = 0 \), implying the root (c) is a solution where the graph of the polynomial intersects the x-axis. Finding the zeros of a polynomial is vital as it helps in constructing its factorized form and aids in analyzing its graph. In the exercise, we examine whether \( c = -\frac{1}{3} \) is a zero of the polynomial \( f(x) \). This is done by applying synthetic division to check if the remainder is zero. If the remainder is zero, it confirms that \( c \) is indeed a zero of the function. Unfortunately, as the exercise demonstrates, the remainder is \(-\frac{2}{3}\), showing that \( c = -\frac{1}{3} \) is not a zero of the polynomial in question. This outcome prompts further exploration of possible zeros and the function's behavior at these points.

Remainder Theorem

The Remainder Theorem is a fundamental concept in algebra which states that the remainder of the division of a polynomial \( f(x) \) by a linear divisor \( x - c \) is \( f(c) \). This theorem provides a straightforward way to evaluate polynomials and test possible zeros.When using synthetic division to divide the polynomial function \( f(x) \) by \( x - c \), the result tells us the remainder alongside other coefficients. If the remainder is zero, \( f(c) = 0 \), confirming that \( c \) is a zero of the polynomial. In the exercise, synthetic division of \( f(x) \) by \( x + \frac{1}{3} \) yields a remainder of \(-\frac{2}{3}\). Therefore, by the Remainder Theorem, \( f(-\frac{1}{3}) = -\frac{2}{3} \), indicating that \(-\frac{1}{3} \) is not a zero of the function. This theorem simplifies the process of testing potential zeros, allowing us to efficiently analyze polynomial functions.

Division Process

The division process in algebra involving polynomials can be simplified using synthetic division, a streamlined method that reduces calculations compared to traditional long division. It is particularly useful with divisors of the form \( x - c \).Synthetic division involves listing only the coefficients of the polynomial and performing arithmetic operations on them with the potential zero \( c \). Here's how the process works:

• Write down the coefficients of the polynomial in order of descending terms.
• Place the possible zero \( c \) outside the synthetic division bracket.
• Bring down the leading coefficient to start the process.
• Multiply \( c \) by the current result, add to the next coefficient, and continue through all coefficients.

The result at the end of this operation provides the remainder of the division and new coefficients for the quotient polynomial.For the exercise provided, \( c = -\frac{1}{3} \) undergoes synthetic division with the polynomial \( 27x^4 - 9x^3 + 3x^2 + 6x + 1 \). This operation involves simple multiplication and addition steps, providing a remainder that confirms where the zero may or may not be—highlighting the ease of use and efficiency of synthetic division.