Question

You use synthetic division to divide \(f(x)\) by \((x-a)\) and find that the remainder equals 15 . Your friend concludes that \(f(15)=a\). Is your friend correct? Explain your reasoning.

Short answer

No, your friend is incorrect. According to the remainder theorem, the remainder of the division of a function \(f(x)\) by \((x-a)\) is equal to \(f(a)\). Here, the remainder is 15 which implies \(f(a) = 15\), not \(f(15) = a\).

Step-by-step solution

Understand The Problem

The key here is to understand the remainder theorem. This theorem states that if a polynomial \(f(x)\) is divided by \((x-a)\), then the remainder is \(f(a)\). In this case, since the remainder is 15, it should be stated that \(f(a)=15\), not \(f(15)=a\). So your friend's conclusion is likely incorrect. We will now review this in more detail to better understand the reasoning.

Apply the Remainder Theorem Correctly

In order to apply the remainder theorem correctly, it has to be remembered that if a function \(f(x)\) is divided by \((x-a)\) with a result of a remainder of 15, this signifies that \(f(a)=15\). The rule is not reversible - this factorization does not indicate that \(f(15)=a\). This is important to understand and remember when assessing your friend's conclusion.

Evaluate the Friend's Conclusion

Given that the correct application of the remainder theorem states that for a polynomial \(f(x)\) divided by \((x-a)\), the remainder is \(f(a)\), we can say that if the remainder is 15, then \(f(a) = 15\). Your friend's conclusion that \(f(15) = a\) is therefore incorrect, as it does not adhere to the rules of the remainder theorem.

Key concepts

Synthetic Division

When dealing with polynomial functions, one efficient way to perform division is through synthetic division. This method is a simplified version of long division and specifically applies to dividing a polynomial by a binomial of the form \(x-a\). Synthetic division reduces the complexity and the amount of work, particularly when you're only interested in finding the quotient or the remainder.

Here's a quick rundown: To perform synthetic division you write down the coefficients of the polynomial, create a row for the synthetic operation using the value \(a\), and then proceed with the operation that ultimately gives the remainder. This technique is particularly useful when applying the remainder theorem, as it provides a quick determination of \(f(a)\) when dividing by \(x-a\). Even though this method is streamlined, it's crucial to remember details like placeholder zero coefficients for any missing powers in the polynomial to avoid errors.

Polynomial Functions

Polynomial functions are algebraic expressions that involve sums of powers of a variable. Typically, they are of the form \(f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_2x^2 + a_1x + a_0\), where each \(a_i\) represents coefficients and \(n\) is a non-negative integer termed the degree of the polynomial. The significant characteristic of these functions is that they are continuous and smooth, meaning they have no gaps, jumps, or sharp corners when graphed.

Understanding polynomial functions is essential because they are foundational to calculus, and they appear frequently in real-world problems. Their behavior is predictable: the degree of the polynomial indicates how many roots (solutions) it has, and the leading coefficient can give insights into its end behavior.

Factoring Polynomials

Factoring polynomials is a crucial skill in algebra that involves breaking down a polynomial into a product of simpler polynomials that, when multiplied together, give back the original polynomial. This is similar to finding the prime factors of a number. The reason factoring is essential is it simplifies polynomial expressions and solves polynomial equations, making it easier to analyze and understand them.

Common methods of factoring involve taking out a greatest common factor, using special product formulas, grouping, or using techniques such as the factor theorem. The factor theorem, which is closely connected to the remainder theorem, states that \(x-a\) is a factor of the polynomial \(f(x)\) if and only if \(f(a)=0\). This insight can save immense time and guides us to focus on specific values of the variable that make the polynomial equal to zero.

Algebraic Expressions

Algebraic expressions are the building blocks of algebra. They consist of numbers, variables, and operations (such as addition, subtraction, multiplication, and division) but do not contain an equality sign. Algebraic expressions can be as simple as a single term or as complex as a large polynomial or rational expression.

They are fundamental because they represent mathematical ideas symbolically, allowing us to generalize and manipulate mathematical relationships. Working with algebraic expressions is like playing with the pieces of a puzzle — we arrange and rearrange the pieces (the numbers and variables) under the rules of algebra to solve for unknowns or to simplify the expression. Developing skills in manipulating these expressions is critical for solving more complex equations and understanding higher-level math.