Use The Remainder Theorem To Find The Remainder

The Remainder Theorem offers a shortcut to determine the remainder of a polynomial division without actually performing the lengthy division process. It’s a powerful tool, especially when dealing with complex polynomials or divisors. Mastering the Remainder Theorem unlocks a deeper understanding of polynomial behavior and their relationships.

Understanding the Remainder Theorem

The Remainder Theorem states: If a polynomial p(x) is divided by (x - a), then the remainder is p(a). In simpler terms, to find the remainder when dividing by (x-a), you simply substitute 'a' into the polynomial p(x). The result of that substitution is the remainder.

A More Formal Explanation

Let's delve a little deeper into the algebraic foundation. When you divide a polynomial p(x) by another polynomial d(x) (the divisor), you get a quotient q(x) and a remainder r(x). This relationship is expressed as:

p(x) = d(x) * q(x) + r(x)

The Remainder Theorem focuses on the specific case where the divisor d(x) is of the form (x - a). In this case, the equation becomes:

p(x) = (x - a) * q(x) + r(x)

Notice that the remainder r(x) must be a constant value (a number without any 'x' terms). This is because the degree of the remainder must always be less than the degree of the divisor. Since (x - a) has a degree of 1, the remainder must have a degree of 0, meaning it's a constant.

Now, if we substitute x = a into the equation, we get:

p(a) = (a - a) * q(a) + r(a) p(a) = (0) * q(a) + r(a) p(a) = r(a)

Since r(x) is a constant, r(a) is simply the value of the remainder, r. Therefore:

p(a) = r

This is the Remainder Theorem in action! Evaluating the polynomial p(x) at x = a directly gives you the remainder when p(x) is divided by (x - a).

Why is This Useful?

The Remainder Theorem is extremely useful for several reasons:

• Efficiency: It saves time. Instead of performing polynomial long division (which can be tedious), you just substitute a value and evaluate.
• Checking Division: You can quickly check if a proposed factor (x - a) actually divides a polynomial evenly. If p(a) = 0, then (x - a) is a factor (this is closely related to the Factor Theorem).
• Finding Roots: By testing different values of 'a', you can potentially find roots (zeros) of the polynomial, which are values of 'x' that make p(x) = 0.
• Problem Solving: It simplifies many algebra problems, particularly those involving polynomial division and factorization.

Step-by-Step Guide to Using the Remainder Theorem

Here's a breakdown of how to apply the Remainder Theorem:

1. Identify the Polynomial: Recognize the polynomial p(x) that you're working with. This will be the expression being divided.
2. Identify the Divisor: Determine the divisor, which should be in the form (x - a). Pay close attention to the sign. If the divisor is (x + 3), then a = -3. The divisor must be in the form (x - a) where the coefficient of 'x' is 1.
3. Find the Value of 'a': Solve for 'a' from the divisor. If the divisor is x - a, then 'a' is simply the number being subtracted from x. If the divisor is in the form (x + b), then rewrite it as (x - (-b)), and a = -b.
4. Substitute 'a' into p(x): Replace every instance of 'x' in the polynomial p(x) with the value of 'a' you found.
5. Evaluate p(a): Simplify the expression to calculate the value of p(a). This value is the remainder.
6. State the Remainder: Clearly state that the value you calculated in step 5 is the remainder when p(x) is divided by (x - a).

Examples of Applying the Remainder Theorem

Let's work through some examples to illustrate the process:

Example 1:

• Problem: Find the remainder when p(x) = x³ - 2x² + 5x - 7 is divided by (x - 2).

• Solution:

  1. p(x) = x³ - 2x² + 5x - 7
  2. Divisor: (x - 2)
  3. a = 2
  4. p(2) = (2)³ - 2(2)² + 5(2) - 7
  5. p(2) = 8 - 8 + 10 - 7 = 3
  6. The remainder is 3.

Example 2:

• Problem: What is the remainder when p(x) = 2x⁴ + x³ - 3x + 1 is divided by (x + 1)?

• Solution:

  1. p(x) = 2x⁴ + x³ - 3x + 1
  2. Divisor: (x + 1). Rewrite as (x - (-1)).
  3. a = -1
  4. p(-1) = 2(-1)⁴ + (-1)³ - 3(-1) + 1
  5. p(-1) = 2(1) - 1 + 3 + 1 = 2 - 1 + 3 + 1 = 5
  6. The remainder is 5.

Example 3:

• Problem: Determine the remainder when p(x) = x⁵ - 32 is divided by (x - 2).

• Solution:

  1. p(x) = x⁵ - 32
  2. Divisor: (x - 2)
  3. a = 2
  4. p(2) = (2)⁵ - 32
  5. p(2) = 32 - 32 = 0
  6. The remainder is 0. (This means (x - 2) is a factor of x⁵ - 32).

Example 4: Dealing with More Complex Polynomials

• Problem: Find the remainder when p(x) = 4x⁶ - 7x⁴ + 3x² - 9 is divided by (x + 2)

• Solution:

  1. p(x) = 4x⁶ - 7x⁴ + 3x² - 9
  2. Divisor: (x + 2). Rewrite as (x - (-2))
  3. a = -2
  4. p(-2) = 4(-2)⁶ - 7(-2)⁴ + 3(-2)² - 9
  5. p(-2) = 4(64) - 7(16) + 3(4) - 9 = 256 - 112 + 12 - 9 = 147
  6. The remainder is 147.

Example 5: Fractions and the Remainder Theorem

The Remainder Theorem works even when 'a' is a fraction.

• Problem: Find the remainder when p(x) = 6x³ + 5x² - 2x + 1 is divided by (x - 1/2).

• Solution:

  1. p(x) = 6x³ + 5x² - 2x + 1
  2. Divisor: (x - 1/2)
  3. a = 1/2
  4. p(1/2) = 6(1/2)³ + 5(1/2)² - 2(1/2) + 1
  5. p(1/2) = 6(1/8) + 5(1/4) - 1 + 1 = 3/4 + 5/4 = 8/4 = 2
  6. The remainder is 2.

Common Pitfalls and How to Avoid Them

• Incorrect Sign: The most common mistake is getting the sign of 'a' wrong. Remember that the divisor must be in the form (x - a). If you have (x + b), then a = -b.
• Incorrect Substitution: Double-check that you've substituted 'a' correctly into every instance of 'x' in the polynomial.
• Arithmetic Errors: Be careful with your arithmetic, especially when dealing with negative numbers or fractions. Use a calculator if needed.
• Trying to Use it with Divisors That Aren't (x - a): The Remainder Theorem only works when dividing by a linear expression of the form (x - a). If you're dividing by something like (x² + 1), you'll need to use polynomial long division.
• Forgetting the Order of Operations: Remember PEMDAS/BODMAS (Parentheses/Brackets, Exponents/Orders, Multiplication and Division, Addition and Subtraction) when evaluating p(a).

Connection to the Factor Theorem

The Remainder Theorem is closely linked to the Factor Theorem. The Factor Theorem is a special case of the Remainder Theorem. It states:

• (x - a) is a factor of the polynomial p(x) if and only if p(a) = 0.

In other words, if the remainder when dividing p(x) by (x - a) is zero, then (x - a) divides p(x) evenly and is therefore a factor of p(x).

Example:

We saw in Example 3 that when p(x) = x⁵ - 32 is divided by (x - 2), the remainder is 0. Therefore, according to the Factor Theorem, (x - 2) is a factor of x⁵ - 32.

Proof of the Remainder Theorem

While the previous explanation provided a solid understanding, let's look at a more formal proof.

Let p(x) be a polynomial and let (x - a) be the divisor. Then, by the division algorithm, there exist unique polynomials q(x) (the quotient) and r(x) (the remainder) such that:

p(x) = (x - a)q(x) + r(x)

Where the degree of r(x) is less than the degree of (x - a). Since (x - a) has degree 1, r(x) must have degree 0, meaning r(x) is a constant, which we can simply denote as r. Thus, the equation becomes:

p(x) = (x - a)q(x) + r

Now, substitute x = a into the equation:

p(a) = (a - a)q(a) + r p(a) = (0)q(a) + r p(a) = r

Therefore, the remainder r is equal to p(a), proving the Remainder Theorem.

Applications Beyond Basic Algebra

While the Remainder Theorem is primarily taught in introductory algebra courses, it has applications in more advanced areas of mathematics:

• Abstract Algebra: The Remainder Theorem is a specific instance of more general theorems about polynomial rings and ideals in abstract algebra.
• Numerical Analysis: In numerical analysis, the Remainder Theorem can be used to approximate the roots of polynomials and to develop efficient algorithms for polynomial evaluation.
• Coding Theory: Polynomials over finite fields are used extensively in coding theory for error detection and correction. The Remainder Theorem can be helpful in analyzing and manipulating these polynomials.

Practice Problems

To solidify your understanding, try these practice problems:

1. Find the remainder when p(x) = x⁴ - 3x³ + 2x² - 5x + 1 is divided by (x - 3).
2. Determine the remainder when p(x) = 2x⁵ + x² - 8 is divided by (x + 2).
3. What is the remainder when p(x) = x³ - 6x² + 11x - 6 is divided by (x - 1)? What does this tell you about (x - 1)?
4. Find the remainder when p(x) = 4x³ - 8x + 5 is divided by (x + 1/2).
5. Is (x - 5) a factor of p(x) = x³ - 7x² + 7x + 15? Use the Remainder Theorem to find out.

Conclusion

The Remainder Theorem provides a powerful and efficient method for finding the remainder of polynomial division when the divisor is of the form (x - a). It saves time, offers a way to check division, and is closely related to the Factor Theorem, which helps in finding roots of polynomials. By understanding the underlying principles and practicing with examples, you can master this valuable algebraic tool and expand your understanding of polynomial behavior. Remember to pay close attention to the sign of 'a', substitute carefully, and double-check your arithmetic. With practice, the Remainder Theorem will become a valuable part of your mathematical toolkit.