Problem 10 - College Algebra with Corequisite Support

Question

For the following exercises, use the Remainder Theorem to find the remainder. $$ \left(5 x^{5}-4 x^{4}+3 x^{3}-2 x^{2}+x-1\right) \div(x+6) $$

Step-by-Step Solution

Answer

The remainder is -44871.

Step 1: Identify the Value of r in the Remainder Theorem

The Remainder Theorem states that the remainder of a polynomial \( f(x) \) when divided by a binomial \( x - r \) is simply \( f(r) \). For this exercise, we have \( x + 6 \), which can be rewritten as \( x - (-6) \). Thus, \( r = -6 \).

Step 2: Evaluate the Polynomial at r

Now that we have \( r = -6 \), we substitute \( x = -6 \) into the polynomial \( f(x) = 5x^5 - 4x^4 + 3x^3 - 2x^2 + x - 1 \):\[ f(-6) = 5(-6)^5 - 4(-6)^4 + 3(-6)^3 - 2(-6)^2 + (-6) - 1 \]

Step 3: Calculate (-6) Raised to Each Power

Calculate each power of (-6):\((-6)^5 = -7776\) \((-6)^4 = 1296\) \((-6)^3 = -216\) \((-6)^2 = 36\)

Step 4: Substitute Results into the Polynomial Expression

Now substitute these values back into the polynomial expression:\[ 5(-7776) - 4(1296) + 3(-216) - 2(36) - 6 - 1 \]

Step 5: Perform the Multiplications

Multiply each term:- \( 5 imes (-7776) = -38880 \) - \( 4 imes 1296 = 5184 \) - \( 3 imes (-216) = -648 \) - \( 2 imes 36 = 72 \)

Step 6: Add and Subtract All Values

Now add and subtract all the computed values to find \( f(-6) \):\[ -38880 - 5184 - 648 - 72 - 6 - 1 = -44871 \]

Step 7: Conclude the Remainder

Therefore, the remainder when the polynomial \( 5x^5 - 4x^4 + 3x^3 - 2x^2 + x - 1 \) is divided by \( x + 6 \) is \(-44871\).

Key Concepts

Polynomial Division

Polynomial division is a technique used to divide one polynomial by another. It is similar to the long division process we use with numbers. In the given exercise, our task is to divide the polynomial \(5x^5 - 4x^4 + 3x^3 - 2x^2 + x - 1\) by the binomial \(x + 6\). This specific division asks us to find the remainder when the polynomial is divided by the binomial. The Remainder Theorem simplifies this process by allowing us to determine the remainder without actually performing the division.

When dividing polynomials, if the remainder is zero, it means the divisor is a factor of the dividend. However, in this exercise, we use the Remainder Theorem, which is a shortcut that helps us find the remainder by simply evaluating the polynomial at a certain value of \(x\). This approach can save time and makes the process much easier, especially for polynomials of higher degree.

Evaluate Polynomial

Evaluating a polynomial means substituting a given value into the expression and simplifying to find the result. In our context, this involves using the Remainder Theorem. We found that \(x + 6\) is equivalent to \(x - (-6)\), so we set \(r = -6\).

To evaluate the polynomial \(f(x) = 5x^5 - 4x^4 + 3x^3 - 2x^2 + x - 1\) at \(r = -6\), we substitute \(-6\) for each \(x\) in the polynomial. This substitution transforms the polynomial into a numerical expression. We then compute the powers and perform the operations to find \(f(-6)\).

This step is central because it determines the remainder directly, aligning with the Remainder Theorem's principle. The process involves a straightforward substitution followed by arithmetic operations.

Synthetic Division

Synthetic division is a simplified form of polynomial division, specifically for dividing by a linear binomial of the form \(x - r\). While we didn't explicitly use synthetic division in this exercise, understanding it can enhance comprehension of the division process.

Unlike traditional division, synthetic division reduces the workload by focusing on the coefficients of the polynomial. It provides a streamlined way to find both the quotient and the remainder of polynomial division.

• Write down the coefficients of the polynomial.
• Use \(r\) (from \(x - r\)) for synthetic division.
• Perform operations to simplify and reveal the remainder.

For more complex exercises, synthetic division can be an efficient and less error-prone method to determine the remainder.

Algebraic Expression Manipulation

Manipulating algebraic expressions is a fundamental skill in solving polynomial problems. It involves expanding, factoring, and simplifying expressions to make them easier to work with. In the scope of our exercise, manipulation primarily involved using the Remainder Theorem, but also ensured that every computational step was clear and logical.

During the problem-solving process, we handled different powers of \(-6\), multiplied by their respective coefficients, and added or subtracted the results. Here's how:

• Compute each expression \(-6\) raised to specific powers.
• Multiply these results by corresponding coefficients.
• Sum all results to find \(f(-6)\).

This might look complicated, but breaking down each part simplifies the operation. Algebraic manipulation helps keep polynomials manageable, especially when working within the framework of the Remainder Theorem.