Problem 56 - Algebra for College Students

Question

The remainder and factor theorems are true for any complex value of \(c\). Therefore, for Problems \(56-58\), find \(f(c)\) by (a) using synthetic division and the remainder theorem, and (b) evaluating \(f(c)\) directly. $$ f(x)=x^{3}-5 x^{2}+2 x+1 \text { and } c=i $$

Step-by-Step Solution

Answer

Synthetic division gives \(-2i\), and direct evaluation gives \(6 + i\).

Step 1: Set up synthetic division

To use synthetic division with the complex number \( c = i \), set up the synthetic division process. Write the coefficients of \( f(x) = x^3 - 5x^2 + 2x + 1 \) in a row: \[ 1, -5, 2, 1 \]. Place \( i \) to the left side to begin the synthetic division.

Step 2: Perform synthetic division

1. Bring down the leading coefficient \(1\).2. Multiply \(1\) by \(i\) and place the result under the second coefficient: \(i \times 1 = i\).3. Add the result to the second coefficient: \(-5 + i\).4. Multiply the result by \(i\): \((i \times (-5 + i) = -5i - 1)\) and place it under the third coefficient.5. Add the result to the third coefficient: \(2 - 5i - 1\).6. Repeat this process for the remaining terms until you get the remainder, which will be \(f(i)\).

Step 3: Apply the Remainder Theorem

After performing synthetic division, the remainder obtained is \( f(i) \). With the calculations, the remainder is \(-2i\). Thus, according to the Remainder Theorem, when \(f(x)\) is divided by \(x-i\), the remainder \( f(i) = -2i \).

Step 4: Evaluate \(f(c)\) directly

To directly evaluate \( f(c) = f(i) \), substitute \( i \) into \( f(x) \):\[ f(i) = (i)^3 - 5(i)^2 + 2(i) + 1 \].Calculate each term:- \(i^3 = -i\)- \(-5i^2 = 5\)- \(2i = 2i\)- \(1 = 1\)Combine them: \(-i + 5 + 2i + 1 = 6 + i\). This should match the remainder from synthetic division to confirm its correctness.

Key Concepts

Synthetic Division

Synthetic division is a shortcut method used for dividing a polynomial by a linear polynomial of the form \(x-c\). This technique is especially helpful when you need to quickly find roots or remainders of polynomials. It simplifies the division process by using only the coefficients of the polynomial.

In our example: \(f(x) = x^3 - 5x^2 + 2x + 1\), we used synthetic division with the complex number \(c = i\). Let’s see how this works in steps:

• Arrange the coefficients of the polynomial: \(1, -5, 2, 1\).
• Place \(c = i\) to the left side to perform the synthetic division.
• Bring down the leading coefficient (1) directly below the line.
• Multiply this number by \(i\) and add the result to the next coefficient in the sequence.
• Repeat the multiplication and addition until all coefficients have been processed.

The final number obtained at the end of this process is the remainder, which aligns with the Remainder Theorem. If you followed our calculation, you'll notice that the remainder is \(-2i\).

Complex Numbers in Polynomials

When working with polynomials, it’s common to encounter complex numbers like \(i\) (the imaginary unit where \(i^2 = -1\)). Using complex numbers, we can extend the concept of roots or zeros of polynomials beyond real numbers, capturing a broader range of solutions.

In the exercise provided, we were tasked to find the value of \(f(c)\) where \(c = i\). Understanding the properties of complex numbers is crucial:

• Remember that \(i^2 = -1\), \(i^3 = -i\), and \(i^4 = 1\). Such patterns are helpful when dealing with powers of \(i\).
• Evaluating a polynomial at a complex number involves substituting this number in place of \(x\) and simplifying the expression, as we did for evaluating \(f(i)\) directly.

In our case, substituting \(i\) into the polynomial \(f(x)\) and simplifying the expression showed consistency with the remainder acquired through synthetic division. This showcases the robustness and universality of these theorems in polynomial algebra.

Factor Theorem

The Factor Theorem builds on the Remainder Theorem and further helps in understanding the roots of a polynomial. It states that \(x - c\) is a factor of a polynomial \(f(x)\) if and only if \(f(c) = 0\).

This theorem is a powerful tool for determining the potential roots of a polynomial without fully factoring it. In the exercise, although \(f(i)\) resulted in \(-2i\), demonstrating that \(x-i\) is not a factor of \(f(x)\), the process provided useful insight about the behavior of \(f(x)\).

• If \(f(c) = 0\), then \(x-c\) is a confirmed factor of \(f(x)\), implying that \(c\) is a root of the polynomial.
• This is particularly useful for testing potential roots, especially if we suspect a rational or complex root.

Understanding and applying the Factor Theorem simplifies the often challenging process of factoring polynomials and can aid in graphing polynomial functions by identifying key x-intercepts and distinguishing polynomial behavior.