Problem 31 - Precalculus : Building Concepts and Connections

Question

One zero of each polynomial is given. Use it to express the polynomial as a product of linear factors over the complex numbers. You may have already factored some of these polynomials into linear and irreducible quadratic factors in the previous group of exercises. $$x^{3}-2 x^{2}+x-2 ; \text { zero: } x=2$$

Step-by-Step Solution

Answer

The polynomial \(x^{3}-2 x^{2}+x-2\) factors to \((x-2)(x+1)(x-1)\) over the complex numbers.

Step 1: Set up Synthetic Division

Set up the polynomial for synthetic division using 2 as a zero. The coefficients of the polynomial \(x^{3}-2 x^{2}+x-2\) are 1, -2, 1, -2 respectively.

Step 2: Perform Synthetic Division

Use synthetic division to substitute the root \(x=2\) into the polynomial. So, basically divide the polynomial by the factor \(x-2\). At the end of synthetic division, the row of numbers at the bottom of the setup are the coefficients of the quotient polynomial.

Step 3: Write Out the Factored Form

The factored form of the polynomial is the divisor and the quotient. The divisor is \(x-2\), since \(x=2\) is a zero, and the quotient obtained from the synthetic division is the polynomial of degree 2, \(x^2-1\). Therefore, the polynomial can be expressed as a product of linear factors over the complex numbers as: \((x-2)(x^2-1)\). We can further factor \(x^2-1\) to \((x+1)(x-1)\). Therefore, the polynomial \(x^{3}-2 x^{2}+x-2\) factors to \((x-2)(x+1)(x-1)\).

Key Concepts

Synthetic Division

Synthetic division is a simplified method of dividing polynomials that works specifically when dividing by linear factors of the form \(x - c\). Here, \(c\) is a constant and is typically a root or zero of the polynomial. This efficient technique involves fewer steps and less writing compared to the long division method, making it an excellent tool for polynomial division.

To perform synthetic division, you need the coefficients of the polynomial and the zero that you're testing (in this case, \(x = 2\)). Arrange the coefficients in descending order of their degrees. Now, initiate the synthetic division by following these steps:

• Bring down the leading coefficient as it is.
• Multiply the root \(c\) with the number you just brought down and write the result under the next coefficient.
• Add this result to the next coefficient and write the sum below the line.
• Repeat this process until all coefficients have been used.

By the end of this process, the bottom row provides the coefficients of the quotient polynomial. The remainder should be zero if \(c\) is a true root, confirming the division was performed correctly.

Complex Numbers

When dealing with polynomials, complex numbers can become useful, especially when the zeros or roots are not real numbers. Complex numbers are numbers that have a real part and an imaginary part, and they are written in the form \(a + bi\) where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, satisfying \(i^2 = -1\).

In the context of polynomial factorization, being able to factor over the complex numbers means that you are accounting for all possible roots, real or otherwise. When a polynomial is factored completely over the complex numbers, it can be broken down into linear factors of the form \(x - c\), where \(c\) could be either a real or a complex number.

In this exercise, even though all the roots (\(x-2\), \(x+1\), \(x-1\)) are real, factoring over the complex numbers ensures the polynomial is expressed in its most fundamental form, incorporating all possible types of roots.

Linear Factors

Linear factors are expressions of the form \(x - c\), where \(c\) is a constant and represents a root of the polynomial. When a polynomial is expressed as a product of linear factors, it means it is broken down into the simplest building blocks that correspond directly to its roots.

The importance of expressing a polynomial as a product of linear factors is that it reveals all possible solutions or zeros of the polynomial. Each factor represents a solution where the polynomial equals zero. Thus, if one factor is \(x - 2\), then 2 is a solution where the polynomial's value equals zero.

In this exercise, once we used synthetic division to simplify the polynomial to \((x-2)(x^2-1)\), we further factored \((x^2-1)\) into \((x+1)(x-1)\). This final expression \((x-2)(x+1)(x-1)\) demonstrates the polynomial completely decomposed into linear factors, presenting a clear picture of all its roots over the complex numbers.