Problem 24 - Precalculus : Building Concepts and Connections

Question

One zero of each polynomial is given. Use it to express the polynomial as a product of linear and irreducible quadratic factors. $$x^{3}-x^{2}+4 x-4 ; \text { zero: } x=1$$

Step-by-Step Solution

Answer

The polynomial \(x^{3}-x^{2}+4x-4\) can be expressed as a product of a linear factor and an irreducible quadratic factor as: \(x^{3}-x^{2}+4x-4 = (x-1)(x^{2}+4)\).

Step 1: Given

The given polynomial is \(x^{3}-x^{2}+4x-4\) and one of its roots (zeroes) is \(x=1\). This means that \(x-1\) is one of its factors.

Step 2: Perform the Polynomial Division

Divide the original polynomial \(x^{3}-x^{2}+4x-4\) by \(x-1\) (which is obtained from the known root). This can be done via long division or synthetic division. The result is a quotient of \(x^{2}+4\).

Step 3: Check for Further Factoring

Check the resulting quotient \(x^{2}+4\) if it can be factored further. In this case, since there are no real roots for \(x^{2}+4=0\), the quotient is already an irreducible quadratic.

Step 4: Express the Polynomial as a Product

The original polynomial can therefore be expressed as a product of a linear factor \(x-1\) and the irreducible quadratic \(x^{2}+4\) as \(x^{3}-x^{2}+4x-4 = (x-1)(x^{2}+4)\)

Key Concepts

Polynomial Division

When we talk about polynomial division, we refer to the process of dividing a polynomial by another polynomial. Think of it much like long division with numbers, but instead of just dealing with plain digits, we're dealing with algebraic expressions.

For instance, a key step in our exercise is to divide the original cubic polynomial by one of its factors that we obtain from a known root. When we divide the given polynomial, \(x^3 - x^2 + 4x - 4\), by \(x - 1\), we end up with a simpler polynomial, a quadratic in this case, which is \(x^2 + 4\). This process effectively breaks down complex polynomials into simpler, more manageable pieces and is particularly useful for finding other factors or roots.

It is essential for students to understand the mechanics of polynomial division, as it serves as the foundation for other operations like factorization and solving polynomial equations.

Irreducible Quadratic Factors

Though some quadratic expressions can be factored into two linear factors, there are others that cannot be broken down using real numbers. These are called irreducible quadratic factors.

In the given exercise, after dividing our cubic polynomial by a linear factor, we are left with \(x^2 + 4\). We then need to determine if this quadratic can be further factored. To find out, we look for real roots of the equation \(x^2 + 4 = 0\). However, since no real numbers square to give -4, there are no real solutions to this equation. Therefore, \(x^2 + 4\) is an irreducible quadratic factor over the real numbers.

Recognizing when a quadratic is irreducible is an important skill in simplifying and solving polynomial expressions, especially when dealing with higher-degree polynomials.

Roots of Polynomial

The roots of a polynomial are the solutions to the polynomial equation set equal to zero. In simpler terms, if you plug in a root into the polynomial, the entire expression will equate to zero. In our exercise, we are given that \(x = 1\) is a root of \(x^3 - x^2 + 4x - 4\).

Once a root is known, it corresponds to a linear factor of the polynomial. If \(x = r\) is a root, then \(x - r\) is a factor. The Fundamental Theorem of Algebra tells us a polynomial of degree \(n\) will have \(n\) roots (some of which may be complex or repeated). Identifying roots is a cornerstone of polynomial algebra as it allows the polynomial to be expressed as a product of its factors, making it easier to analyze and solve.

Synthetic Division

Like long polynomial division, synthetic division is another technique used to divide polynomials. However, it's a shortcut method that simplifies the process and is particularly handy when dividing by a linear factor.

In the solution to our exercise, synthetic division could be used instead of long division to divide \(x^3 - x^2 + 4x - 4\) by \(x - 1\) to get \(x^2 + 4\). It involves less writing and fewer calculations, making it quicker and reducing the potential for errors. It's essential to learn this method as it streamlines polynomial division and is especially useful for checking roots, analyzing polynomial functions, and simplifying expressions.