Problem 53 - College Algebra

Question

How does the linear factorization of \(f(x),\) that is, $$f(x)=a_{n}\left(x-c_{1}\right)\left(x-c_{2}\right) \cdots\left(x-c_{n}\right)$$ show that a polynomial equation of degree \(n\) has \(n\) roots?

Step-by-Step Solution

Answer

The linear factorization of a polynomial function demonstrates that a polynomial of degree \(n\) has \(n\) roots because every linear term in the factorization equates to a solution, or root, of the polynomial equation.

Step 1: Understanding Polynomial Linear Factorization

Any polynomial function of degree \(n\) can be factored into a product of \(n\) linear terms. This is called the linear factorization. Each linear term in the product corresponds to a root of the polynomial equation. Consequently, a polynomial of degree \(n\) has \(n\) linear terms, implying that it has \(n\) roots.

Step 2: Looking at the Formula

Looking at the given form of a polynomial \(f(x)\), it's clear that it's factored into linear terms: \(f(x) = a_n \cdot (x-c_1) \cdot (x-c_2) \cdot ... \cdot (x-c_n)\). Here, \(c_1, c_2, ..., c_n\) are the roots of the polynomial equation.

Step 3: Connecting Roots and Factors

Each root corresponds to one factor in the equation. For instance, when \(x = c_1\), \(f(x) = a_n \cdot (x-c_1) = 0\). This is because any number multiplied by zero equals zero. The same relation exists between \(x = c_2\), \(x = c_3\), ..., and \(x = c_n\), proving that each factor corresponds to a root and thus a polynomial of degree \(n\) has \(n\) roots.