91Ó°ÊÓ

Factoring polynomials is like breaking down a math problem into smaller, simpler parts. It involves expressing a polynomial as the product of its factors. A polynomial is any expression involving variables that have various exponents. For example, the expression \(x^2 - 2x + 1\) in this exercise is a polynomial.

The goal is to "factor" this expression, which means to write it as a product of simpler expressions. In this example, the polynomial \(x^2 - 2x + 1\) factors perfectly into \((x-1)^2\).

This factoring process is crucial because it simplifies calculations, particularly when dealing with partial fraction decomposition. It transforms a complex denominator into more manageable parts which are either linear factors or quadratic factors. Here, it turned into a simple square of a binomial, which makes it easier to work with.

A perfect square trinomial is a specific type of polynomial that can be written as the square of a binomial. It follows the general pattern of \(a^2 - 2ab + b^2 = (a-b)^2\) or \(a^2 + 2ab + b^2 = (a+b)^2\).

In our problem, \(x^2 - 2x + 1\) fits this pattern because it can be rewritten as \((x-1)^2\). Breaking it down:

• The first term \(x^2\) is \(x\) squared.
• The last term \(1\) is \(1\) squared.
• The middle term \(-2x\) is twice the product of \(x\) and \(1\), showing the characteristic "+2ab" and "-2ab" pattern of a perfect square trinomial.

Recognizing these trinomials quickly allows you to reduce complex algebraic expressions into simpler forms, essential for making partial fraction decomposition manageable.

Linear factors are expressions of the first degree, meaning they contain no exponents other than one. In our scenario, \(x - 1\) is a linear factor.

Linear factors are essential in simplifying complex fractions, like in partial fraction decomposition. When you have a quadratic that is a perfect square trinomial such as \((x-1)^2\), you end up with repeated linear factors. Each factor corresponds to a piece in the solution puzzle.

In the partial fraction decomposition, the linear factor \(x-1\) leads to a fraction component \(\frac{A}{x-1}\) and its repetition \(\frac{B}{(x-1)^2}\). Understanding these single-degree expressions makes tackling larger, more complicated algebraic expressions more approachable.

Repeated roots occur when a specific root appears more than once in a polynomial. In our example, since \((x-1)^2\) is a factor, the root \(x=1\) is repeated twice. A repeated root affects how we break down the polynomial using partial fraction decomposition.

When you deal with repeated roots like \((x-1)^2\), you must consider all powers of these roots up to their multiplicity when decomposing the fraction. In this exercise, we consider both \(\frac{A}{x-1}\) and \(\frac{B}{(x-1)^2}\) to capture the effects of having \(x-1\) as a repeated root.

Recognizing and understanding repeated roots ensure that the decomposition captures all necessary elements of the original expression, leading to an accurate simplification.