Problem 26 - Calculus Early Transcendentals: Pearson New International Edition

Question

$$ \text { perform the indicated operations and simplify. } $$ $$ \frac{2 x-2 x^{2}}{x^{3}-2 x^{2}+x} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{-2}{x-1} \), for \( x \neq 0, \pm 1 \).

Step 1: Factor the Numerator

The numerator of the expression is \( 2x - 2x^2 \). We factor out the greatest common factor, which is \( 2x \). This gives us: \[ 2x(1-x) \].

Step 2: Factor the Denominator

The denominator of the expression is \( x^3 - 2x^2 + x \). We can factor by grouping. First, factor out \( x \) from the entire expression: \[ x(x^2 - 2x + 1) \]. The quadratic \( x^2 - 2x + 1 \) can be further factored as a perfect square: \( (x-1)^2 \). Thus, the factored form is: \[ x(x-1)^2 \].

Step 3: Simplify the Expression

We now have: \[ \frac{2x(1-x)}{x(x-1)^2} \]. Recognizing that \( 1-x \) can be written as \( -(x-1) \), we rewrite the numerator: \[ 2x(1-x) = -2x(x-1) \]. Thus, the expression becomes: \[ \frac{-2x(x-1)}{x(x-1)^2} \].

Step 4: Cancel Common Factors

Now, we can cancel \( x \) and \( (x-1) \) from the numerator and denominator: \[ \frac{-2x(x-1)}{x(x-1)^2} = \frac{-2}{x-1} \text{ for } x eq 0, \pm 1. \]This is because setting \( x = 0 \) or \( x = \pm 1 \) would cause the original denominator to become zero.

Key Concepts

Factoring Polynomials

Factoring polynomials is like breaking down a mathematical expression into simpler "pieces" known as factors. Each factor is a polynomial in its own right. Imagine factoring as trying to find the "ingredients" that, when combined, form the original polynomial expression.

In the context of our problem, the numerator is originally given as \(2x - 2x^2\). Here, you can spot that both terms share a common element, which is \(2x\). By taking \(2x\) as a common factor out of each term, we simplify the expression to \(2x(1-x)\). This means \(2x\) and \(1-x\) multiply together to give \(2x - 2x^2\).

For the denominator \(x^3 - 2x^2 + x\), notice it can also be expressed through factors. First, \(x\) is factored from each term: \(x(x^2 - 2x + 1)\). The part \(x^2 - 2x + 1\) is a perfect square, meaning it can be further factored to \((x-1)^2\). Together, these make the polynomial easier to manipulate and understand.

Algebraic Fractions

An algebraic fraction is similar to a standard fraction, but it includes variables within its numerator and/or denominator. These fractions rely on algebraic expressions, and simplifying them is vital to easier manipulation and problem-solving.

In our example, \(\frac{2x - 2x^2}{x^3 - 2x^2 + x}\), both the numerator and the denominator are algebraic expressions. Simplifying them involves factoring, as seen previously, which can significantly reduce the complexity of the fraction.

During simplification, you may come across expressions like \(1-x\), which can be written as \(-(x-1)\). This transformation is a useful trick, allowing common factors within algebraic fractions to be more easily identified and canceled. The key is to maintain equivalent expressions even as their form changes for simplification.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. Simplifying these expressions often involves both factoring and canceling common terms.

In the given exercise, the expression \(\frac{2x(1-x)}{x(x-1)^2}\) is a rational expression. Once each part is factored, the rational expression can be simplified by canceling out common factors. Here, \(x\) and \((x-1)\) appear in both the numerator and denominator. By realizing that \(1-x\) is \(-(x-1)\), the factors \(x\) and \((x-1)\) are canceled, simplifying the expression to \(\frac{-2}{x-1}\).

It's crucial to note that rational expressions can have restrictions. For instance, values that make any denominator zero are excluded from the solution's valid domain. In this case, \(x eq 0, \pm1\), since these values would make the original denominator zero, thus undefined.