Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Perform the indicated operations and simplify.\(\frac{x^{3}-8}{x+1} \cdot \frac{x^{2}-1}{x^{3}-3 x^{2}+2 x}\)

Step-by-Step Solution

Answer

The simplified form of the given expression is \(\frac{(x -2)(x^2 + 2x + 4)(x-1)}{(x^3 - 3x^2 + 2x)}\).

Step 1: Factorize the Terms

Rewrite each term in a factorized form. This involves rewriting \(x^3 - 8\) as \((x -2)(x^2 + 2x + 4)\) and \(x^2 - 1\) as \((x-1)(x+1)\). The denominators don't factorize nicely, so we keep them as they are.

Step 2: Apply Fraction Multiplication Rule

Now, we apply the multiplication rule which states that the multiplication of fractions is done by multiplying the numerators together for the new numerator and the denominators together for the new denominator. This would give: \(\frac{(x -2)(x^2 + 2x + 4)(x-1)(x+1)}{(x+1)(x^3 - 3x^2 + 2x)}\).

Step 3: Simplify the Terms

Simplify the fraction by canceling out the common terms in the numerator and the denominator. Here we can cancel out \(x+1\) in the numerator and the denominator. Therefore the simplified form becomes: \(\frac{(x -2)(x^2 + 2x + 4)(x-1)}{(x^3 - 3x^2 + 2x)}\). Make sure the denominator doesn't have terms that add up to zero as this would make the fraction undefined.

Key Concepts

Factoring Polynomials

The process of factoring polynomials is essential in simplifying algebraic fractions. It involves breaking down a complex expression into a product of simpler factors. For instance, when we have a cubic polynomial like (x^3 - 8), it can be factored into (x - 2)(x^2 + 2x +4). This is because (x - 2)(x^2 + 2x + 4) represents the product of a linear factor (x - 2) and a quadratic factor (x^2 + 2x + 4) that, when multiplied, give us the original polynomial.
Factoring polynomials is usually done by finding the greatest common factor, using methods like grouping, or applying special factoring formulas such as the difference of squares or cubes. The difference of squares, for instance, was used to factor (x^2 - 1) into (x - 1)(x + 1), indicating that two binomials can be multiplied to get back the original quadratic polynomial. These techniques are not just mathematical tricks; they allow us to simplify complex algebraic fractions by canceling common factors, which we'll discuss in another section.

Multiplication of Fractions

When multiplying fractions, the rule is straightforward: multiply the numerators together to find the new numerator and multiply the denominators together to find the new denominator. For our exercise, this means taking the factored forms of the numerators, (x - 2)(x^2 + 2x + 4) and (x - 1)(x + 1), and multiplying them. Similarly, we do the same for the denominators, even though in this case, one of them does not factorize neatly.
Through multiplication, the complex-looking initial expression simplifies significantly. It's tempting to think of this process as simply squashing everything together, but it's important to keep track of what's what—after all, simplification might be possible, and multiplying correctly sets the stage for us to potentially cancel common factors later on in the problem-solving process.

Canceling Common Factors

Once you've factored your polynomials and multiplied your fractions, you may find that both the numerator and denominator share some common terms.
Canceling common factors comes into play here. It's a critical step in simplification, which involves dividing both the numerator and denominator by the same term, effectively 'canceling' it out from the fraction. In the exercise, we observe that (x+1) appears in both the numerator and the denominator and can therefore be canceled. Remember, we can only cancel factors—terms that are multiplied—not terms that are added or subtracted.
Canceling correctly is vital; not only does it simplify the expression, but it ensures that you don't lose any solutions or mistakenly cancel terms that should remain. Always check your simplified fraction to make sure you haven't created any terms in the denominator that could make the expression undefined.