Simplify This Complex Algebraic Fraction

Are you ready to tackle a challenging algebraic expression? This article will guide you through simplifying the following expression: $\frac{x^2-16}{2 x+8} \frac{x^3-2 x^2+x}{x^2+3 x-4}$. We'll break down each step, making it easier to understand and solve. Let's dive in and simplify this fraction step by step, so you can confidently arrive at the correct answer among the options provided: A. $\frac{x(x-4)(x-1)}{2(x+4)}$, B. $\frac{\pi x-11}{2}$, C. $\frac{(x+4)(x-4)}{2 x(x-1)}$, D. $\frac{(x-4)(x-1)}{2 x(x+4)}$.

Understanding the Problem: Simplifying Algebraic Fractions

Simplifying algebraic fractions is a fundamental skill in mathematics that involves reducing complex expressions to their simplest forms. This process often requires factoring polynomials and canceling out common terms in the numerator and denominator. The given expression, $\frac{x^2-16}{2 x+8} \frac{x^3-2 x^2+x}{x^2+3 x-4}$, is a product of two rational expressions. To simplify it, we need to factor each polynomial individually. This involves recognizing common factoring patterns such as the difference of squares ($a^2 - b^2 = (a-b)(a+b)$), common monomial factors, and trinomials. Once factored, we can identify and cancel out any factors that appear in both the numerator and the denominator. This cancellation is a crucial step that significantly reduces the complexity of the expression. It's important to remember that we can only cancel factors, not terms. For example, in an expression like $\frac{a+b}{a+c}$, we cannot cancel 'a' because it is part of a sum, not a factor. However, in an expression like $\frac{ab}{ac}$, we can cancel 'a' because it is a common factor. The goal is to manipulate the expression algebraically so that the numerator and denominator share as few common factors as possible, ideally leaving only constants or single variables. This process not only simplifies the expression but also helps in understanding its behavior, such as its domain (the values of x for which the expression is defined) and its roots. Mastering this technique is essential for further studies in algebra, calculus, and other related fields.

Step-by-Step Solution

Let's break down the simplification of the expression $\frac{x^2-16}{2 x+8} \frac{x^3-2 x^2+x}{x^2+3 x-4}$ into manageable steps. Our goal is to factor each part of the expression first.

Step 1: Factor the Numerator of the First Fraction ($x^2-16$)

The term $x^2-16$ is a classic example of the difference of squares. It can be factored using the formula $a^2 - b^2 = (a-b)(a+b)$. Here, $a=x$ and $b=4$. Therefore, $x^2-16$ factors into $(x-4)(x+4)$.

Step 2: Factor the Denominator of the First Fraction ($2x+8$)

In the expression $2x+8$, we can see a common factor of 2. Factoring out 2, we get $2(x+4)$.

Step 3: Factor the Numerator of the Second Fraction ($x^3-2x^2+x$)

Here, we can first factor out a common factor of $x$. This leaves us with $x(x^2-2x+1)$. Now, we need to factor the quadratic expression inside the parentheses, $x^2-2x+1$. This is a perfect square trinomial, which factors into $(x-1)^2$. So, the factored form of the numerator is $x(x-1)^2$ or $x(x-1)(x-1)$.

Step 4: Factor the Denominator of the Second Fraction ($x^2+3x-4$)

This is a quadratic trinomial. We need to find two numbers that multiply to -4 and add to 3. These numbers are 4 and -1. So, $x^2+3x-4$ factors into $(x+4)(x-1)$.

Step 5: Rewrite the Entire Expression with Factored Terms

Now, let's substitute the factored forms back into the original expression:

$\frac{(x-4)(x+4)}{2(x+4)} \frac{x(x-1)(x-1)}{(x+4)(x-1)}$

Step 6: Cancel Out Common Factors

Now we look for common factors in the numerators and denominators that can be cancelled.

• We have an $(x+4)$ in the numerator of the first fraction and in the denominator of the first fraction. These cancel out.
• We have an $(x-1)$ in the numerator of the second fraction and in the denominator of the second fraction. These cancel out.
• We have another $(x+4)$ in the denominator of the second fraction. This does not cancel with anything else in the numerator.
• We have an $(x-4)$ in the numerator of the first fraction. This does not cancel with anything else in the denominator.
• We have an $x$ in the numerator of the second fraction. This does not cancel with anything else in the denominator.

After canceling the common factors, we are left with:

$\frac{(x-4)}{2} \frac{x(x-1)}{(x+4)}$

Step 7: Combine the Remaining Terms

Finally, multiply the remaining terms in the numerators and denominators:

Numerator: $(x-4) \times x(x-1) = x(x-4)(x-1)$

Denominator: $2 \times (x+4) = 2(x+4)$

So, the simplified expression is: $\frac{x(x-4)(x-1)}{2(x+4)}$.

Comparing with the Options

Let's compare our simplified expression $\frac{x(x-4)(x-1)}{2(x+4)}$ with the given options:

A. $\frac{x(x-4)(x-1)}{2(x+4)}$ - This matches our result. B. $\frac{\pi x-11}{2}$ - This contains $\pi$ and is linear, clearly not our result. C. $\frac{(x+4)(x-4)}{2 x(x-1)}$ - This has different factors in the numerator and denominator. D. $\frac{(x-4)(x-1)}{2 x(x+4)}$ - This has different factors in the numerator and denominator.

Therefore, the correct option is A.

Conclusion

We have successfully simplified the complex algebraic expression $\frac{x^2-16}{2 x+8} \frac{x^3-2 x^2+x}{x^2+3 x-4}$ by factoring each polynomial and canceling out common factors. This systematic approach ensures accuracy. Remember, mastering algebraic simplification is key to solving more advanced mathematical problems. Practice is crucial, so don't hesitate to work through more examples.

For further exploration into algebraic manipulation and problem-solving, you can visit Khan Academy, a fantastic resource for learning and reinforcing mathematical concepts.