Problem 41

Question

Simplify each expression. $$ \frac{x^{3}+7 x^{2}}{x^{2}+5 x-14} $$

Step-by-Step Solution

Verified
Answer
The simplified expression is \(\frac{x^2}{x - 2}\).
1Step 1: Factor the Numerator
The numerator of the fraction is \(x^3 + 7x^2\). Start by factoring this expression. Both terms have a common factor, which is \(x^2\). Thus, \(x^3 + 7x^2 = x^2(x + 7)\).
2Step 2: Factor the Denominator
The denominator of the fraction is \(x^2 + 5x - 14\). This can be factored by looking for two numbers that multiply to \(-14\) and add up to \(5\). These numbers are \(7\) and \(-2\). Thus, the denominator factors into \((x + 7)(x - 2)\), making the expression \(x^2 + 5x - 14 = (x + 7)(x - 2)\).
3Step 3: Cancel Common Factors
Now that both the numerator and the denominator are factored, look for common factors to cancel. Both have the factor \((x + 7)\). Cancel \((x + 7)\) from both the numerator and the denominator.
4Step 4: Write the Simplified Expression
After canceling the common factor \((x + 7)\), the expression simplifies to \(\frac{x^2}{x - 2}\). This is the simplified form of the given expression.

Key Concepts

Factoring PolynomialsSimplifying ExpressionsRational Functions
Factoring Polynomials
Factoring polynomials is like unlocking a secret code in algebra.
Think of it as breaking down a complicated expression into simpler pieces, or 'factors'. In our original problem, the first step was to factor both the numerator and the denominator.

🔑 **Why Factor?**
  • It makes complex expressions easier to handle.
  • It allows us to see hidden relationships between terms.
  • It gives us the opportunity to simplify expressions.
In the numerator, we have a polynomial expression, \(x^3 + 7x^2\). Here, both terms have a common factor of \(x^2\). We can factor this out to get: \(x^2(x + 7)\). This means that \(x^3 + 7x^2\) is made up of \(x^2\) multiplied by \(x + 7\).

Then, in the denominator, \(x^2 + 5x - 14\) needs two numbers that multiply to \(-14\) and add to \(5\). In this case, those numbers are \(7\) and \(-2\), so our factorization becomes \((x + 7)(x - 2)\). Factoring both parts is crucial for the next steps of simplifying expressions.
Simplifying Expressions
Simplifying expressions means making them as neat and tidy as possible, similar to organizing your room so it's easy to find what you need. After factoring the polynomials, simplification becomes possible if there are any common factors between the numerator and the denominator.

🤔 **How Do We Simplify?**
  • Identify common factors.
  • Cancel those common factors out.
  • Rewrite the expression without the canceled factors.
In the given exercise, after factoring, \(x^2(x + 7)\) over \((x + 7)(x - 2)\), we noticed a common factor, \(x + 7\), in both the numerator and the denominator. Since \(x + 7\) appears in both, it can be canceled.

Once \(x + 7\) is canceled, the expression simplifies to \(\frac{x^2}{x - 2}\), which is a much simpler form to work with. Simplifying expressions makes them more straightforward and easier to understand or use in further calculations.
Rational Functions
A rational function is essentially a fraction but made up of polynomials.
They appear frequently in algebra because they describe ratios of quantities. In this exercise, the rational function started as \(\frac{x^3 + 7x^2}{x^2 + 5x - 14}\). Simplifying this function allows us to better understand its behavior.

📊 **Why Study Rational Functions?**
  • They arise naturally in real-world problems, like calculating rates and densities.
  • They often model relationships like speed and time.
  • Understanding their simplification can clarify what a function is doing without extra "noise" from unnecessary terms.
Once the simplification is done and we are left with \(\frac{x^2}{x - 2}\), it becomes clearer to evaluate or graph this expression. The asymptotes, zeroes, and other characteristics of the function are easier to deduce from its simplified form.

Knowing how to handle rational functions is a powerful tool because it opens the door to solving complex algebraic problems effortlessly.