Problem 75 - Intermediate Algebra

Question

Perform the operations and simplify. $$ \frac{p^{3}-q^{3}}{p^{2}-q^{2}} \cdot \frac{q^{2}+p q}{p^{3}+p^{2} q+p q^{2}} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{q^{2}+pq}{p(p+q)} \).

Step 1: Simplify the first fraction

The first fraction is \( \frac{p^{3}-q^{3}}{p^{2}-q^{2}} \). We recognize that \( p^{3} - q^{3} \) is a difference of cubes and can be factored using the formula \( a^{3} - b^{3} = (a-b)(a^{2}+ab+b^{2}) \). So, \( p^{3} - q^{3} = (p-q)(p^{2}+pq+q^{2}) \). The denominator is a difference of squares: \( p^{2} - q^{2} = (p-q)(p+q) \). Thus, the fraction simplifies to:\[\frac{(p-q)(p^{2} + pq + q^{2})}{(p-q)(p+q)}\]By canceling common terms \( (p-q) \), we simplify to:\[\frac{p^{2} + pq + q^{2}}{p+q}\]

Step 2: Simplify the second fraction

The second fraction is \( \frac{q^{2}+p q}{p^{3}+p^{2} q+p q^{2}} \). Notice that the denominator \( p^{3} + p^{2} q + pq^{2} \) can be factored by taking \( p \) common:\[p(p^{2} + pq + q^{2})\]Since the numerator is already simplified with no common factors with the denominator \( q^{2} + pq \), this fraction becomes:\[\frac{q^{2}+p q}{p (p^{2}+pq+q^{2})}\]

Step 3: Multiplication of the two fractions

Multiply the simplified fractions from Step 1 and Step 2:\[\frac{p^{2} + pq + q^{2}}{p+q} \cdot \frac{q^{2}+p q}{p (p^{2}+pq+q^{2})}\]When multiplying, we observe that \( p^{2} + pq + q^{2} \) exists in both the numerator and the denominator, hence these terms cancel each other out:\[\frac{q^{2}+pq}{p(p+q)}\]

Step 4: Final simplification

After cancellation, we have \( \frac{q^{2}+pq}{p(p+q)} \). This cannot be simplified further, as there are no common terms left to cancel between the numerator and the denominator. Hence, this is our final simplified expression.

Key Concepts

Factoring Polynomials

Polynomials are algebraic expressions made up of terms involving variables and coefficients. Sometimes, these expressions can be rewritten as products of simpler polynomials, a process known as factoring. Factoring is crucial because it simplifies polynomial expressions and helps us solve mathematical equations more efficiently.

To factor a polynomial, one can look for common factors in the terms of the expression, or apply special formulas such as the

• difference of cubes formula: \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \).
• sum or difference of squares formula: \( a^2 - b^2 = (a-b)(a+b) \).

Applying these formulas makes it much easier to manipulate and solve polynomial equations. In the exercise, we've used the difference of cubes and difference of squares to simplify complex expressions into manageable parts.

Difference of Cubes

The difference of cubes is a polynomial pattern that arises when you subtract one cubic expression from another. This pattern is expressed in the formula: \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \).
This formula helps break down cubic expressions into a product of a linear factor and a quadratic factor. For example, when factoring \( p^3 - q^3 \), we identify \( a = p \) and \( b = q \). Applying the formula gives us:

• First, the linear factor: \( (p-q) \).
• Then, the quadratic factor: \( (p^2 + pq + q^2) \).

By understanding and applying the difference of cubes, we manage to streamline complex fraction operations into simpler expressions.

Difference of Squares

The difference of squares is another essential polynomial pattern that involves subtracting one perfect square term from another. The formula to express this difference is: \( a^2 - b^2 = (a-b)(a+b) \).
This is commonly used in algebra to simplify expressions, as it provides two linear factors which are easier to work with compared to quadratic or cubic terms.

In the exercise, the term \( p^2 - q^2 \) is recognized as a difference of squares. By setting \( a = p \) and \( b = q \), the expression factors into the product \( (p-q)(p+q) \).
This allows for simplification through cancellation and can be crucial in reducing higher-order polynomial fractions.

Simplification Steps

Simplification steps are foundational in solving algebraic expressions, involving structured strategies to make expressions more manageable. The process typically involves:

• Recognizing patterns such as the difference of cubes or squares.
• Factoring to rewrite expressions in simpler forms.
• Cancelling common terms between numerators and denominators.
• Performing operations, like multiplication, on simplified expressions.

In our exercise, we started by simplifying the fractions individually. This was done by factoring numerators and denominators using recognizable patterns, followed by cancelling common expressions.

Through the multiplication of these simplified parts, we were then able to further cancel common terms, eventually arriving at a simplified expression. Having a clear understanding of these steps helps ensure accuracy and efficiency in problem-solving.