Problem 48

Question

For the following exercises, factor the polynomials. $$ 9 y(3 y-13)^{\frac{1}{5}}-2(3 y-13)^{\frac{6}{5}} $$

Step-by-Step Solution

Verified
Answer
The factorized expression is \((3 y - 13)^{\frac{1}{5}} (3 y + 26)\).
1Step 1: Recognize Common Factor
Observe both terms in the polynomial: \(9 y(3 y-13)^{\frac{1}{5}}\) and \(-2(3 y-13)^{\frac{6}{5}}\).Both terms share a common factor of \((3y-13)^{\frac{1}{5}}\). Identify and factor this common factor out of the expression.
2Step 2: Factor Out the Common Factor
Factor out \((3 y - 13)^{\frac{1}{5}}\) from the polynomial:\[9 y (3 y - 13)^{\frac{1}{5}} - 2 (3 y - 13)^{\frac{6}{5}} = (3 y - 13)^{\frac{1}{5}} \left(9 y - 2 (3 y - 13)\right)\]Here, we have extracted the common factor from each term, leaving the expression in the parentheses to simplify further.
3Step 3: Simplify the Remaining Expression
Inside the parentheses, simplify the expression \(9 y - 2 (3 y - 13)\):- Distribute \(-2\): - \(-2 \times 3y = -6y\) - \(-2 \times (-13) = 26\) Thus, we have:\[9y - 6y + 26 = 3y + 26\]
4Step 4: Write the Factored Expression
Combine the factored term with the simplified expression:\[(3 y - 13)^{\frac{1}{5}} (3 y + 26)\]This is the factorization of the given polynomial.

Key Concepts

Common FactorExponents in PolynomialsSimplifying Expressions
Common Factor
When working with polynomials, identifying a common factor is key to simplifying expressions. A common factor is an expression that is present in each term of a polynomial. In this exercise, we see the terms
  • \(9y(3y-13)^{\frac{1}{5}}\)
  • \(-2(3y-13)^{\frac{6}{5}}\)
Both share a piece that looks the same: \((3y-13)^{\frac{1}{5}}\). This means that \((3y-13)^{\frac{1}{5}}\) is a common factor. By spotting and factoring out this common factor, we significantly simplify the expression, making the equation easier to work with. Always remember that finding a common factor is a strategic step in simplifying and rewriting polynomial expressions in a more manageable form. By factoring it out, we are essentially breaking down the polynomial into simpler, separate parts. This not only helps in manual calculations but is also useful for understanding the structure of the expression.
Exponents in Polynomials
Exponents appear frequently in polynomials, defining the degree of power to which a term is raised. In this scenario, we encounter fractional exponents, which can often pose a challenge. In this problem, we see terms like
  • \((3y-13)^{\frac{1}{5}}\)
  • \((3y-13)^{\frac{6}{5}}\)
The exponents \(\frac{1}{5}\) and \(\frac{6}{5}\) might look intimidating at first. However, understanding them is crucial. They represent roots: the expression \((3y-13)^{\frac{1}{5}}\) indicates the fifth root of \((3y-13)\). Meanwhile, \((3y-13)^{\frac{6}{5}}\) means the entire expression is first raised to the power of 6, then the fifth root is taken.It's important to remember that when factoring, consistent bases with different exponents allow us to apply properties of exponents. For instance, when we factor out a common exponential term, we're leveraging these properties to simplify complex polynomial expressions. Recognizing these nuances with exponents can profoundly impact solving polynomial equations efficiently.
Simplifying Expressions
After factoring out common factors and dealing with exponents, simplifying expressions is the next essential step. It involves reducing complex terms to their simplest form. For the polynomial in question, after extracting the common factor, we are left with
  • \(9y - 2(3y-13)\)
Simplifying this involves some straightforward arithmetic within the parentheses:
  • First, distribute \(-2\) inside \((3y - 13)\), giving:
    • \(-2 \times 3y = -6y\)
    • \(-2 \times (-13) = 26\)
Then, combine like terms:
  • \(9y - 6y = 3y\)
  • Add the result to 26: \(3y + 26\)
These steps lead us to a simplified expression. Simplifying expressions reduces complexity and clarifies the final factors of the polynomial. It also aids in further calculations or evaluations, illustrating why each step in polynomial factorization is crucial.