Problem 42 - College Algebra with Corequisite Support

Question

For the following exercises, factor the polynomials. \(729 q^{3}+1331\)

Step-by-Step Solution

Answer

The polynomial \(729 q^{3}+1331\) is factored as \((9q + 11)(81q^2 - 99q + 121)\).

Step 1: Identify the Form

The given polynomial is a sum of cubes: \(729q^3 + 1331\). Recognize that this fits the form of \(a^3 + b^3\), where \(a = 9q\) and \(b = 11\) since \(729=(9)^3\) and \(1331=(11)^3\).

Step 2: Recall the Formula for Sum of Cubes

The formula for factoring the sum of cubes is \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\).

Step 3: Substitute the Values into the Sum of Cubes Formula

Substitute \(a = 9q\) and \(b = 11\) into the sum of cubes formula: \[(9q)^3 + (11)^3 = (9q + 11)((9q)^2 - 9q \cdot 11 + 11^2)\].

Step 4: Calculate Each Part

- \((9q)^2 = 81q^2\)- \(9q \cdot 11 = 99q\)- \(11^2 = 121\)Thus the polynomial is \( (9q + 11)(81q^2 - 99q + 121)\).

Step 5: Write the Factored Form

Therefore, the given polynomial \(729 q^{3}+1331\) is factored as \((9q + 11)(81q^2 - 99q + 121)\).

Key Concepts

Sum of Cubes

The sum of cubes is an essential concept when factoring polynomial expressions. It involves polynomials that can be expressed in the form \(a^3 + b^3\). In our exercise, the polynomial \(729q^3 + 1331\) fits this form. We must recognize that \(729q^3\) is equivalent to \((9q)^3\) and \(1331\) is equivalent to \(11^3\). Thus, fundamentally, we have \(a = 9q\) and \(b = 11\) in the expression. Understanding that numbers like 729 and 1331 are perfect cubes allows students to easily apply this technique. When you spot such expressions, it directs you to tackle them using the sum of cubes factoring strategy. By recognizing this structure, it simplifies the problem significantly, turning it into an easier algebraic task. A sum of cubes can be decomposed using the formula:

• \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

This formula is elegant and powerful, transforming complex polynomial expressions into more manageable forms.

Factoring Techniques

Factoring techniques are a cornerstone in solving polynomial expressions efficiently. For expressions like the sum of cubes, knowing the right factoring formula streamlines the process. In our example, after identifying \(a = 9q\) and \(b = 11\), we use the sum of cubes formula:

• \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\)

Once identified, substitute \(a\) and \(b\) into this formula, transforming the polynomial \[(9q)^3 + 11^3 = (9q + 11)(81q^2 - 99q + 121)\].This approach of substitution is crucial. It's like solving a puzzle; once you have the correct pieces (\(a\) and \(b\)), the rest follows with methodical calculation:

• \((9q)^2 = 81q^2\)
• \(9q \cdot 11 = 99q\)
• \(11^2 = 121\)

With these calculations, the polynomial is efficiently decomposed into: \((9q + 11)(81q^2 - 99q + 121)\). Factoring polynomials not only simplifies the problem but is also a fundamental skill in algebra, allowing further calculations and simplifications of expressions.

Polynomial Expressions

Polynomial expressions are made up of terms with varying degrees of exponents. The expression \(729q^3 + 1331\) from our exercise is a 3rd-degree polynomial because of the highest exponent, which is \(q^3\). Understanding the nature of polynomials is key for factoring and transforming them.Different kinds of polynomials require different approaches. For instance:

• Linear polynomials (degree 1): \(ax + b\)
• Quadratic polynomials (degree 2): \(ax^2 + bx + c\)
• Cubic polynomials (degree 3): \(ax^3 + bx^2 + cx + d\)

Our polynomial is a sum of cubes, a specific type within cubic polynomials. Recognizing it helps in choosing the correct strategy to factor it. This recognition and transformation into its simplest form make complex algebra problems manageable.By understanding polynomial structure and degree, we navigate through algebra with ease, applying distinct techniques effectively based on the type of polynomial presented.