Problem 59 - College Algebra with Corequisite Support

Question

For the following exercises, factor the polynomials completely. \(\left(32 x^{3}+48 x^{2}-162 x-243\right)^{-1}\)

Step-by-Step Solution

Answer

The reciprocal of the factored form is \(\frac{1}{(2x + 3)(4x - 9)(4x + 9)}\).

Step 1: Identify and Arrange Polynomial

The given polynomial is \(32x^3 + 48x^2 - 162x - 243\). Our task is to factor this polynomial completely.

Step 2: Group Terms

We will group the terms in pairs: \((32x^3 + 48x^2) + (-162x - 243)\). This makes it easier to factor by grouping.

Step 3: Factor Out the GCF from Each Group

For the first group \(32x^3 + 48x^2\), the greatest common factor (GCF) is \(16x^2\), so we factor that out: \(16x^2(2x + 3)\). For the second group \(-162x - 243\), the GCF is \(-81\), so we factor that out: \(-81(2x + 3)\).

Step 4: Factor by Grouping

Now we have \(16x^2(2x + 3) - 81(2x + 3)\). Notice \(2x + 3\) is common in both terms. Factor \((2x + 3)\) out: \((2x + 3)(16x^2 - 81)\).

Step 5: Factor the Difference of Squares

The expression \(16x^2 - 81\) is a difference of squares, as it can be written as \((4x)^2 - 9^2\). This factors into \((4x - 9)(4x + 9)\).

Step 6: Write the Fully Factored Form

Using our factors from Steps 4 and 5, the polynomial \(32x^3 + 48x^2 - 162x - 243\) can be completely factored as: \((2x + 3)(4x - 9)(4x + 9)\).

Step 7: Identify the Role of the Inverted Exponent

Originally, the polynomial was enclosed in a parenthesis with a raised exponent \((-1)\), meaning it is asking for the reciprocal of the factored polynomial. Thus, the reciprocal is: \(\frac{1}{(2x + 3)(4x - 9)(4x + 9)}\).

Key Concepts

Difference of Squares

A difference of squares is a useful pattern that simplifies the process of factoring certain types of quadratic expressions. To recognize a difference of squares, look for two squared terms separated by a subtraction sign. In the expression from the exercise, we have a difference of squares in the term \(16x^2 - 81\).
Here’s how this works:

• Identify each square component. For \(16x^2\), it is \((4x)^2\), and for \(81\), it is \(9^2\).
• The difference of squares formula is \(a^2 - b^2 = (a - b)(a + b)\).
• Applying this formula, \((4x)^2 - 9^2\) transforms to \((4x - 9)(4x + 9)\).

This technique significantly changes complex-looking expressions into simpler, more usable forms, enabling further numeral or algebraic operations if needed.

Greatest Common Factor

The Greatest Common Factor (GCF) is a critical part of factoring polynomials, as it allows simplification of expressions by removing the largest shared factor. To find the GCF of each group in our polynomial, analyze the coefficients and variables of each term.
In the exercise, two pairs were identified:

• For the first group, \(32x^3 + 48x^2\), the GCF is \(16x^2\) because 16 divides both coefficients, and \(x^2\) is the highest power of \(x\) common to both terms.
• We factor this to \(16x^2(2x + 3)\).
• For the second group, \(-162x - 243\), the GCF is \(-81\), leading to \(-81(2x + 3)\).

Factoring out the GCF simplifies the polynomial and sets the stage for using techniques like factoring by grouping.

Factoring by Grouping

Factoring by grouping is an effective strategy used primarily for polynomials with four terms. This technique groups terms into pairs, facilitating easier factorization.
In the provided exercise, the polynomial \(32x^3 + 48x^2 - 162x - 243\) was grouped as follows:

• The first pair: \(32x^3 + 48x^2\)
• The second pair: \(-162x - 243\)

Once these groups are identified and the GCF is factored out from each, terms become manageable.

• After factoring the GCF from each pair, the expression is \(16x^2(2x + 3) - 81(2x + 3)\).
• The common factor here is \(2x + 3\), which can be factored out resulting in \((2x + 3)(16x^2 - 81)\).

This ultimately simplifies the polynomial to a point where other factoring methods, such as the difference of squares, can be applied to complete the process.