Problem 42 - Elementary and Intermediate Algebra

Question

Factor expression. Factor out any GCF first. \(3 a c^{4}-243 a\)

Step-by-Step Solution

Answer

The expression factors to \(3a(c^2 + 9)(c + 3)(c - 3)\).

Step 1: Identify the GCF

First, identify the greatest common factor (GCF) of the terms in the expression. In this case, the terms are \(3ac^4\) and \(-243a\). The GCF is \(3a\) because it is the largest factor common to both terms.

Step 2: Factor out the GCF

Write the expression as a product of the GCF and the remaining terms. Factoring out \(3a\) from each term gives:\[3a(c^4 - 81)\]

Step 3: Recognize the Difference of Squares

The expression inside the parentheses \((c^4 - 81)\) is a difference of squares. Recall that \(c^4 - 81\) can be rewritten as \((c^2)^2 - (9)^2\). This matches the form \(A^2 - B^2\), where \(A = c^2\) and \(B = 9\).

Step 4: Apply the Difference of Squares Formula

Apply the difference of squares factorization formula \(A^2 - B^2 = (A + B)(A - B)\):\[(c^2 + 9)(c^2 - 9)\]

Step 5: Factor Again

Notice that \(c^2 - 9\) is also a difference of squares: \((c)^2 - (3)^2\). Apply the formula again:\[(c + 3)(c - 3)\]

Step 6: Write the Final Factorization

Substitute back into the expression to write the fully factored form:\[3a(c^2 + 9)(c + 3)(c - 3)\]

Key Concepts

Greatest Common Factor

Factoring expressions becomes simpler when you begin by identifying the Greatest Common Factor (GCF). The GCF is the highest number or variable that divides all terms of an expression without a remainder. It's essential because factoring it out reduces the complexity of the expression.

Let's consider our expression: \(3ac^4 - 243a\). Look at the coefficients and variables:

• Coefficients are 3 and 243.
• Variables include \(a\), appearing in both terms.

The greatest integer dividing both 3 and 243 is 3. Since \(a\) is present in both terms, the GCF is \(3a\).

By factoring \(3a\) out, we reduce the expression to \(3a(c^4 - 81)\). This simplification makes the next steps easier to handle, emphasizing the GCF's role in organizing terms and simplifying expressions.

Difference of Squares

The Difference of Squares formula is a powerful tool in algebra for factoring specific types of expressions. It follows the principle: \(A^2 - B^2 = (A + B)(A - B)\).

In the expression \(c^4 - 81\), notice the subtraction between two square terms:

• \((c^2)^2 = c^4\)
• \(9^2 = 81\)

This matches the form \(A^2 - B^2\) with \(A = c^2\) and \(B = 9\). By applying the Difference of Squares formula, you rewrite it as \((c^2 + 9)(c^2 - 9)\).

This approach breaks down complex squares into simpler, manageable products, showcasing the elegance and efficiency of the formula.

Algebraic Factorization

Algebraic Factorization involves breaking down expressions into products of simpler factors. It often involves recognizing patterns, like the difference of squares or trinomials, to simplify expressions.

After applying the difference of squares to \(c^4 - 81\), resulting in \((c^2 + 9)(c^2 - 9)\), the process doesn't stop there. Notably, \(c^2 - 9\) is itself a difference of squares:

• Here, \(c^2 = (c)^2\)
• \(9 = (3)^2\)

Now, apply the Difference of Squares formula again: \((c + 3)(c - 3)\).

Thus, the complete factorization becomes \(3a(c^2 + 9)(c + 3)(c - 3)\), fully breaking down the original expression into its simplest factors. This strategy highlights how recognizing patterns and systematic factoring can transform and simplify complex algebraic expressions.