Problem 51 - Beginning and Intermediate Algebra

Question

Factor completely. $$12 c^{3}+15 c^{2}-18 c$$

Step-by-Step Solution

Answer

The completely factored expression of \(12c^3 + 15c^2 - 18c\) is \(3c(c + 2)(4c - 3)\).

Step 1: Identify the Greatest Common Factor (GCF)

In the given expression \(12c^3 + 15c^2 - 18c\), we can see that the coefficients have a common factor of 3 and each term has a variable factor of c. Therefore, the GCF is \(3c\).

Step 2: Factor out the GCF

Divide each term by the GCF and factor it out to get: \[3c(4c^2 + 5c - 6)\]

Step 3: Factor the quadratic expression

We now need to factor the quadratic expression \(4c^2 + 5c - 6\). Since this is a quadratic expression of the form \(ax^2 + bx + c\), we will look for two binomials that multiply to the given expression. We look for two numbers whose product is equal to the product of the first and last terms (4 * -6 = -24) and whose sum is equal to the middle term's coefficient, which is 5. Those two numbers are 8 and -3, as \(8 * (-3) = -24\) and \(8 + (-3) = 5\). Now, we can rewrite the middle term, as: \[4c^2 + 8c - 3c - 6\]

Step 4: Factor by grouping

Group the terms in pairs and factor each pair: \[4c(c + 2) - 3(c + 2)\] Now, factor out the common factor \((c+2)\) to get: \[(c + 2)(4c - 3)\]

Step 5: Write the complete factorization

Now that we have factored both the GCF and the quadratic expression, we can write the complete factorization as: \[3c(c + 2)(4c - 3)\] So, the completely factored expression of \(12c^3 + 15c^2 - 18c\) is \(3c(c + 2)(4c - 3)\).

Key Concepts

Greatest Common Factor

To begin factoring a polynomial, identifying the Greatest Common Factor (GCF) is a crucial step. The GCF is the largest factor shared by all terms in a polynomial. It can include both numerical and variable components.
For example, in the expression \(12c^3 + 15c^2 - 18c\), the GCF involves both numbers and letters: 3 and \(c\). Here's why:

• **Numerically**: The numbers 12, 15, and 18 all share a factor of 3.
• **Variable**: Each term has at least one \(c\) as a variable factor.

Once the GCF is identified, it is factored out of each term. This simplifies the polynomial to make further factoring easier.
This process reduced the original expression to \(3c(4c^2 + 5c - 6)\), setting the stage for further steps.

Quadratic Expression

After factoring out the GCF, the next task is to focus on the quadratic expression. A quadratic expression typically takes the form \(ax^2 + bx + c\), similar to what we see in \(4c^2 + 5c - 6\). Here, 'a', 'b', and 'c' are coefficients in the expression.
The goal is to break down this type of polynomial into the product of two binomials. For \(4c^2 + 5c - 6\), it requires finding two numbers that multiply to \(-24\) (product of coefficient of \(c^2\) and constant term, \(4\times-6\)) and add up to 5 (coefficient of the \(c\) term).
These numbers are 8 and -3. By rewriting the middle term using these numbers, we modify the quadratic into four terms: \(4c^2 + 8c - 3c - 6\).
This adjustment allows the polynomial to be grouped and factored further.

Factor by Grouping

Once the quadratic expression is broken down, the next step is to factor by grouping. This method involves organizing the terms into pairs and factoring each pair separately.
Using our example \(4c^2 + 8c - 3c - 6\), we group them into two parts: \(4c^2 + 8c\) and \(-3c - 6\).
Here's how you do it:

• **First Pair**: Factor \(4c\) out of \(4c^2 + 8c\), resulting in \(4c(c + 2)\).
• **Second Pair**: Factor \(-3\) out of \(-3c - 6\), resulting in \(-3(c + 2)\).

Notice that both groups contain the common factor \((c + 2)\). Pull this common factor out front, leaving us with \((c + 2)(4c - 3)\).
This completes the factoring of the quadratic expression. Combining this with the earlier factored GCF, the entire expression is \(3c(c + 2)(4c - 3)\).
Factoring by grouping is all about simplifying complex expressions by strategically pairing and reducing terms.