Problem 65 - Beginning and Intermediate Algebra

Question

Factor completely. $$8 c^{2} d^{3}+4 c^{2} d^{2}-60 c^{2} d$$

Step-by-Step Solution

Answer

The completely factored expression is \(2c^2d(2(2d - 5)(d + 3))\).

Step 1: Identify Common Factors

Observe that each term in the expression has a factor of \(2c^2d\).

Step 2: Factor out the Greatest Common Factor (GCF)

Factor \(2c^2d\) from each term in the expression: \begin{align*} 8c^2d^3 + 4c^2d^2 - 60c^2d &= 2c^2d(4d^2 + 2d - 30) \end{align*}

Step 3: Simplify the expression within parentheses

In the parentheses, we have a quadratic expression. Let's first check if there are any common factors amongst the coefficients. The terms within the parentheses are: \[4d^2 + 2d - 30\] All three coefficients are divisible by 2. We can simplify further by dividing each term by 2: \begin{align*} 4d^2 + 2d - 30 &= 2(2d^2 + d - 15) \end{align*}

Step 4: Factor the quadratic expression

Now factor the quadratic expression \(2d^2 + d - 15\). We are looking for two numbers that multiply to -30 (the product of the leading coefficient and the constant term) and add up to 1 (the middle coefficient). These numbers are 6 and -5. We can rewrite the quadratic expression as: \begin{align*} 2d^2 + d - 15 &= 2d^2 + 6d - 5d - 15 \end{align*}

Step 5: Apply factoring by grouping

Now, let's factor by grouping: \begin{align*} 2d^2 + 6d - 5d - 15 &= (2d^2 + 6d) - (5d + 15) \\ &= 2d(d + 3) - 5(d + 3) \end{align*}

Step 6: Factor the common factor from the groups

Both terms now have a common factor of \((d+3)\). Factor it out: \begin{align*} 2d(d+3) - 5(d+3) &= (2d - 5)(d + 3) \end{align*}

Step 7: Write the completely factored expression

Now, substitute the factored quadratic expression back into the overall expression: \begin{align*} 2c^2d(4d^2 + 2d - 30) &= 2c^2d(2(2d^2 + d - 15)) \\ &= 2c^2d(2(2d - 5)(d + 3)) \end{align*} The completely factored expression is: \[2c^2d(2(2d - 5)(d + 3))\]

Key Concepts

Greatest Common Factor

The Greatest Common Factor (GCF) is the largest term that can evenly be divided from all terms within a polynomial.When factoring polynomials, finding the GCF is the first essential step.To uncover the GCF, identify all the common variables and constants shared by each term.
For example, in the expression \(8c^2d^3 + 4c^2d^2 - 60c^2d\), each term contains the factor \(2c^2d\).By factoring this out, the expression becomes \(2c^2d(4d^2 + 2d - 30)\).Here's why the GCF is important:

• It simplifies complex polynomials.
• Reduces the powers of variables, making subsequent factoring easier.
• Also helps in reducing possibilities of calculation errors.

It's key to always factor the GCF first before moving on to other factoring methods.

Factoring by Grouping

Factoring by grouping is a method used for polynomials where terms can be grouped into pairs, and a common factor is then factored out of each group.This method is often used when dealing with expressions involving four terms.
Consider an expression from our problem: \(2d^2 + 6d - 5d - 15\). To factor by grouping, split the expression into two groups: \((2d^2 + 6d)\) and \((-5d - 15)\).
Next, factor out the GCF from each group:

• From the first group: \(2d(d + 3)\)
• From the second group: \(-5(d + 3)\)

Both groups now contain the common factor \((d + 3)\).Finally, factor \((d + 3)\) from both groups to get \((2d - 5)(d + 3)\).This method effectively simplifies polynomials by leveraging common factors within grouped terms.

Quadratic Expression

Quadratic expressions are polynomials of the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).These expressions can often be factored further into simpler binomial terms.
An important part of solving or factoring quadratic expressions is to look for two numbers that multiply to \(a \cdot c\) and add to \(b\).
In our example, the quadratic expression \(2d^2 + d - 15\) needs to be factored.We need to find numbers that multiply to \(-30\) (i.e., \(2 \cdot -15\)) and add to \(1\).These numbers are \(6\) and \(-5\).Rewrite the middle term using these numbers: \(2d^2 + 6d - 5d - 15\).Now it can be factored further using the grouping method.Quadratic expressions require a targeted approach for effective factoring.

Algebraic Expressions

Algebraic expressions consist of variables, coefficients, and constants combined using addition, subtraction, multiplication, and division.Mastering the art of manipulating these expressions is key in algebra.
When tasked with factoring an algebraic expression, like \(8c^2d^3+4c^2d^2-60c^2d\), the goal is to break them down into products of simpler expressions.
Polynomials, a specific type of algebraic expression, often require strategies like finding the GCF or factoring by grouping to simplify.Success in handling algebraic expressions relies on understanding core concepts:

• Recognizing common factors,
• Using methods like factoring by grouping,
• Applying knowledge of quadratic forms,
• Understanding how to rearrange and regroup terms efficiently.

By honing these skills, handling any algebraic expression becomes significantly easier.