Problem 26 - Elementary Algebra

Question

For the following problems, use the grouping method to factor the polynomials. Some polynomials may not. be factorable using the grouping method. $$ 4 a^{3} b c-14 a^{2} b c^{3}+10 a b c^{2}-35 b c^{4} $$

Step-by-Step Solution

Answer

Question: Use the grouping method to factor the polynomial \(4a^3bc - 14a^2bc^3 + 10abc^2 - 35bc^4\). Answer: \((2a^2bc + 5bc^2)(2a - 7c^2)\)

Step 1: Group the terms in pairs

First, group the terms in pairs. We'll have two pairs: $$ (4a^3bc - 14a^2bc^3) + (10abc^2 - 35bc^4) $$

Step 2: Determine the Greatest Common Factors (GCF) for each pair

Now, we'll find the greatest common factors for each pair of terms: For the first pair \((4a^3bc - 14a^2bc^3)\), the GCF is \(2a^2bc\). For the second pair \((10abc^2 - 35bc^4)\), the GCF is \(5bc^2\).

Step 3: Factor out the GCFs for each pair

Factor out the GCFs from each pair of terms: $$ 2a^2bc(2a - 7c^2) + 5bc^2(2a - 7c^2) $$

Step 4: Determine the overall GCF

Now, we can see that both pairs have a common factor of \((2a - 7c^2)\). Thus, we can factor this out of both pairs of terms:

Step 5: Factor the polynomial using the grouping method

Factor the polynomial: $$ (2a^2bc + 5bc^2)(2a - 7c^2) $$ So the factored form of the polynomial using the grouping method is: $$ (2a^2bc + 5bc^2)(2a - 7c^2) $$

Key Concepts

Grouping Method

Polynomials can sometimes be broken down into simpler expressions using a method known as the "Grouping Method." This approach is especially useful when a straightforward factorization isn't immediately obvious. The idea is to separate a polynomial into smaller groups, where each group can be individually factored. In simpler terms, think of it as slicing a cake into smaller pieces before you serve it.Here's how it works:

• **Identify Pairs**: Start by dividing the polynomial into two pairs of terms. In the example \(4a^3bc - 14a^2bc^3 + 10abc^2 - 35bc^4\), the terms are grouped as \((4a^3bc - 14a^2bc^3)\) and \((10abc^2 - 35bc^4)\).
• **Factor Each Group**: Treat each pair as if it were its own mini problem, and factor out the greatest common factor (GCF) from each.
• **Combine with New Common Factor**: Often, these factored pairs will reveal a common expression, which can then be factored out to simplify the original polynomial.

Using steps, like in this example, makes solving complicated polynomial problems less daunting.

Greatest Common Factor

Finding the *Greatest Common Factor* (GCF) is an essential skill when factoring polynomials. The GCF is the largest number or expression that divides two or more numbers or terms without leaving a remainder.Here are the steps to identifying the GCF:

• **Look for Common Terms**: The GCF may consist of a numerical factor (like 2 or 5) as well as variables (like \(a^2\), \(b\), or \(c\)); both elements are required.
• **Choose the Lowest Power of Variables**: When factoring out common variables, select the lowest power present in all terms. In the provided problem, for \(4a^3bc - 14a^2bc^3\), the GCF was \(2a^2bc\), using the smallest exponent for each variable across terms.

By identifying the GCF correctly, you simplify the process of factoring larger expressions, and make the larger problem more readable and manageable.

Factoring Polynomials

Factoring polynomials involves breaking down a complex expression into simpler pieces or factors that, when multiplied, give back the original polynomial. Think of it like reverse engineering.Key points for factoring:

• **Different Methods**: There are different strategies, such as grouping, used depending on the form of the polynomial.
• **The Role of GCF**: Often, the first step is identifying any Greatest Common Factor as it simplifies the polynomial before further steps.
• **End Goal**: A fully factored polynomial means you've expressed it as a product of irreducible polynomials. For instance, given \( (2a^2bc + 5bc^2)(2a - 7c^2) \), neither of the factors can be broken down further.

Understanding factoring polynomials helps in solving equations and simplifying algebraic expressions, making it a fundamental algebraic skill.