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Magazine Chapter 6: Problem 53
Factor. See Example 5 or Example 7. $$ 15 a+8 a^{3}-26 a^{2} $$
Short Answer
Expert verified
The expression factors to \(a(4a - 3)(2a - 5)\).
Step by step solution
01
Arrange Terms in Standard Form
First, rearrange the given expression in descending order of powers of \( a \). This helps to clearly see the structure of the polynomial. The expression is initially \(15a + 8a^3 - 26a^2\). Rearranging gives: \[8a^3 - 26a^2 + 15a\]
02
Identify the Greatest Common Factor (GCF)
Next, determine if there is any common factor in all the terms of the polynomial. In this case, the terms are \(8a^3, -26a^2,\) and \(15a\). The common factor in all these terms is \(a\). Factor \(a\) out of each term: \[a(8a^2 - 26a + 15)\]
03
Factor the Quadratic Expression
Now we need to factor the quadratic expression \(8a^2 - 26a + 15\). We'll use the method of factoring by grouping, which requires finding two numbers that multiply to \(8 \times 15 = 120\) and add to \(-26\). These numbers are \(-20\) and \(-6\) since \(-20 \times -6 = 120\) and \(-20 + -6 = -26\). Re-expand the middle term using these factors:\[8a^2 - 20a - 6a + 15\]
04
Apply Factoring by Grouping
Regroup the expression into two pairs and factor each pair:- For \(8a^2 - 20a\), factor out \(4a\), getting \(4a(2a - 5)\). - For \(-6a + 15\), factor out \(-3\), getting \(-3(2a - 5)\).Now the expression becomes:\[4a(2a - 5) - 3(2a - 5)\]
05
Factor Out the Common Binomial Factor
Notice that \((2a - 5)\) is a common factor in both terms of the expression. Factor \((2a - 5)\) out:\[(4a - 3)(2a - 5)\]Now we have factored the original quadratic expression completely as \((4a - 3)(2a - 5)\).
06
Combine All Factors
Combine the GCF from Step 2 with the factored quadratic from Step 5 to write the fully factored form of the original expression:\[a(4a - 3)(2a - 5)\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding the Greatest Common Factor (GCF)
The Greatest Common Factor (GCF) is a crucial tool when factoring polynomials. It's the largest factor that divides each term of a polynomial without leaving a remainder. Identifying the GCF simplifies expressions and is the first step in factoring them.
To find the GCF, examine each term of the polynomial to see what is shared:
To find the GCF, examine each term of the polynomial to see what is shared:
- List the factors of each term. For example, for the polynomial terms \(8a^3, -26a^2,\) and \(15a\), the common factor is \(a\).
- Identify the highest power of any common factors. In our case, \(a\) is the common factor with the lowest power being just \(a^1\). Thus, \(a\) is your GCF.
Exploring Quadratic Expressions
Quadratic expressions are polynomials of degree two. They take the form \(ax^2 + bx + c\). Recognizing and factoring quadratics is a cornerstone of algebra. Our expression, after extracting the GCF, is \(8a^2 - 26a + 15\). This is a quadratic expression, where you seek two numbers that multiply to \(ac\) and add to \(b\).
Steps to tackle quadratic expressions include:
Steps to tackle quadratic expressions include:
- Multiply \(a\) and \(c\), then find two numbers whose product is \(ac\) and sum is \(b\). In our example: \(8 \times 15 = 120\), and \(-26\) are the numbers \(-20\) and \(-6\).
- Split the middle term (\(-26a\)) into two parts using these numbers. This changes the expression to \(8a^2 - 20a - 6a + 15\).
- Use factoring by grouping to simplify it further, a method we'll explore next.
Mastering Factoring by Grouping
Factoring by grouping is an effective strategy for polynomials that involve multiple terms. This method is especially useful when dealing with four-term polynomials, which can arise from tweaking quadratic expressions.
Here is how you can factor by grouping:
Here is how you can factor by grouping:
- Look at pairs of terms in the expanded polynomial form, \(8a^2 - 20a - 6a + 15\).
- Group terms to make factoring easier: \((8a^2 - 20a)\) and \((-6a + 15)\).
- Factor out the greatest common factor from each group separately. From \(8a^2 - 20a\), factor out \(4a\) to get \(4a(2a - 5)\). From \(-6a + 15\), factor out \(-3\) to get \(-3(2a - 5)\).
- Notice that each group now contains the common binomial \((2a - 5)\). Factor this out to simplify further, resulting in \((4a - 3)(2a - 5)\).
