Factorization Common Factor Practice Problems & Examples

Master factorization by extracting common factors with step-by-step practice problems. Learn to convert addition expressions to multiplication form easily.

📚Master Common Factor Extraction with Interactive Practice
  • Extract common numerical factors from algebraic expressions like 2A + 4B
  • Factor out common variables from terms such as 4X + CX = X(4 + C)
  • Apply exponent laws to factor expressions like Z⁵ + 3Z⁷ = Z⁵(1 + 3Z²)
  • Handle multi-term expressions with three or more summands efficiently
  • Factor expressions in parentheses like 3A(B-5) + 8(B-5)
  • Work with opposite signs in brackets using the distributive property

Understanding Factorization: Common factor extraction

Complete explanation with examples

Factorization

The factorization we do by extracting the common factor is our way of modifying the way the exercise is written, that is, from an expression with addition to an expression with multiplication.

For example, the expression
2A+4B2A + 4B
is composed of two terms and a plus sign. We can factor it by excluding the largest common term.
In this case it is 2 2 .

We will write it as follows:
​​​​​​​2A+4B=2×(A+2B)​​​​​​​2A + 4B = 2\times (A + 2B)

Since both terms ( A A and B B ) were multiplied by 2 2 we could "extract" it. The remaining expression is written in parentheses and the common factor (the 2 2 ) is kept out.
In this way we went from having two terms in an addition operation to having a multiplication. This procedure is called factorization.

A - Factorization

You can also apply the distributive property to do a reverse process as needed.
In certain cases we will prefer to have a multiplication and in others an addition.

Detailed explanation

Practice Factorization: Common factor extraction

Test your knowledge with 36 quizzes

Break down the expression into basic terms:

\( 5x^2 + 10 \)

Examples with solutions for Factorization: Common factor extraction

Step-by-step solutions included
Exercise #1

Break down the expression into basic terms:

4x2+6x 4x^2 + 6x

Step-by-Step Solution

To break down the expression4x2+6x 4x^2 + 6x into its basic terms, we need to look for a common factor in both terms.

The first term is 4x2 4x^2 , which can be rewritten as 4xx 4\cdot x\cdot x .

The second term is6x 6x , which can be rewritten as 23x 2\cdot 3\cdot x .

The common factor between the terms is x x .

Thus, the expression can be broken down into 4x2+6x 4\cdot x^2 + 6\cdot x , and further rewritten with common factors as 4xx+6x 4\cdot x\cdot x + 6\cdot x .

Answer:

4xx+6x 4\cdot x\cdot x+6\cdot x

Exercise #2

Simplify the expression:

5x3+3x2 5x^3 + 3x^2

Step-by-Step Solution

To simplify the expression 5x3+3x2 5x^3 + 3x^2 , we can break it down into basic terms:

The term 5x3 5x^3 can be written as 5xxx 5 \cdot x \cdot x \cdot x .

The term3x2 3x^2 can be written as 3xx 3 \cdot x \cdot x .

Thus, the expression simplifies to5xxx+3xx 5 \cdot x \cdot x \cdot x + 3 \cdot x \cdot x .

Answer:

5xxx+3xx 5\cdot x\cdot x\cdot x + 3\cdot x\cdot x

Exercise #3

Break down the expression into basic terms:

2x2 2x^2

Step-by-Step Solution

The expression 2x2 2x^2 can be factored and broken down into the following basic terms:

  • The coefficient 2 2 remains as it is since it is already a basic term.
  • The term x2 x^2 can be broken down into xx x \cdot x .
  • Therefore, the entire expression can be written as 2xx 2 \cdot x \cdot x .

This breakdown helps in understanding the multiplicative nature of the expression.

Among the provided choices, the correct one that matches this breakdown is choice 2: 2xx 2\cdot x\cdot x .

Answer:

2xx 2\cdot x\cdot x

Exercise #4

Rewrite using basic components:

8x24x 8x^2 - 4x

Step-by-Step Solution

To rewrite the expression 8x24x 8x^2 - 4x using its basic components, we'll follow these steps:

  • Step 1: Identify the greatest common factor of the terms.
  • Step 2: Factor each term using the greatest common factor.

Let's go through each step:

Step 1: Recognize that both terms 8x2 8x^2 and 4x 4x contain x x as a common factor.
Moreover, the numerical coefficients 8 and 4 have a common factor of 4.

Step 2: Factor the expression:
- 8x2 8x^2 can be expressed as 8xx 8 \cdot x \cdot x .
- 4x 4x can be written as 4x 4 \cdot x .

Bringing them together, we can rewrite the expression:

8x24x=8xx4x 8x^2 - 4x = 8 \cdot x \cdot x - 4 \cdot x .

Thus, the solution to the problem is 8xx4x 8\cdot x\cdot x-4\cdot x .

Answer:

8xx4x 8\cdot x\cdot x-4\cdot x

Exercise #5

Break down the expression into basic terms:

3y3 3y^3

Step-by-Step Solution

To break down the expression 3y3 3y^3 into its basic terms, we understand the components of the expression:

3is a constant multiplier 3 \, \text{is a constant multiplier}

y3 y^3 can be rewritten as yyy y \cdot y \cdot y

Thus, 3y3 3y^3 can be decomposed into 3yyy 3 \cdot y \cdot y \cdot y .

Answer:

3yyy 3\cdot y\cdot y \cdot y

Frequently Asked Questions

What is factorization by common factor in algebra?

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Factorization by common factor is the process of converting an expression with addition into multiplication form by extracting the largest common term. For example, 2A + 4B becomes 2(A + 2B) by extracting the common factor 2.

How do you find the greatest common factor in algebraic expressions?

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To find the greatest common factor: 1) Identify common numerical coefficients, 2) Find common variables with the lowest exponents, 3) Combine both parts. For 6A³ + 9A⁵, the GCF is 3A³.

What's the difference between factoring and the distributive property?

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Factoring extracts common factors to go from addition to multiplication (2A + 4B = 2(A + 2B)). The distributive property does the reverse, expanding multiplication into addition (2(A + 2B) = 2A + 4B).

How do you factor expressions with variables and exponents?

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Use the exponent rule aᵐⁿ = aᵐ × aⁿ to break down terms. For Z⁵ + 3Z⁷, rewrite as Z⁵ + 3Z⁵ × Z², then extract Z⁵ to get Z⁵(1 + 3Z²).

Can you factor expressions with more than two terms?

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Yes, you can factor multi-term expressions by finding the greatest common factor among all terms. For 3A³ + 6A⁵ + 9A⁴, extract 3A³ to get 3A³(1 + 2A² + 3A).

How do you handle expressions with parentheses when factoring?

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When the same expression in parentheses appears in multiple terms, treat it as the common factor. For 3A(B-5) + 8(B-5), extract (B-5) to get (B-5)(3A + 8).

What do you do when expressions have opposite signs in brackets?

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Use the fact that (X-4) = -1(4-X). For 3(X-4) + X(4-X), rewrite as 3(X-4) - X(X-4), then factor out (X-4) to get (X-4)(3-X).

How can I check if my factorization is correct?

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Use the distributive property to expand your factored form back to the original expression. If you get the same starting expression, your factorization is correct.

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