Break down the expression into basic terms: $$ 5m $$

$$ $$$$ m\cdot m\cdot m\cdot m\cdot m $$

Break down the expression into basic terms: $$ 3x^2 + 2x $$

$$ 3\cdot x^2 + x + 2\cdot x $$

Break down the expression into basic terms: $$ 4x^2 + 3x $$

$$ $$$$ 7\cdot x\cdot x\cdot x $$

Rewrite using basic components: $$ 6z^2 + z $$

$$ 6\cdot z\cdot z + 1\cdot z $$

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

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

Since both terms ( $A$ and $B$ ) were multiplied by $2$ we could "extract" it. The remaining expression is written in parentheses and the common factor (the $2$ ) 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.

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

Simplify the expression:

$$ 5x^3 + 3x^2 $$

Examples with solutions for Factorization: Common factor extraction

Step-by-step solutions included

Exercise #1

Break down the expression into basic terms:

$2x^2$

Step-by-Step Solution

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

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

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: $2\cdot x\cdot x$ .

Answer:

$2\cdot x\cdot x$

Exercise #2

Break down the expression into basic terms:

$6x$

Step-by-Step Solution

To solve this problem, we'll clearly delineate the expression $6x$ as follows:

The number 6 is the coefficient.
The letter $x$ is the variable.
These two components are connected by multiplication, represented as $6 \cdot x$ .

Thus, the expression $6x$ is equivalent to $6 \cdot x$ , where 6 is multiplied by $x$ .

Examining the choice options:

Choice 1: $6\cdot x$ is correct because it represents the expression as a product of the coefficient and the variable.
Choice 2: $x \cdot x \cdot x \cdot x \cdot x \cdot x$ represents a repeated multiplication of $x$ , not applicable here.
Choice 3: $\frac{6}{x}$ represents division, not the required approach.
Choice 4: Incorrect, as $6x$ can indeed be expressed as $6 \cdot x$ .

Therefore, the best breakdown of the expression is $6 \cdot x$ , matching choice 1.

Answer:

$6\cdot x$

Exercise #3

Break down the expression into basic terms:

$8y^2$

Step-by-Step Solution

To break down the expression $8y^2$ , we identify the basic components. The expression $y^2$ is a shorthand for $y \times y$ . Therefore, $8y^2$ can be decomposed as $8 \cdot y \cdot y$ .

Answer:

$8\cdot y\cdot y$

Exercise #4

Break down the expression into basic terms:

$3a^3$

Step-by-Step Solution

To break down the expression $3a^3$ , we recognize that $a^3$ means $a \times a \times a$ . Therefore, $3a^3$ can be decomposed as $3 \cdot a\cdot a\cdot a$ .

Answer:

$3 \cdot a\cdot a\cdot a$

Exercise #5

Break down the expression into basic terms:

$8y$

Step-by-Step Solution

To break down the expression $8y$ , we can see it as the multiplication of $8$ and $y$ :

$8y = 8 \cdot y$

This shows the expression as a product of two factors, $8$ and $y$ .

Answer:

$8\cdot y$

Frequently Asked Questions

Everything you need to know about Factorization: Common factor extraction

What is factorization by common factor in algebra?

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?

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?

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?

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?

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?

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?

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?

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.