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Unlocking Algebraic Expressions: How to Multiply Out and Simplify \(4(x + 3) + m(x + 3)\)

A step-by-step guide to expanding and simplifying algebraic terms with clarity.

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Simplifying algebraic expressions is a fundamental skill in mathematics. It involves rewriting an expression in its most compact and clear form. Let's explore how to "multiply out of the bracket and simplify" the expression \(4(x + 3) + m(x + 3)\).

Key Insights

  • The Distributive Property is Fundamental: This property, \(a(b + c) = ab + ac\), is the primary tool for multiplying out terms from brackets.
  • Combining Like Terms is Crucial for Simplification: After expanding, terms with the same variable raised to the same power (or constant terms) are grouped together.
  • Multiple Paths to Simplification: The expression can be simplified by first distributing each multiplier or by first factoring out a common binomial term. Both methods yield equivalent results.

The Cornerstone: Understanding the Distributive Property

The distributive property is a key algebraic concept that allows us to multiply a single term by a sum or difference of terms enclosed in parentheses. In its general form, it's expressed as:

\(a(b + c) = ab + ac\)

This means that the term 'a' outside the parenthesis is distributed (multiplied) to each term ('b' and 'c') inside the parenthesis. This principle is essential for expanding bracketed expressions before simplification.

Mathematical formulas on a chalkboard illustrating algebraic concepts

Algebraic expressions often involve applying fundamental properties like distributivity.


Step-by-Step Simplification: Multiplying Out and Combining

Let's break down the expression \(4(x + 3) + m(x + 3)\) step by step.

Step 1: Apply the Distributive Property to Each Term

We have two terms involving parentheses: \(4(x + 3)\) and \(m(x + 3)\). We'll apply the distributive property to each one.

Expanding the First Term: \(4(x + 3)\)

Multiply 4 by each term inside the first parenthesis:

\(4 \times x = 4x\)

\(4 \times 3 = 12\)

So, \(4(x + 3) = 4x + 12\).

Expanding the Second Term: \(m(x + 3)\)

Similarly, multiply \(m\) by each term inside the second parenthesis:

\(m \times x = mx\)

\(m \times 3 = 3m\)

So, \(m(x + 3) = mx + 3m\).

Step 2: Combine the Expanded Terms

Now, substitute these expanded forms back into the original expression:

\(4(x + 3) + m(x + 3) = (4x + 12) + (mx + 3m)\)

Since we are adding, we can remove the parentheses:

\(4x + 12 + mx + 3m\)

Step 3: Group and Simplify Like Terms

"Like terms" are terms that contain the same variable raised to the same power. Constants are also like terms with each other.

  • Terms with \(x\): We have \(4x\) and \(mx\). These can be combined by factoring out \(x\):

    \(4x + mx = (4 + m)x\)

  • Constant and \(m\) terms: We have \(12\) (a constant) and \(3m\) (a term with \(m\)). These are not like terms with each other (unless \(m\) is a specific number that makes \(3m\) a constant that can be combined with 12, but \(m\) is treated as a variable here). So, they remain as \(12 + 3m\).

Putting it all together, the simplified expression is:

\((4 + m)x + 12 + 3m\)

This can also be written as \((m + 4)x + (3m + 12)\) for conventional ordering.


Alternative Approach: Factoring Out the Common Binomial

Another way to approach this problem is to notice that both parts of the expression, \(4(x + 3)\) and \(m(x + 3)\), share a common factor of \((x + 3)\). We can factor this common binomial out:

\(4(x + 3) + m(x + 3)\)

Treat \((x + 3)\) as a single entity. If we let \(Y = (x + 3)\), the expression becomes \(4Y + mY\).

Factoring out \(Y\), we get:

\(Y(4 + m)\)

Now, substitute back \(Y = (x + 3)\):

\((x + 3)(4 + m)\)

Or, more conventionally written: \((4 + m)(x + 3)\).

If you expand this factored form using the distributive property (or FOIL method for binomials), you'll arrive at the same result as before:

\((4 + m)(x + 3) = 4(x + 3) + m(x + 3) = 4x + 4(3) + mx + m(3) = 4x + 12 + mx + 3m\)

This confirms both approaches lead to equivalent simplified forms.


Visualizing the Simplification Path

This mindmap illustrates the two main pathways to simplify the expression \(4(x + 3) + m(x + 3)\). It shows the steps involved in both distributing first and factoring first, leading to the equivalent simplified forms.

mindmap root["Simplify: \(4(x + 3) + m(x + 3)\)"] Method1["Method 1: Distribute then Combine"] Step1_1["Apply Distributive Property"] Term1["Distribute 4:
\(4(x + 3) \Rightarrow 4x + 12\)"] Term2["Distribute m:
\(m(x + 3) \Rightarrow mx + 3m\)"] Step1_2["Combine Expanded Terms"] Expr1["Sum: \(4x + 12 + mx + 3m\)"] Step1_3["Group Like Terms"] XTerms["Terms with x:
\(4x + mx \Rightarrow (4 + m)x\)"] Const_M_Terms["Constant and m terms:
\(12 + 3m\)"] Result1["Simplified Form 1:
\((4 + m)x + 12 + 3m\)"] Method2["Method 2: Factor out Common Term \((x+3)\)"] Step2_1["Identify Common Factor"] CommonFactor["\((x+3)\) is common to both \(4(x+3)\) and \(m(x+3)\)"] Step2_2["Factor out \((x+3)\)"] Expr2["\( (x+3)(4+m) \)"] Result2["Simplified Form 2:
\((4 + m)(x + 3)\)"] Equivalence["Equivalence Check"] ExpandR2["Expanding \((4+m)(x+3)\) gives:
\(4x + 12 + mx + 3m\),
matching Form 1."]

Comparing Simplification Strategies

The choice between distributing first or factoring first can depend on the complexity of the expression and personal preference. The radar chart below offers a conceptual comparison of these two strategies based on several aspects of the simplification process for an expression like the one given.

This chart suggests that "Distributing then Combining" is very clear in its application of the distributive property, while "Factoring First" can be more efficient and offer greater step reduction if the common factor is easily identified. Both require a good conceptual understanding and care to avoid errors.


Summary of Expansion and Simplification

The following table summarizes the key transformations in the process of simplifying \(4(x + 3) + m(x + 3)\) by first multiplying out the brackets:

Original Component After Distributive Property
\(4(x + 3)\) \(4x + 12\)
\(m(x + 3)\) \(mx + 3m\)
Full Expression (Expanded) \(4x + 12 + mx + 3m\)
Full Expression (Simplified by Grouping Like Terms) \((4 + m)x + 12 + 3m\)
Full Expression (Simplified by Factoring Common Binomial) \((4 + m)(x + 3)\)

Visual Learning: The Distributive Property in Action

Understanding the distributive property is key to expanding expressions. This video provides a clear explanation and examples of how the distributive property works in algebra, which is directly applicable to the expression we've simplified.

The video "Algebra Basics: The Distributive Property - Math Antics" breaks down the concept \(a(b + c) = ab + ac\), showing how a term outside parentheses multiplies each term inside. This is precisely what we did for \(4(x+3)\) to get \(4x+12\), and for \(m(x+3)\) to get \(mx+3m\).


Frequently Asked Questions (FAQ)

What is the distributive property?

The distributive property states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products. Symbolically: \(a(b + c) = ab + ac\). It's a fundamental rule for expanding expressions in algebra.

What are 'like terms' in algebra?

Like terms are terms in an algebraic expression that have the same variables raised to the same powers. Only the numerical coefficients can differ. For example, \(4x\) and \(mx\) are like terms with respect to \(x\) (they can be combined as \((4+m)x\)). Constants like 12 and 7 are also like terms. Simplifying expressions often involves combining like terms.

Why are there two main forms of the simplified answer?

The two forms, \((4 + m)x + 12 + 3m\) and \((4 + m)(x + 3)\), are algebraically equivalent.

  • \((4 + m)x + 12 + 3m\) is the result of fully distributing all terms and then combining like terms where \(x\) is the primary variable of interest. This form is often called the "expanded and simplified" form.
  • \((4 + m)(x + 3)\) is a "factored" form, achieved by recognizing \((x+3)\) as a common factor. This form can be useful for other algebraic manipulations, like solving equations if the expression were set to zero.
The specific request "multiply out of the bracket and simplify" most directly leads to \((4 + m)x + 12 + 3m\), but understanding the factored form is also valuable.

What if 'm' was a specific number, for example, m = 2?

If \(m\) had a specific numerical value, say \(m=2\), we would substitute this value into the simplified expression. Using \((4 + m)x + 12 + 3m\):

\((4 + 2)x + 12 + 3(2)\)

\(= 6x + 12 + 6\)

\(= 6x + 18\)

Alternatively, using the original expression with \(m=2\):

\(4(x + 3) + 2(x + 3)\)

\(= 4x + 12 + 2x + 6\)

\(= (4x + 2x) + (12 + 6)\)

\(= 6x + 18\)

Both paths yield the same numerical result, demonstrating the consistency of the algebraic simplification.


Recommended Further Exploration


References

mathsolver.microsoft.com
Expand - Microsoft Math Solver

Last updated May 15, 2025
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