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 \(4(x + 3) + m(x + 3)\).
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.
Algebraic expressions often involve applying fundamental properties like distributivity.
Let's break down the expression \(4(x + 3) + m(x + 3)\) step by step.
We have two terms involving parentheses: \(4(x + 3)\) and \(m(x + 3)\). We'll apply the distributive property to each one.
Multiply 4 by each term inside the first parenthesis:
\(4 \times x = 4x\)
\(4 \times 3 = 12\)
So, \(4(x + 3) = 4x + 12\).
Similarly, multiply \(m\) by each term inside the second parenthesis:
\(m \times x = mx\)
\(m \times 3 = 3m\)
So, \(m(x + 3) = mx + 3m\).
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:
\(4x + 12 + mx + 3m\)
"Like terms" are terms that contain the same variable raised to the same power. Constants are also like terms with each other.
\(4x + mx = (4 + m)x\)
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.
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.
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.
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.
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) | \((4 + m)(x + 3)\) |
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 \(mx+3m\).