Match together expressions of equal value$ (a+b)(c+d) $$ (a+c)(b+d) $$ (a+d)(c+b) $a.$ ac+ad+bc+bd $b.$ ac+ab+dc+db $c.$ ab+ad+cb+cd $

All expressions have the same value

Match Equivalent Expressions: (a+b)(c+d) and Its Expanded Forms

Q: How do I expand (a+b)(c+d) correctly?

Use the distributive property: multiply each term in the first binomial by each term in the second. So (a+b)(c+d) = ac + ad + bc + bd. Think of it as FOIL - First, Outer, Inner, Last!

Q: Why aren't all three expressions equivalent?

Because they have different variable arrangements! While (a+b)(c+d) gives ac+ad+bc+bd, the expression (a+c)(b+d) gives ab+ad+cb+cd. The terms are different, so the expressions aren't equal.

Q: Can I use the commutative property to rearrange terms?

Yes! Since multiplication is commutative, $ ab = ba $ and $ cd = dc $. This helps you match equivalent expressions that might have terms in different orders.

Q: What if I get confused with all the variables?

Try substituting simple numbers like a=1, b=2, c=3, d=4. Calculate each expression numerically to see which ones give the same result. This confirms your algebraic work!

Q: How do I know which expanded form matches which binomial?

Expand each binomial systematically and compare term by term. Look for the same variables multiplied together, remembering that order doesn't matter (ac = ca).

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Open parentheses

00:03 Let's open parentheses properly, multiply each factor by each factor

00:13 We'll use the same method for all parentheses

00:29 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

1

Understand the problem

Match together expressions of equal value

$(a+b)(c+d)$
$(a+c)(b+d)$
$(a+d)(c+b)$
a. $ac+ad+bc+bd$
b. $ac+ab+dc+db$
c. $ab+ad+cb+cd$

2

Step-by-step solution

Let's simplify the given expressions, open the parentheses using the extended distribution law:

$(\textcolor{red}{x}+\textcolor{blue}{y})(m+r)=\textcolor{red}{x}m+\textcolor{red}{x}r+\textcolor{blue}{y}m+\textcolor{blue}{y}r$

In the formula template for the above distribution law, we take by default that the operation between the terms inside of the parentheses is addition. Note that the sign preceding the term is an inseparable part of it. Furthermore we will apply the laws of sign multiplication to our expression. We will then open the parentheses using the above formula, where there is an addition operation between all terms.

Proceed to simplify each of the expressions in the given problem, whilst making sure to open the parentheses using the mentioned distribution law, the commutative law of addition and multiplication and combining like terms (if there are like terms in the expression obtained after opening the parentheses):

$(a+b)(c+d) \\ \boxed{ac+ad+bc+bd}\\$
$(a+c)(b+d) \\ \boxed{ab+ad+cb+cd}$
$(a+d)(c+b) \\ \boxed{ac+ab+dc+db}$
In all expressions where we performed a multiplication operation between the expressions in the above parentheses, the result of the multiplication (obtained after applying the mentioned distribution law) yielded an expression where the terms cannot be combined. This is due to the fact that all terms in the resulting expression are different from each other ( All variables in like terms need to be identical and have the same exponent)
After applying the commutative law of addition and multiplication we observe that:
The simplified expression in 1 matches the expression in option A,
The simplified expression in 2 matches the expression in option C,
The simplified expression in 3 matches the expression in option B,

Therefore the correct answer (among the suggested options) is answer C.

3

Final Answer

1-a, 2-b, 3-b

Key Points to Remember

Essential concepts to master this topic

Distribution Rule: Apply (x+y)(m+n) = xm + xn + ym + yn systematically
Technique: For (a+b)(c+d), get ac + ad + bc + bd terms
Check: Verify by expanding all expressions and comparing resulting terms ✓

Common Mistakes

Avoid these frequent errors

Assuming all binomial products are equivalent
Don't think (a+b)(c+d) = (a+c)(b+d) without expanding = wrong matches! Different variable arrangements create different expanded forms. Always expand each expression completely using FOIL or distribution to compare actual terms.

Practice Quiz

Test your knowledge with interactive questions

$$ (x+y)(x-y)= $$

FAQ

Everything you need to know about this question

How do I expand (a+b)(c+d) correctly?

+

Use the distributive property: multiply each term in the first binomial by each term in the second. So (a+b)(c+d) = ac + ad + bc + bd. Think of it as FOIL - First, Outer, Inner, Last!

Why aren't all three expressions equivalent?

+

Because they have different variable arrangements! While (a+b)(c+d) gives ac+ad+bc+bd, the expression (a+c)(b+d) gives ab+ad+cb+cd. The terms are different, so the expressions aren't equal.

Can I use the commutative property to rearrange terms?

+

Yes! Since multiplication is commutative, $ab = ba$ and $cd = dc$ . This helps you match equivalent expressions that might have terms in different orders.

What if I get confused with all the variables?

+

Try substituting simple numbers like a=1, b=2, c=3, d=4. Calculate each expression numerically to see which ones give the same result. This confirms your algebraic work!

How do I know which expanded form matches which binomial?

+

Expand each binomial systematically and compare term by term. Look for the same variables multiplied together, remembering that order doesn't matter (ac = ca).

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Binomial Expansion with Equivalent Algebraic Forms

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