Distributive Property: Simplifying (a+c+d)(a+e) Step-by-Step

Q: How do I keep track of all the terms when expanding?

Use the FOIL-extended method! For $ (a+c+d)(a+e) $, multiply systematically: a with both terms, then c with both terms, then d with both terms.

Q: Why do I get 6 terms instead of 4?

Because you have 3 terms times 2 terms = 6 products! Each of the 3 terms $ (a, c, d) $ must multiply each of the 2 terms $ (a, e) $.

Q: Can I combine like terms in the final answer?

In this problem, all terms are unlike terms (different variable combinations), so they cannot be combined. The final answer stays as 6 separate terms.

Q: What if I miss a term during expansion?

Check your term count! $ (a+c+d)(a+e) $ should give exactly 6 terms. If you have fewer, you missed a multiplication step.

Q: Is there a pattern to remember the distributive property?

Yes! Think "everyone meets everyone" - each term in the first parentheses must shake hands with each term in the second parentheses exactly once.

Binomial Expansion with Three-Term Factors

It is possible to use the distributive property to simplify the expression

$(a+c+d)(a+e)$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solution

00:03 Open parentheses properly, multiply each factor by each factor

00:28 Calculate the products

00:56 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

It is possible to use the distributive property to simplify the expression

$(a+c+d)(a+e)$

Step-by-step solution

To solve this problem using the distributive property, let's expand the given expression $(a+c+d)(a+e)$ step by step.

Step 1: Expand $(a+c+d)(a+e)$ using the distributive property:

Distribute $a$ over $(a+e)$ :
$a(a+e) = a^2 + ae$
Distribute $c$ over $(a+e)$ :
$c(a+e) = ca + ce$
Distribute $d$ over $(a+e)$ :
$d(a+e) = da + de$

Step 2: Combine all distributed terms:

$a^2 + ae + ca + ce + da + de$

Thus, the expression simplifies to $\bm{a^2 + ae + ca + ce + da + de}$ .

Therefore, the solution to the problem is Yes, $(a^2 + ae + ca + ce + da + de)$ , which matches choice 3.

Final Answer

Yes, $a^2+ae+ca+ce+da+de$

Key Points to Remember

Essential concepts to master this topic

Distributive Property: Each term multiplies every term in second factor
Technique: $(a+c+d)(a+e)$ gives 6 terms total: a², ae, ca, ce, da, de
Check: Count terms: 3 terms × 2 terms = 6 total terms ✓

Common Mistakes

Avoid these frequent errors

Forgetting to multiply every term with every other term
Don't just multiply $a \times a = a^2$ and skip combinations = missing terms! This gives incomplete expansions like $a^2+ec+de+da$ instead of all 6 terms. Always ensure each term in the first factor multiplies each term in the second factor.

Practice Quiz

Test your knowledge with interactive questions

It is possible to use the distributive property to simplify the expression below?

What is its simplified form?

$$ (ab)(c d) $$

$$

FAQ

Everything you need to know about this question

How do I keep track of all the terms when expanding?

Use the FOIL-extended method! For $(a+c+d)(a+e)$ , multiply systematically: a with both terms, then c with both terms, then d with both terms.

Why do I get 6 terms instead of 4?

Because you have 3 terms times 2 terms = 6 products! Each of the 3 terms $(a, c, d)$ must multiply each of the 2 terms $(a, e)$ .

Can I combine like terms in the final answer?

In this problem, all terms are unlike terms (different variable combinations), so they cannot be combined. The final answer stays as 6 separate terms.

What if I miss a term during expansion?

Check your term count! $(a+c+d)(a+e)$ should give exactly 6 terms. If you have fewer, you missed a multiplication step.

Is there a pattern to remember the distributive property?

Yes! Think "everyone meets everyone" - each term in the first parentheses must shake hands with each term in the second parentheses exactly once.

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