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

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

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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