Simplify (a+b)·4·(b+2): Applying the Distributive Property

Q: Why can't I just multiply the first terms together?

Because distributive property requires multiplying every term in one binomial by every term in the other! Skipping terms means you're missing parts of the full expansion.

Q: Do I have to multiply in a specific order?

No! You can multiply $ 4 $ with either binomial first. The key is to systematically distribute each result to all terms in the remaining factors.

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

Use a step-by-step approach: First handle one multiplication completely, then move to the next. Write out each distribution clearly before combining like terms.

Q: Can I combine like terms in this problem?

In this case, no like terms exist to combine! Each term has different variables or powers: $ 4ab $, $ 8a $, $ 4b^2 $, and $ 8b $ are all different.

Q: Why does the correct answer say 'Yes' at the beginning?

The question asks "Is it possible to use distributive property?" The answer is "Yes" because we can indeed use it, and here's the result when we do!

Distributive Property with Multiple Factors

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

$(a+b)\cdot4\cdot(b+2)$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Solution

00:03 We'll use the commutative law and arrange the exercise in a convenient way to solve

00:14 Open the parentheses properly, multiply each factor by each factor

00:31 Open the parentheses properly, multiply by each factor

00:54 Calculate the products

01:06 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

$(a+b)\cdot4\cdot(b+2)$

Step-by-step solution

Let's simplify the expression $(a+b)\cdot4\cdot(b+2)$ using the distributive property.

Step 1: Distribute the $4$ across $(b + 2)$ .
$4(b + 2) = 4b + 4 \times 2 = 4b + 8$

Step 2: Now distribute this result $(4b + 8)$ across $(a + b)$ .
$(a+b)(4b+8) = a(4b+8) + b(4b+8)$

Step 3: Apply the distributive property again for both terms.
- For $a(4b+8)$ , we get:
$a \times 4b + a \times 8 = 4ab + 8a$
- For $b(4b+8)$ , we get:
$b \times 4b + b \times 8 = 4b^2 + 8b$

Step 4: Combine all parts.
The expanded expression is:
$4ab + 8a + 4b^2 + 8b$

Therefore, the simplified expression is $\boxed{4ab + 8a + 4b^2 + 8b}$ , and the correct choice is:

Yes, $4ab + 8a + 4b^2 + 8b$ .

Final Answer

Yes, $4ab+8a+4b^2+8b$

Key Points to Remember

Essential concepts to master this topic

Rule: Apply distributive property systematically to all factors in sequence
Technique: First distribute 4 across (b+2), then distribute result across (a+b)
Check: Count terms: $4ab+8a+4b^2+8b$ has 4 terms as expected ✓

Common Mistakes

Avoid these frequent errors

Distributing only to first terms of each binomial
Don't distribute 4 to just 'b' and skip the '+2', getting $4ab+4b$ = missing terms! This ignores parts of the binomials and gives incomplete results. Always distribute to every single term in each binomial factor.

Practice Quiz

Test your knowledge with interactive questions

$(3+20)\times(12+4)=$

FAQ

Everything you need to know about this question

Why can't I just multiply the first terms together?

Because distributive property requires multiplying every term in one binomial by every term in the other! Skipping terms means you're missing parts of the full expansion.

Do I have to multiply in a specific order?

No! You can multiply $4$ with either binomial first. The key is to systematically distribute each result to all terms in the remaining factors.

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

Use a step-by-step approach: First handle one multiplication completely, then move to the next. Write out each distribution clearly before combining like terms.

What if I get a different number of terms than the answer choices?

Count your terms carefully! With two binomials, you should get 4 terms total before combining any like terms. If you have fewer, you missed some distributions.

Can I combine like terms in this problem?

In this case, no like terms exist to combine! Each term has different variables or powers: $4ab$ , $8a$ , $4b^2$ , and $8b$ are all different.

Why does the correct answer say 'Yes' at the beginning?

The question asks "Is it possible to use distributive property?" The answer is "Yes" because we can indeed use it, and here's the result when we do!

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