Expand the Expression: 8x(5+y) Using Distributive Property

Q: Why do I need to multiply 8x by both 5 and y?

The distributive property says you must multiply the outside term by every term inside the parentheses. Think of it like sharing: if you have 8x groups and each group contains (5 + y), you need 8x × 5 plus 8x × y.

Q: What's the difference between 8xy and 8x + y?

$ 8xy $ means 8 times x times y (multiplication), while $ 8x + y $ means 8 times x plus y (addition). These are completely different expressions!

Q: Can I distribute in any order?

Yes! You can distribute $ 8x $ to the 5 first or to the y first. Either way: $ 8x \cdot 5 + 8x \cdot y = 40x + 8xy $.

Q: How can I check if my answer is right?

Try factoring your answer back! If you get $ 40x + 8xy $, factor out $ 8x $: $ 8x(5 + y) $. This should match your original expression!

Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Simply

00:03 We'll open parentheses properly

00:06 We'll make sure the outer term multiplies each of the terms

00:14 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

$8x(5+y)=$

2

Step-by-step solution

To solve this problem, we will simplify the expression $8x(5+y)$ using the distributive property.

Step 1: Recognize the expression as $8x$ multiplied by two terms inside a parenthesis, $5$ and $y$ .
Step 2: Apply the distributive property: $8x(5+y) = 8x \cdot 5 + 8x \cdot y$ .
Step 3: Compute each multiplication:
- $8x \cdot 5 = 40x$
- $8x \cdot y = 8xy$
Step 4: Combine the results from the multiplication steps: $40x + 8xy$ .

Thus, the simplified expression of $8x(5+y)$ is $40x + 8xy$ .

Comparing with provided answer choices, the correct solution corresponds to choice $1: 40x + 8xy$ .

3

Final Answer

$40x+8xy$

Key Points to Remember

Essential concepts to master this topic

Rule: Multiply outside term by each term inside parentheses
Technique: $8x \cdot 5 = 40x$ and $8x \cdot y = 8xy$
Check: Factor out common term: $40x + 8xy = 8x(5 + y)$ ✓

Common Mistakes

Avoid these frequent errors

Only distributing to the first term
Don't multiply $8x$ by just the 5 to get $40x + y$ ! This ignores the y term completely and violates the distributive property. Always multiply the outside term by every single term inside the parentheses.

Practice Quiz

Test your knowledge with interactive questions

Are the expressions the same or not?

$$ 20x $$

$2\times10x$

FAQ

Everything you need to know about this question

Why do I need to multiply 8x by both 5 and y?

+

The distributive property says you must multiply the outside term by every term inside the parentheses. Think of it like sharing: if you have 8x groups and each group contains (5 + y), you need 8x × 5 plus 8x × y.

How do I multiply 8x by y?

+

When multiplying variables, simply write them next to each other: $8x \cdot y = 8xy$ . The coefficient 8 stays in front, and the variables x and y are written together.

What's the difference between 8xy and 8x + y?

+

$8xy$ means 8 times x times y (multiplication), while $8x + y$ means 8 times x plus y (addition). These are completely different expressions!

Can I distribute in any order?

+

Yes! You can distribute $8x$ to the 5 first or to the y first. Either way: $8x \cdot 5 + 8x \cdot y = 40x + 8xy$ .

How can I check if my answer is right?

+

Try factoring your answer back! If you get $40x + 8xy$ , factor out $8x$ : $8x(5 + y)$ . This should match your original expression!

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Expand the Expression: 8x(5+y) Using Distributive Property

Distributive Property with Algebraic Terms

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Step-by-step video solution

Step-by-step written solution

Understand the problem

Step-by-step solution

Final Answer

Key Points to Remember

Common Mistakes

Practice Quiz

FAQ

Why do I need to multiply 8x by both 5 and y?

How do I multiply 8x by y?

What's the difference between 8xy and 8x + y?

Can I distribute in any order?

How can I check if my answer is right?

🌟 Unlock Your Math Potential

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