Simplify a(b+c)(-b-c) Using the Distributive Property

Q: Why can't I just distribute a to both (b+c) and (-b-c) separately?

You could, but it creates more work! You'd get a(b+c) and a(-b-c), then still need to multiply these together. It's much easier to first simplify (b+c)(-b-c), then distribute a.

Q: How do I keep track of all the negative signs?

Work systematically! When expanding (b+c)(-b-c), each term in the first parentheses multiplies each term in the second. Since (-b-c) has all negative terms, every product will be negative.

Q: Is there a pattern I can recognize here?

Yes! The expression (b+c)(-b-c) is actually -(b+c)² in disguise. You can rewrite (-b-c) as -(b+c), giving you (b+c)·(-(b+c)) = -(b+c)².

Q: What if the signs were different, like (b+c)(b+c)?

That would be (b+c)², which expands to b² + 2bc + c². The key difference is that our problem has opposite signs in the parentheses, creating all negative terms.

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

00:26 Calculate the multiplications

00:56 Open parentheses properly, multiply by each factor

01:13 Calculate the multiplications

01:22 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

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

$a(b+c)(-b-c)$

2

Step-by-step solution

To simplify the expression $a(b+c)(-b-c)$ using the distributive property, follow these steps:

Step 1: Apply Distributive Property to $(b+c)(-b-c)$ :
The expression $(b+c)(-b-c)$ can be expanded using the distributive property:
$(b+c)(-b-c) = b(-b-c) + c(-b-c)$ .
Step 2: Simplify Each Part:
Let's simplify each term individually:
- $b(-b) = -b^2$
- $b(-c) = -bc$
- $c(-b) = -bc$
- $c(-c) = -c^2$
So, combining these results:
$(b+c)(-b-c) = -b^2 - bc - bc - c^2 = -b^2 - 2bc - c^2$ .
Step 3: Distribute $a$ Over the Result:
Now, apply a further distribution of $a$ to get:
$a(-b^2 - 2bc - c^2) = a(-b^2) + a(-2bc) + a(-c^2)$ .
Step 4: Simplify:
Perform the distribution:
- $a(-b^2) = -ab^2$
- $a(-2bc) = -2abc$
- $a(-c^2) = -ac^2$
Thus, the expression simplifies to:
$-ab^2 - 2abc - ac^2$ .

Therefore, the simplified expression using the distributive property is $-ab^2 - 2abc - ac^2$ .

Given the multiple-choice options, the correct choice that corresponds to our derived expression is:

Choice 4: Yes, $-ab^2 - 2abc - ac^2$

3

Final Answer

Yes, $-ab^2-2abc-ac^2$

Key Points to Remember

Essential concepts to master this topic

Rule: Apply distributive property step by step with multiple factors
Technique: First expand (b+c)(-b-c) = -b² - 2bc - c²
Check: Multiply a through each term: a(-b² - 2bc - c²) = -ab² - 2abc - ac² ✓

Common Mistakes

Avoid these frequent errors

Trying to distribute all three factors simultaneously
Don't try to distribute a(b+c)(-b-c) all at once = confusion and wrong signs! This creates too many terms to track and leads to sign errors. Always work step by step: first expand (b+c)(-b-c), then distribute the remaining factor a.

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 distribute a to both (b+c) and (-b-c) separately?

+

You could, but it creates more work! You'd get a(b+c) and a(-b-c), then still need to multiply these together. It's much easier to first simplify (b+c)(-b-c), then distribute a.

How do I keep track of all the negative signs?

+

Work systematically! When expanding (b+c)(-b-c), each term in the first parentheses multiplies each term in the second. Since (-b-c) has all negative terms, every product will be negative.

Is there a pattern I can recognize here?

+

Yes! The expression (b+c)(-b-c) is actually -(b+c)² in disguise. You can rewrite (-b-c) as -(b+c), giving you (b+c)·(-(b+c)) = -(b+c)².

What if the signs were different, like (b+c)(b+c)?

+

That would be (b+c)², which expands to b² + 2bc + c². The key difference is that our problem has opposite signs in the parentheses, creating all negative terms.

How can I check if my final answer is correct?

+

Substitute simple values like a=1, b=1, c=1. Your original expression gives 1(1+1)(-1-1) = 1(2)(-2) = -4. Your simplified answer gives -1(1)² - 2(1)(1)(1) - 1(1)² = -1 - 2 - 1 = -4 ✓

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