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

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

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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