Evaluate the Expression: What is (a+3b)²-(3b-a)²?

Q: Why don't the squared terms just cancel out to give zero?

While $ a^2 $ and $ 9b^2 $ do cancel, the middle terms don't! You get $ +6ab $ from the first expansion and $ -6ab $ from the second, so they add to give $ 12ab $.

Q: Can I use the difference of squares formula directly?

Not directly! The difference of squares $ a^2 - b^2 = (a+b)(a-b) $ works when you have perfect squares. Here, neither $ (a+3b)^2 $ nor $ (3b-a)^2 $ is a simple variable squared.

Q: What if I expand (3b-a)² as (3b)² - a² instead?

That's wrong! $ (3b-a)^2 \neq (3b)^2 - a^2 $. You must use the full formula: $ (x-y)^2 = x^2 - 2xy + y^2 $, giving $ 9b^2 - 6ab + a^2 $.

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

Write it step by step: $ (a^2 + 6ab + 9b^2) - (9b^2 - 6ab + a^2) $. Distribute the negative sign to get $ a^2 + 6ab + 9b^2 - 9b^2 + 6ab - a^2 $, then combine like terms.

Q: Is there a shortcut for this type of problem?

Yes! Notice that $ (a+3b)^2 - (3b-a)^2 $ has the form $ (A)^2 - (B)^2 $ where $ A = a+3b $ and $ B = 3b-a $. Use $ (A)^2 - (B)^2 = (A+B)(A-B) $ to get $ (2a)(6b) = 12ab $!

Difference of Squares with Algebraic Expansion

$(a+3b)^2-(3b-a)^2=\text{?}$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Simply

00:03 We'll use the abbreviated multiplication formulas to open all parentheses

00:23 We'll solve the multiplications and squares

00:47 Negative times positive is always negative

00:52 Negative times negative is always positive

01:06 Let's group terms

01:14 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

$(a+3b)^2-(3b-a)^2=\text{?}$

Step-by-step solution

To solve this problem, we will follow these steps:

Step 1: Expand $(a+3b)^2$ .
Step 2: Expand $(3b-a)^2$ .
Step 3: Subtract the result of step 2 from step 1.

Now, let's work through the calculations:

Step 1: Expand $(a+3b)^2$
Using the formula $(x + y)^2 = x^2 + 2xy + y^2$ , we let $x = a$ and $y = 3b$ to get:
$(a+3b)^2 = a^2 + 2 \cdot a \cdot 3b + (3b)^2$
$= a^2 + 6ab + 9b^2$ .

Step 2: Expand $(3b-a)^2$
Again using the squaring formula, letting $x = 3b$ and $y = -a$ , we have:
$(3b-a)^2 = (3b)^2 - 2 \cdot 3b \cdot a + a^2$
$= 9b^2 - 6ab + a^2$ .

Step 3: Perform the subtraction
We subtract the expansion of $(3b-a)^2$ from $(a+3b)^2$ :
$(a^2 + 6ab + 9b^2) - (9b^2 - 6ab + a^2)$
$= a^2 + 6ab + 9b^2 - 9b^2 + 6ab - a^2$
= $12ab$ .

The solution to the problem is $12ab$ , which corresponds to choice 2.

Final Answer

$12ab$

Key Points to Remember

Essential concepts to master this topic

Formula: Expand (x+y)² = x² + 2xy + y² carefully
Technique: (a+3b)² = a² + 6ab + 9b²; (3b-a)² = 9b² - 6ab + a²
Check: Verify cancellation: a² cancels, 9b² cancels, leaving 12ab ✓

Common Mistakes

Avoid these frequent errors

Incorrectly expanding squared terms
Don't forget the middle term 2xy in (x+y)² = x² + y² instead of x² + 2xy + y²! This omits crucial terms like 6ab. Always include the middle term: (a+3b)² = a² + 2(a)(3b) + (3b)² = a² + 6ab + 9b².

Practice Quiz

Test your knowledge with interactive questions

Declares the given expression as a sum

$$ (7b-3x)^2 $$

FAQ

Everything you need to know about this question

Why don't the squared terms just cancel out to give zero?

While $a^2$ and $9b^2$ do cancel, the middle terms don't! You get $+6ab$ from the first expansion and $-6ab$ from the second, so they add to give $12ab$ .

Can I use the difference of squares formula directly?

Not directly! The difference of squares $a^2 - b^2 = (a+b)(a-b)$ works when you have perfect squares. Here, neither $(a+3b)^2$ nor $(3b-a)^2$ is a simple variable squared.

What if I expand (3b-a)² as (3b)² - a² instead?

That's wrong! $(3b-a)^2 \neq (3b)^2 - a^2$ . You must use the full formula: $(x-y)^2 = x^2 - 2xy + y^2$ , giving $9b^2 - 6ab + a^2$ .

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

Write it step by step: $(a^2 + 6ab + 9b^2) - (9b^2 - 6ab + a^2)$ . Distribute the negative sign to get $a^2 + 6ab + 9b^2 - 9b^2 + 6ab - a^2$ , then combine like terms.

Is there a shortcut for this type of problem?

Yes! Notice that $(a+3b)^2 - (3b-a)^2$ has the form $(A)^2 - (B)^2$ where $A = a+3b$ and $B = 3b-a$ . Use $(A)^2 - (B)^2 = (A+B)(A-B)$ to get $(2a)(6b) = 12ab$ !

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Evaluate the Expression: What is (a+3b)²-(3b-a)²?

Difference of Squares with Algebraic Expansion

❤️ Continue Your Math Journey!

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 don't the squared terms just cancel out to give zero?

Can I use the difference of squares formula directly?

What if I expand (3b-a)² as (3b)² - a² instead?

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

Is there a shortcut for this type of problem?

🌟 Unlock Your Math Potential

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