Expand and Simplify: (3y+4a)² - 9(y-2a)(y+2a)

Algebraic Expansion with Difference of Squares

(3y+4a)29(y2a)(y+2a)=? (3y+4a)^2-9(y-2a)(y+2a)=\text{?}

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:00 Simply
00:08 We'll use the shortened multiplication formulas to open the parentheses
00:39 We'll square each factor in the multiplication
01:07 We'll properly open parentheses and multiply by each factor
01:16 We'll group the terms
01:37 We'll factor each multiplication with factor 4
01:45 We'll find the common factor and take it out of parentheses
01:59 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

(3y+4a)29(y2a)(y+2a)=? (3y+4a)^2-9(y-2a)(y+2a)=\text{?}

2

Step-by-step solution

To solve the problem, we will follow these steps:

  • Step 1: Expand the square of the binomial (3y+4a)(3y + 4a).
  • Step 2: Simplify the product of the binomials 9(y2a)(y+2a)9(y-2a)(y+2a).
  • Step 3: Subtract the second result from the first and simplify.

Let's work through each step:
Step 1: The expression (3y+4a)2(3y + 4a)^2 can be expanded as follows:

(3y+4a)2=(3y)2+2(3y)(4a)+(4a)2=9y2+24ay+16a2(3y + 4a)^2 = (3y)^2 + 2(3y)(4a) + (4a)^2 = 9y^2 + 24ay + 16a^2.

Step 2: Simplify the expression 9(y2a)(y+2a)9(y-2a)(y+2a). This uses the difference of squares formula where (y2a)(y+2a)=y2(2a)2(y-2a)(y+2a) = y^2 - (2a)^2:

(y2a)(y+2a)=y24a2(y-2a)(y+2a) = y^2 - 4a^2.

Then multiply by 9:

9(y24a2)=9y236a29(y^2 - 4a^2) = 9y^2 - 36a^2.

Step 3: Now subtract the second result from the first:

(9y2+24ay+16a2)(9y236a2)=9y2+24ay+16a29y2+36a2(9y^2 + 24ay + 16a^2) - (9y^2 - 36a^2) = 9y^2 + 24ay + 16a^2 - 9y^2 + 36a^2.

Combine and simplify:

24ay+52a224ay + 52a^2.

Factoring out 4a4a from the expression:

4a(6y+13a)4a(6y + 13a).

Therefore, the solution to the problem is: 4a(6y+13a)4a(6y + 13a).

3

Final Answer

4a(6y+13a) 4a(6y+13a)

Key Points to Remember

Essential concepts to master this topic
  • Binomial Squares: Use (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2 formula
  • Difference of Squares: (y2a)(y+2a)=y24a2 (y-2a)(y+2a) = y^2 - 4a^2 simplifies quickly
  • Check: Factor final answer 24ay+52a2=4a(6y+13a) 24ay + 52a^2 = 4a(6y + 13a)

Common Mistakes

Avoid these frequent errors
  • Forgetting to distribute the negative sign when subtracting
    Don't write (9y2+24ay+16a2)(9y236a2) (9y^2 + 24ay + 16a^2) - (9y^2 - 36a^2) as 9y2+24ay+16a29y236a2 9y^2 + 24ay + 16a^2 - 9y^2 - 36a^2 = wrong signs! The negative distributes to ALL terms in parentheses. Always change 9y2 -9y^2 AND (36a2)=+36a2 -(-36a^2) = +36a^2 .

Practice Quiz

Test your knowledge with interactive questions

Solve:

\( (2+x)(2-x)=0 \)

FAQ

Everything you need to know about this question

Why can't I just expand everything without using special formulas?

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You absolutely can! But using difference of squares (ab)(a+b)=a2b2 (a-b)(a+b) = a^2 - b^2 saves time and reduces errors. It's faster than expanding (y2a)(y+2a) (y-2a)(y+2a) term by term.

How do I know when to factor out common terms at the end?

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Look for a common factor in all terms of your final answer. Here, both 24ay 24ay and 52a2 52a^2 are divisible by 4a 4a , so factor it out!

What if I get different terms that don't seem to cancel?

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That's normal! In this problem, the 9y2 9y^2 terms cancel perfectly, but we're left with different terms involving a a . Just combine like terms carefully.

Can I check my answer by substituting numbers?

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Yes! Try y=1,a=1 y = 1, a = 1 . Original expression: (3+4)29(12)(1+2)=259(1)(3)=52 (3+4)^2 - 9(1-2)(1+2) = 25 - 9(-1)(3) = 52 . Your answer: 4(1)(6+13)=76 4(1)(6+13) = 76 ... Wait, that should be 4(1)(6(1)+13(1))=4(19)=76 4(1)(6(1)+13(1)) = 4(19) = 76 . Check your arithmetic!

Why does the answer have no y² terms?

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The y2 y^2 terms cancelled out! We had +9y2 +9y^2 from the first part and 9y2 -9y^2 from the second part, so they subtract to zero. This is perfectly normal in algebra!

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