Expand and Simplify: Solving (x+y)²-(x-y)²+(x-y)(x+y)

Algebraic Expansion with Difference of Squares

(x+y)2−(x−y)2+(x−y)(x+y)=? (x+y)^2-(x-y)^2+(x-y)(x+y)=\text{?}

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations
00:11 First, let's simplify.
00:14 We will use shorter multiplication formulas to expand the brackets.
01:00 Next, we'll reduce what we can. Looking good so far!
01:07 Remember, negative times positive gives a negative result.
01:12 And negative times negative is always positive.
01:23 Now, let's gather the like terms together.
01:36 And that's the solution to our question! Great job!

Step-by-step written solution

Follow each step carefully to understand the complete solution
1

Understand the problem

(x+y)2−(x−y)2+(x−y)(x+y)=? (x+y)^2-(x-y)^2+(x-y)(x+y)=\text{?}

2

Step-by-step solution

To solve this problem, we'll follow these steps:

  • Step 1: Expand (x+y)2(x+y)^2
  • Step 2: Expand (x−y)2(x-y)^2
  • Step 3: Rearrange and simplify the entire expression

Now, let's work through each step:
Step 1: We expand (x+y)2(x+y)^2 using the formula for the square of a sum:
(x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2.

Step 2: We expand (x−y)2(x-y)^2 using the formula for the square of a difference:
(x−y)2=x2−2xy+y2(x-y)^2 = x^2 - 2xy + y^2.

Step 3: Substitute these expansions back into the original expression: (x+y)2−(x−y)2+(x−y)(x+y)(x+y)^2-(x-y)^2+(x-y)(x+y) becomes: (x2+2xy+y2)−(x2−2xy+y2)+(x−y)(x+y).(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) + (x-y)(x+y).

First, simplify (x2+2xy+y2)−(x2−2xy+y2)(x^2 + 2xy + y^2) - (x^2 - 2xy + y^2):
– x2+2xy+y2−x2+2xy−y2=4xy.x^2 + 2xy + y^2 - x^2 + 2xy - y^2 = 4xy.

Next, consider (x−y)(x+y)(x-y)(x+y):
By using the identity for difference of squares: (x−y)(x+y)=x2−y2(x-y)(x+y) = x^2 - y^2.

Thus, combining our results gives:
4xy+x2−y2=x2+4xy−y2.4xy + x^2 - y^2 = x^2 + 4xy - y^2.

Therefore, the solution to the problem is x2+4xy−y2x^2 + 4xy - y^2.

3

Final Answer

x2+4xy−y2 x^2+4xy-y^2

Key Points to Remember

Essential concepts to master this topic
  • Pattern: Use (a+b)2=a2+2ab+b2 (a+b)^2 = a^2 + 2ab + b^2 and (a−b)2=a2−2ab+b2 (a-b)^2 = a^2 - 2ab + b^2
  • Technique: Recognize (x−y)(x+y)=x2−y2 (x-y)(x+y) = x^2 - y^2 difference of squares identity
  • Check: Combine like terms carefully: 4xy+x2−y2=x2+4xy−y2 4xy + x^2 - y^2 = x^2 + 4xy - y^2 ✓

Common Mistakes

Avoid these frequent errors
  • Forgetting to distribute the negative sign when subtracting squared terms
    Don't expand (x+y)2−(x−y)2 (x+y)^2 - (x-y)^2 as x2+2xy+y2−x2−2xy+y2=2y2 x^2 + 2xy + y^2 - x^2 - 2xy + y^2 = 2y^2 ! The negative distributes to ALL terms in the second expansion. Always change signs: (x2+2xy+y2)−(x2−2xy+y2)=x2+2xy+y2−x2+2xy−y2=4xy (x^2 + 2xy + y^2) - (x^2 - 2xy + y^2) = x^2 + 2xy + y^2 - x^2 + 2xy - y^2 = 4xy .

Practice Quiz

Test your knowledge with interactive questions

Choose the expression that has the same value as the following:

\( (x+y)^2 \)

FAQ

Everything you need to know about this question

Why do I need to expand each squared term separately?

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Expanding each term like (x+y)2 (x+y)^2 and (x−y)2 (x-y)^2 separately helps you see all the terms clearly. This prevents sign errors when you subtract the second expansion from the first.

How do I remember the difference of squares formula?

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Think of it as "first squared minus last squared": (x−y)(x+y)=x2−y2 (x-y)(x+y) = x^2 - y^2 . The middle terms +xy +xy and −xy -xy always cancel out!

What's the most important step in this problem?

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The most critical step is correctly handling the negative sign when subtracting (x−y)2 (x-y)^2 . Remember: minus a negative becomes positive, so −(−2xy)=+2xy -(-2xy) = +2xy .

Can I solve this problem in a different order?

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Yes! You could recognize that (x+y)2−(x−y)2 (x+y)^2 - (x-y)^2 is a difference of squares pattern and use the identity a2−b2=(a+b)(a−b) a^2 - b^2 = (a+b)(a-b) first. Both methods work!

How do I check my final answer?

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Substitute simple values like x=1,y=1 x=1, y=1 into both the original expression and your answer. If (1+1)2−(1−1)2+(1−1)(1+1)=4 (1+1)^2-(1-1)^2+(1-1)(1+1) = 4 and 12+4(1)(1)−12=4 1^2+4(1)(1)-1^2 = 4 , you're correct!

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