HSA.SSE.A.2: Using Structure to Rewrite Expressions
In plain English: HSA.SSE.A.2 is the Common Core algebra standard that asks students to use the structure of an expression to identify ways to rewrite it. Students recognize patterns such as a common factor, a difference of squares or a perfect square trinomial, for example seeing x⁴ - y⁴ as (x²)² - (y²)². It is usually taught in Algebra I and extended to more patterns in Algebra II.
Use the structure of an expression to identify ways to rewrite it. For example, see x4 - y4 as (x2)2 - (y2)2, thus recognizing it as a difference of squares that can be factored as (x2 - y2)(x2 + y2).
Common Core State Standards for Mathematics · Domain: Seeing Structure in Expressions (SSE) · Cluster: Interpret the structure of expressions Also written as HSA-SSE.A.2 or A-SSE.2 · Official standard
In this lesson, students learn to look at an expression before they start manipulating it and ask what shape it has. They recognize familiar patterns, such as a difference of two squares, a perfect square trinomial, a common factor or a quadratic hiding inside a higher power, and use those patterns to rewrite the expression in an equivalent form. A standard example is seeing x4 - y4 as (x2)2 - (y2)2, a difference of squares.
The key habit is treating a piece of an expression, like x2, 3x or (x + 5), as a single object that fits a known pattern. Students also use structure for mental arithmetic, such as 482 - 422 = 6 × 90, which shows that these patterns are about numbers, not only about letters.
Learning Objectives
By the end of this lesson, students will be able to:
Recognize the difference of squares, perfect square trinomial and common factor patterns, including when the squared quantities are themselves expressions such as 3x, x2 or (x + 5)
Rewrite an expression in quadratic form by substituting a single variable for a repeated chunk, then factor and substitute back
Factor an expression completely by applying patterns more than once, such as x4 - 81 = (x - 3)(x + 3)(x2 + 9)
Use the structure of numerical expressions to compute efficiently, such as 1012 - 992
Verify that a rewritten expression is equivalent by multiplying back or by substituting a value
Prior Knowledge Required
Students should already be comfortable with:
Applying properties of operations to factor and expand linear expressions 7.EE.A.1
Using the properties of integer exponents, such as (x2)2 = x48.EE.A.1
Multiplying binomials and factoring simple quadratics with leading coefficient 1
Identifying terms, factors and coefficients of an expression HSA.SSE.A.1
"Compute 21 × 19, 32 × 28 and 45 × 55 in your head. Then look at your three answers. What shortcut did you use, or could you have used?"
Collect strategies. Guide students to see that 21 × 19 = (20 + 1)(20 - 1) = 400 - 1 = 399, 32 × 28 = (30 + 2)(30 - 2) = 900 - 4 = 896, and 45 × 55 = (50 - 5)(50 + 5) = 2500 - 25 = 2475. Write the general pattern (a + b)(a - b) = a2 - b2 on the board and point out that reading it right to left is a way to factor.
Direct Instruction20 minutes
Show Diagram 1 to justify the difference of squares with area, then introduce a routine for reading structure (Diagram 2):
Look for a common factor and factor it out first.
Count the terms. Two terms: is it a difference of squares (or, in Algebra II, a sum or difference of cubes)? Three terms: is it a perfect square trinomial, or a quadratic in some chunk?
Name the chunks. Write each square as (something)2. If a chunk repeats, call it u.
Rewrite, then check. Multiply back to confirm the forms are equivalent, and look for any factor that can be rewritten again.
Work these five examples, asking students to name the pattern before any algebra is written:
Numbers with structure
Compute 482 - 422 without squaring either number.
Equation: It is a difference of squares: (48 - 42)(48 + 42) = 6 × 90 = 540.
Difference of squares, used twice
Rewrite x4 - 81 as a product.
Equation: (x2)2 - 92 = (x2 - 9)(x2 + 9) = (x - 3)(x + 3)(x2 + 9). The factor x2 + 9 is a sum of squares and does not factor over the real numbers.
Perfect square trinomial
Rewrite 9x2 - 30x + 25.
Equation: The first and last terms are (3x)2 and 52, and the middle term is -2(3x)(5). So 9x2 - 30x + 25 = (3x - 5)2.
Quadratic in form
Rewrite x4 - 13x2 + 36.
Equation: Let u = x2: u2 - 13u + 36 = (u - 4)(u - 9) = (x2 - 4)(x2 - 9) = (x - 2)(x + 2)(x - 3)(x + 3).
Pairs sort eight expressions by structure before rewriting any of them: 49 - y2, x2 + 14x + 49, 5x2 - 45, x6 - 1, a4 - 10a2 + 9, (x - 1)2 - 25, x(y + 2) - 3(y + 2) and x2 + 25. For each one, pairs write the pattern name and the chunk ("a difference of squares with a = 7 and b = y"), then rewrite it. Circulate and listen for students who try to factor x2 + 25 as (x + 5)(x - 5) or (x + 5)2. Ask them to multiply back. For x6 - 1, accept (x3 - 1)(x3 + 1) in Algebra I and ask Algebra II students to continue with the cube patterns.
Independent Practice15 minutes
Students complete six problems: two numerical (such as 997 × 1003 and 752 - 252), two that need a pattern applied twice (such as 3x4 - 48), and two quadratic-in-form or chunk problems (such as (x + 2)2 + 3(x + 2) - 10). For every problem, students write one line naming the structure they saw before they write the rewritten form, and they check one answer by substituting x = 2 into both forms.
Closure5-10 minutes
Exit ticket: "Rewrite x4 - y4 as a product of three factors, and explain what structure you saw at each step." The expected answer is (x2 - y2)(x2 + y2) = (x - y)(x + y)(x2 + y2), with the explanation that x4 - y4 is (x2)2 - (y2)2 and that x2 - y2 is a second difference of squares.
Differentiation Strategies
For Struggling Students
Provide a list of perfect squares (1 to 144) and of squared monomials (4x2, 9x2, x4, 25y2) to make patterns easier to spot
Have students rewrite each square in the form (something)2 before factoring, for example 25x2 = (5x)2
Use the area model in Diagram 1 with paper cut-outs so students can physically move piece P
Start with numeric examples, then move to one variable, then to chunks
For Advanced Students
Factor x6 - 64 two ways, first as a difference of squares and then as a difference of cubes, and show the results agree
Show that the product of two consecutive odd numbers is always one less than a perfect square, using (n - 1)(n + 1) = n2 - 1
Rewrite x4 + 4 by adding and subtracting 4x2 to create a difference of squares: (x2 + 2)2 - (2x)2
Assessment Guidance
What to Look For
Ask students to name the structure before they rewrite. A student who writes "difference of squares, a = x2, b = 9" has shown the reasoning this standard targets, even before finishing the factoring. Watch for three errors: factoring a sum of squares such as x2 + 9 as if it were a difference, stopping after one step when a factor can be rewritten again, and forgetting a common factor at the start (for example, factoring 3x2 - 12 as (3x - 6)(x + 2) instead of 3(x - 2)(x + 2); both are equivalent, but the second shows the structure more completely).
02
Classroom Activities
3 Activities
1
Cut and Rearrange
15 minPairs
Pairs build the area model for a difference of squares with grid paper, so they can see why a2 - b2 = (a + b)(a - b) instead of memorizing it.
Procedure
Cut out a 9-by-9 square of grid paper, then cut a 4-by-4 square from one corner. Record the remaining area: 81 - 16 = 65.
Cut the remaining L-shape into two rectangles, 5 by 4 and 9 by 5, and rearrange them into one rectangle. It measures 13 by 5, and 13 × 5 = 65.
Repeat with a 10-by-10 square and a 3-by-3 corner, and predict the final rectangle (13 by 7) before cutting.
Write the general rule in terms of a and b.
Discussion Questions
Why is one side of the new rectangle a + b and the other a - b?
Why does this model not work for a2 + b2?
Modification for Distance Learning
Use an online geometry tool with draggable rectangles, or have students draw the pieces on graph paper and photograph their rearrangement.
2
Structure Sort
20 minGroups of 3-4
Groups sort 12 expression cards into categories by structure, then rewrite each one. Several cards look alike but have different structure.
Look-alikes that do not fit: x2 + 36, x2 - 12x + 35, x2 + 20x + 25, 16x2 - 8
Procedure
Groups sort all 12 cards, then rewrite every card that fits a pattern and verify one by multiplying back.
For each look-alike, groups write one sentence explaining why it does not fit. For example, 16x2 - 8 is not a difference of squares with whole-number coefficients, but it does have the common factor 8: 8(2x2 - 1).
3
Mental Math Challenge
15 minIndividual then share
Students race to compute products and differences using structure, then explain their shortcuts. This shows that algebraic structure is a property of numbers.
Problems
99 × 101 = 1002 - 1 = 9,999
632 - 372 = 26 × 100 = 2,600
522 = (50 + 2)2 = 2,500 + 200 + 4 = 2,704
9982 - 4 = (998 - 2)(998 + 2) = 996,000
Share-Out
Volunteers write their shortcut in algebraic form on the board, for example (a - b)(a + b) = a2 - b2 with a = 100 and b = 1.
Partner Variation
Each student writes two new mental math problems that use a pattern and trades with a partner, who must solve them and name the pattern.
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Why a² - b² = (a + b)(a - b)
Remove a b-by-b corner from an a-by-a square, then move piece P next to piece Q. The result is a rectangle with sides a + b and a - b. Drawn to scale with a = 7 and b = 3.
Diagram 2: A Checklist for Reading Structure
Students ask these questions in order before rewriting. The last step matters: after one rewrite, a factor such as x2 - 4 may have structure of its own.
04
Homework Assignment
~30 min
HSA.SSE.A.2 Homework: Seeing Structure to Rewrite
Directions: For each problem, first write the structure you see (for example, "difference of squares with a = 2x and b = 9y"). Then rewrite the expression as a product, or compute the value. Check one problem in each part by multiplying back or by substituting a number.
Part 1: Spot the Pattern (Problems 1-3)
Rewrite 25x2 - 49y2 as a product of two binomials.
Rewrite 4x2 + 28x + 49 as the square of a binomial. Explain how you know the middle term fits.
Compute 1012 - 992 without a calculator and without squaring either number. Show the structure you used.
Part 2: Hidden Structure (Problems 4-6)
Rewrite 16x4 - 1 as a product of three factors. Explain why one factor cannot be rewritten further using real numbers.
Rewrite x4 + 5x2 - 36 by letting u = x2. Give the final answer in terms of x, with every factor that can be rewritten again rewritten.
Rewrite (x + 3)2 - 5(x + 3) + 6 by treating x + 3 as a single quantity. Then simplify each factor, and check your answer by substituting x = 1 into both forms.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Structure Named
Pattern and its parts identified correctly
Pattern named, parts missing or wrong
No structure identified
Rewriting
Correct equivalent form
Correct method with one algebra error
Incorrect or not equivalent
Completeness
Every factor that can be rewritten has been
Stopped one step early
Not attempted
Verification
Check by multiplying back or substituting shown
Check started but not finished
No check
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which expression is equivalent to x2 - 64?
Answer: C
x2 - 64 = x2 - 82, a difference of squares, so it equals (x - 8)(x + 8). Choice A expands to x2 - 16x + 64, which has a middle term. Choice B halves 64 instead of taking its square root.
Question 2 of 20 · Multiple Choice
Which expression is a perfect square trinomial?
Answer: A
x2 + 10x + 25 = (x + 5)2, since 25 = 52 and 10x = 2(x)(5). In choice C, the first and last terms are squares, but the middle term would need to be 10x, not 5x. Choice D is a difference of squares, not a trinomial.
Question 3 of 20 · Multiple Choice
Which is equivalent to 36x2 - 60x + 25?
Answer: B
36x2 = (6x)2, 25 = 52 and -60x = -2(6x)(5), so the expression is (6x - 5)2. Choice A has the wrong sign on the middle term. Choice C is 36x2 - 25, which has no middle term. Choice D halves 36 instead of taking its square root.
Question 4 of 20 · Multiple Choice
Which is the complete factorization of 81x4 - y4?
Answer: D
See 81x4 - y4 as (9x2)2 - (y2)2 = (9x2 - y2)(9x2 + y2), then factor 9x2 - y2 = (3x)2 - y2 again. The factor 9x2 + y2 is a sum of squares and stops there. Choice A squares the first factor instead of multiplying it by 9x2 + y2. Choice C equals (9x2 - y2)2, the same error.
Question 5 of 20 · Multiple Choice
What is the value of 742 - 262?
Answer: A
(74 - 26)(74 + 26) = 48 × 100 = 4,800. Choice B is 482, which comes from treating 742 - 262 as (74 - 26)2. Choice D doubles the correct answer.
Question 6 of 20 · Multiple Choice
Which substitution turns x6 - 7x3 - 8 into a quadratic expression?
Answer: C
x6 = (x3)2, so with u = x3 the expression is u2 - 7u - 8 = (u - 8)(u + 1) = (x3 - 8)(x3 + 1). Choice A gives u3 - 7x3 - 8, which still has x3 and is not quadratic in u.
Question 7 of 20 · Multiple Choice
Which is the complete factorization of 2x2 - 72?
Answer: B
Factor out 2 first: 2(x2 - 36), then use the difference of squares: 2(x - 6)(x + 6). Choice C factors out 2 from only the first term. Choice A expands to 2x2 - 24x + 72, which has a middle term. Choice D expands to 2x2 + 18x - 72.
Question 8 of 20 · Multiple Choice
Which is equivalent to (x + 5)2 - 9?
Answer: D
Treat x + 5 as one quantity: (x + 5)2 - 32 = (x + 5 - 3)(x + 5 + 3) = (x + 2)(x + 8). Choice A uses 9 instead of its square root 3. Choice C comes from squaring x + 5 as x2 + 25 and forgetting the middle term.
Question 9 of 20 · Multiple Choice
Which expression cannot be rewritten as a product of two binomials using a difference of squares?
Answer: A
x2 + 49 is a sum of squares. No product of real binomials gives it, since (x + 7)(x - 7) = x2 - 49 and (x + 7)2 has a middle term. Choice C is (7 - x)(7 + x), and choice D is (x2 - 7)(x2 + 7).
Question 10 of 20 · Multiple Choice
Which is equivalent to 2x(x - 7) + 3(x - 7)?
Answer: C
Treat (x - 7) as a single common factor: 2x(x - 7) + 3(x - 7) = (2x + 3)(x - 7). Choice B multiplies the coefficients 2x and 3 instead of adding them. Choice D counts the common factor twice.
Question 11 of 20 · Multiple Choice
(Algebra II) Which is equivalent to x3 - 27?
Answer: B
x3 - 27 = x3 - 33, a difference of cubes: a3 - b3 = (a - b)(a2 + ab + b2). With a = x and b = 3, that is (x - 3)(x2 + 3x + 9). Choice C has the sign pattern of a sum of cubes. Choice A expands to x3 - 9x2 + 27x - 27.
Question 12 of 20 · Multiple Choice
What is the greatest common factor of 8x3 + 12x2?
Answer: D
The greatest common factor of 8 and 12 is 4, and the highest power of x in both terms is x2, so 8x3 + 12x2 = 4x2(2x + 3). Choices A and B are common factors but not the greatest. Choice C is the least common multiple.
Question 13 of 20 · Multiple Choice
Which is equivalent to (a + b)2 - (a - b)2?
Answer: C
Treat a + b and a - b as single quantities in a difference of squares: [(a + b) - (a - b)][(a + b) + (a - b)] = (2b)(2a) = 4ab. Choice D is the sum (a + b)2 + (a - b)2. Choice A comes from assuming the squares are equal.
Question 14 of 20 · Multiple Choice
Which is equivalent to x2 - 6x + 9 - y2?
Answer: A
The first three terms are (x - 3)2, so the expression is (x - 3)2 - y2, a difference of squares: (x - 3 - y)(x - 3 + y). Choice D reads x2 - 6x + 9 as (x + 3)2. Choice B drops the square on y.
Question 15 of 20 · Short Answer
Rewrite 50x2 - 8 as a product, factoring completely. Name each structure you used.
Common factor 2: 50x2 - 8 = 2(25x2 - 4). Then 25x2 - 4 = (5x)2 - 22 is a difference of squares, so the answer is 2(5x - 2)(5x + 2).
Question 16 of 20 · Short Answer
Rewrite x4 - 26x2 + 25 as a product of four binomials.
Let u = x2: u2 - 26u + 25 = (u - 1)(u - 25) = (x2 - 1)(x2 - 25). Each factor is a difference of squares: (x - 1)(x + 1)(x - 5)(x + 5).
Question 17 of 20 · Short Answer
Compute 596 × 604 without a calculator. Show the structure you used.
Use a difference of squares to show that (x + 1)2 - (x - 1)2 = 4x.
Treat x + 1 and x - 1 as single quantities: [(x + 1) - (x - 1)][(x + 1) + (x - 1)] = (2)(2x) = 4x. No squaring is needed.
Question 19 of 20 · Short Answer
Rewrite 9(x - 2)2 - 16 as a product of two binomials, and simplify each factor.
9(x - 2)2 = [3(x - 2)]2 and 16 = 42, so the expression is [3(x - 2) - 4][3(x - 2) + 4] = (3x - 6 - 4)(3x - 6 + 4) = (3x - 10)(3x - 2).
Question 20 of 20 · Short Answer
A student notices that x2 - 6x - 16 = (x - 3)2 - 25. Verify this, then use the structure of the right side to factor the expression.
(x - 3)2 - 25 = x2 - 6x + 9 - 25 = x2 - 6x - 16, so the two forms are equivalent. The right side is a difference of squares with a = x - 3 and b = 5: (x - 3 - 5)(x - 3 + 5) = (x - 8)(x + 2).
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does using the structure of an expression mean?
It means looking at the overall shape of an expression, not just its individual symbols, and recognizing a pattern you know. For example, x4 - y4 can be seen as (x2)2 - (y2)2, a difference of two squares. Once you see the pattern, you know a way to rewrite it: (x2 - y2)(x2 + y2).
Is this standard only about factoring?
No. Factoring is a frequent use, but the standard is about any rewriting that structure suggests. That includes computing 482 - 422 as 6 × 90, recognizing a common binomial factor, and rewriting x2 - 4x - 12 as (x - 2)2 - 16 to see a difference of squares. Choosing a form to reveal a property, such as zeros or a maximum, is the focus of HSA.SSE.B.3.
Why can't x² + 9 be factored the same way as x² - 9?
Multiplying (x + 3)(x - 3) gives x2 - 9, and (x + 3)2 gives x2 + 6x + 9. No pair of real binomials multiplies to x2 + 9, because x2 + 9 is never zero for a real number x, while a product of real linear factors is zero somewhere. A sum of squares like this does not factor over the real numbers.
What does quadratic in form mean?
An expression is quadratic in form when it looks like au2 + bu + c for some chunk u. For x4 - 13x2 + 36, the chunk is u = x2. For (2x + 1)2 - 7(2x + 1) + 10, the chunk is u = 2x + 1. Replacing the chunk with a single letter makes the structure easy to see, and you substitute back at the end.
How do students know when they are finished factoring?
Check each factor for structure of its own. After x4 - 81 = (x2 - 9)(x2 + 9), the factor x2 - 9 is another difference of squares, so it can be rewritten again. The factor x2 + 9 cannot be rewritten using real numbers. A question will usually say "factor completely" when it expects every step.
What are the common mistakes students make on this standard?
Factoring a sum of squares as if it were a difference
Writing (a - b)2 for a2 - b2
Missing a common factor at the start
Stopping after one rewrite when a factor has more structure
Forgetting to substitute back after using u for a chunk
Mixing up the signs in the sum and difference of cubes patterns (Algebra II)
Are sum and difference of cubes part of HSA.SSE.A.2?
Usually in Algebra II rather than Algebra I. The standard does not list specific patterns. In many Algebra I courses the focus is on common factors, differences of squares and perfect square trinomials. Algebra II courses usually add sums and differences of cubes and more complex polynomial and rational expressions. On this page, cube problems are marked Algebra II.
How is HSA.SSE.A.2 tested?
Common question types ask for an equivalent form of an expression, a complete factorization, or a quick numerical computation that uses a pattern. On the digital SAT, recognizing and producing equivalent expressions is part of the Advanced Math domain, and structure often gives a much faster route than expanding.
How can students check that two forms are really equivalent?
Multiply the factored form back out, or substitute the same value into both forms. For example, at x = 2, x4 - 13x2 + 36 = 16 - 52 + 36 = 0 and (x - 2)(x + 2)(x - 3)(x + 3) = 0. One matching value does not prove equivalence, but a mismatch proves an error, and checking two or three values catches most mistakes.
How does this standard connect to later courses?
Rewriting by structure is used to find zeros of quadratics and polynomials (HSA.SSE.B.3 and HSA.APR.B.3), to simplify rational expressions, and to prove polynomial identities (HSA.APR.C.4). In calculus, students regularly rewrite expressions this way before they can evaluate limits or derivatives.
07
Related Standards
6 standards
These standards connect to HSA.SSE.A.2: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
7.EE.A.1Prerequisite
Apply properties of operations to add, subtract, factor, and expand linear expressions
