Equivalent Expressions and the Distributive Property
A.SSE.2
Read the standard text
A.SSE.2 Use the structure of an expression to identify ways to rewrite it. For example, to factor 3x(x − 5) + 2(x − 5), students should recognize that the "x − 5" is common to both expressions being added, so it simplifies to (3x + 2)(x − 5); or see x^4 − y^4 as (x^2)^2 − (y^2)^2, thus recognizing it as a difference of squares that can be factored as (x^2 − y^2)(x^2 + y^2).
Section 1 · LearnLearn the idea
In this lesson you will: Use the distributive property forwards to expand and backwards to factor; Spot a common factor, including a common binomial factor; Recognize a difference of two squares as a structure; Decide whether two expressions are equivalent, and justify the decision.
Equivalent means equal for every value
Two expressions are equivalent when they produce the same number for every input. That is a strong claim, and it is not established by one lucky substitution — 3x and x+2 agree at x=1 and nowhere else.
What does establish it is a chain of legal rewrites. Distributing, combining like terms and reordering a sum all change how an expression looks without changing what it equals. Substitution works in the other direction: a single disagreement disproves equivalence on the spot.
The distributive property, read backwards
Everyone learns a(b+c)=ab+ac from left to right. Read it from right to left and it is a factoring rule: if every term contains the same factor, that factor can be pulled out front.
The factor pulled out does not have to be a number. In 3x(x-5)+2(x-5) both terms contain the parentheses (x-5), so it comes out whole and leaves (3x+2) behind.
Structures worth spotting on sight
Rewriting is easier when you recognize the shape of what is in front of you. A handful of shapes cover most of what Ohio asks in Algebra 1, and the last one in the table is worth memorizing in both directions.
A difference of two squares, a^2-b^2, always factors as (a+b)(a-b). A sum of two squares does not factor at all over the real numbers, which is a difference worth being sure about.
Common mistake — Losing a term that becomes a 1
Asked to factor 7x+7, a student writes 7(x).
The thinking is half right. The 7 does come out of both terms, and after it leaves the first term there is an x standing there. The second term, though, was 7 all by itself, and when its 7 leaves, something has to remain.
What remains is 1, because 7=7*1. The correct factoring is 7(x+1). The same slip turns 5x^2+5x into 5x(x) instead of 5x(x+1).
How to catch it: Multiply your answer back out. 7(x) is 7x, which is missing a term, so the factoring is wrong before you write another line. Distributing back is not optional busywork — it is the only check this kind of problem has.
Section 2 · DoPractice this standard
These published practice sets are scored on the student site and cover A.SSE.2 directly.