Solving Equivalent Expressions Questions on the SAT

What these questions look like on the SAT

Equivalent expressions is one of three skills College Board groups under Advanced Math, next to nonlinear equations and nonlinear functions. Advanced Math is about 35% of SAT Math, or 13 to 15 questions, shared among the three. This is the rewriting skill: you change an expression’s form without changing its value.

The framework describes it as making “strategic use of algebraic structure and the properties of operations to identify and create equivalent expressions.” It names the factoring you need (“finding a common factor, rewriting binomials that represent a difference of two squares, and rewriting trinomials as the product of two binomials”), then adds “rewriting simple rational expressions, rewriting expressions with rational exponents in radical form,” other polynomial factoring, and the ability to “fluently add, subtract, and multiply polynomials.” Its sample question for the skill, labeled “Rewrite a rational expression,” subtracts two fractions with different binomial denominators.

Math is two adaptive 35-minute modules of 22 questions, with about 75% multiple choice and about 25% student-produced response (SPR). On a multiple-choice version you compare four forms. An SPR version has no choices, so you might type a constant that makes two forms match. The built-in Desmos calculator is available for every question.

College Board’s Skills Insight ties skill statements to Math score bands. These are the ones about rewriting expressions, quoted without a note that limits part of the top one to the SAT:

The ladder climbs from combining like terms to rational and radical forms to “using structure.” The 420–460 and 680–800 statements also mention solving a formula for a variable of interest. The framework’s detailed list files that under nonlinear equations, but it uses the same rewriting moves. For K = ½mv² with v > 0: multiply by 2 (2K = mv²), divide by m (v² = 2K/m), then take the positive square root: v = √(2K/m).

The method: Expand, Factor, or Test

Two expressions are equivalent when they produce the same value for every valid input. On the SAT you prove that by algebra (expand or factor until both sides match) or, on multiple choice, by testing smart numbers and then confirming.

Use Expand, Factor, or Test. If one side is factored and the other isn’t, expand the factored form (or factor the expanded form). If you’re matching choices, expand your expression once and compare. If algebra is messy, plug in a convenient value of x (avoid 0 and 1 when those make every choice look the same) to eliminate options, then verify the survivor.

Go faster without getting sloppy

Memorize the patterns you see constantly: (a+b)² = a² + 2ab + b², (a−b)² = a² − 2ab + b², (a+b)(a−b) = a² − b². Factoring out a GCF first often reveals one of those. For rationals, factor numerator and denominator and cancel common factors — don’t cancel terms across a plus sign.

Adding, subtracting and multiplying polynomials is the other fluency the framework names. To subtract a polynomial, subtract every term: distribute the minus sign first, then combine like terms. To multiply, every term of one factor multiplies every term of the other. For example, (x − 2)(x² + 3x − 1) = x³ + 3x² − x − 2x² − 6x + 2 = x³ + x² − 7x + 2. Six partial products, then combine. Count them so none goes missing.

Rational expressions: common denominators

A rational expression is a fraction with a polynomial on top and bottom. The rules are the ones you use with numbers, applied to factors. To simplify one fraction, factor completely and cancel common factors: (x² − 9)/(x² + 3x) = (x − 3)(x + 3)/(x(x + 3)) = (x − 3)/x for x ≠ 0 and x ≠ −3. To add or subtract, build a common denominator first. That’s the move in College Board’s sample question, and it’s the one to practice.

When the top has a higher degree than the bottom, rewrite the numerator so it contains the denominator. For (x² + 5x + 7)/(x + 2): since (x + 2)(x + 3) = x² + 5x + 6, the numerator is (x + 2)(x + 3) + 1, so the fraction equals x + 3 + 1/(x + 2). When the choices look like a polynomial plus a small fraction, this is the move.

Exponents and radicals

Exponent questions test a short list of rules. Learn them as a set, because wrong choices are built from mixing them up:

• x^a · x^b = x^(a+b) and x^a / x^b = x^(a−b): same base, add or subtract exponents.
• (x^a)^b = x^(ab): a power of a power multiplies.
• x^(−a) = 1/x^a: a negative exponent means reciprocal, not a negative number.
• x^(m/n) = ⁿ√(x^m): the denominator is the root, the numerator is the power. So x^(1/2) = √x and x^(2/3) = ∛(x²).

Worked example: for x > 0, write ∛(x²) · √x as a single power of x. Convert each radical: ∛(x²) = x^(2/3) and √x = x^(1/2). Multiply by adding exponents: 2/3 + 1/2 = 4/6 + 3/6 = 7/6. The product is x^(7/6), which is also ⁶√(x⁷). Check with x = 64: ∛4096 · √64 = 16 · 8 = 128, and 64^(7/6) = 2⁷ = 128 ✓.

Radicals simplify by pulling out perfect squares: √(18x²) = √(9x²) · √2 = 3x√2 when x ≥ 0. The condition matters, because √(x²) is really |x|. That’s why these questions so often say x > 0.

When structure hides the match

Sometimes the SAT writes an expression to look different on purpose: 2(x − 4) + 3(x − 4) is 5(x − 4) by combining like groups. Factor out the common binomial before expanding everything.

Two more structure moves come up at the top of the ladder. First, treat a repeated expression as one unit. (x + 3)² − 16 is a difference of squares with a = x + 3 and b = 4, so it factors as (x + 3 − 4)(x + 3 + 4) = (x − 1)(x + 7) without expanding anything. The same idea factors x⁴ − 16 as (x² − 4)(x² + 4), and then (x − 2)(x + 2)(x² + 4).

Second, match coefficients. If two polynomials are equal for every x, their coefficients match term by term: the x² terms, the x terms and the constants. That turns “find the constant” questions into one-line equations. Use two of the matches to find the constant and the third to check it.

Check equivalence with Desmos

The built-in Desmos calculator turns “is this the same expression?” into a picture, which makes it a strong check on multiple choice.

• Overlay the graphs. Type y = (2x² + 7x + 8)/(x + 2) and then each choice. An equivalent expression draws exactly on top of the original. Zoom out once to make sure two curves don’t split apart off-screen.
• Evaluate at an awkward number. Define f(x) as the original and g(x) as a choice, then type f(1.37) and g(1.37). Ugly inputs make accidental ties very unlikely, which is the weakness of testing 0 or 1 by hand.
• Slide a constant. Type y = (2x + k)(x − 3), add a slider for k, and drag until the graph lands on y = 2x² + 4x − 30. Then confirm by matching coefficients.

Two cautions. A graph can’t show you a single missing point well, so (x² − 4)/(x − 2) and x + 2 will look identical even though the first is undefined at x = 2. Read the restriction in the question. And when the answer must be a specific form, such as factored or with a radical, the graph confirms equivalence but doesn’t pick the form. That part is still algebra.

Spot the traps in the choices

Wrong choices often expand with a sign error on the middle term, factor incompletely, or cancel illegally in a fraction. Another trap: two expressions that match at x = 2 but aren’t identical — always prefer an algebraic match when you can. Here are the patterns behind most wrong choices:

• The missing middle term. (x − 5)² = x² − 10x + 25, not x² + 25 or x² − 25. Squaring a sum never just squares each piece.
• Cancelling terms instead of factors. (x + 6)/6 is not x + 1, and (x² + 4)/(x + 2) doesn’t simplify at all, because x² + 4 doesn’t factor.
• A half-distributed minus sign. In a subtraction, every term of the second polynomial changes sign, constants included.
• Mixed-up exponent rules. (x³)² = x⁶ but x³ · x² = x⁵; x^(−2) = 1/x², not −x²; x^(1/2) = √x, not x/2.
• Test-value coincidences. 0 and 1 make many different expressions equal. If two choices survive a test, try a second value.

Practice until it’s automatic

Work each one on paper before you reveal it. These are the moves that should take seconds, not minutes.

Practice questions

Here are three Equivalent Expressions questions from our problem sets, from easier to harder, in the same format as the Digital SAT. Answer each one to see the full explanation.

Quick checks

Shorter questions on the same skills. Question 7 combines a rational expression with matching coefficients.

Frequently asked questions