Equivalent expressions

Advanced Math is about 35 percent of the Math section — 13 to 15 of the 44 questions — which ties it with Algebra for the largest domain. That weighting is College Board's own, from the published assessment framework. How those questions divide across the domain's three skill points is not published; prep-industry counts put Equivalent Expressions at roughly a third of them. Treat the 35 percent as official and the split as an estimate.

What distinguishes this skill point is that most of its items carry no story. No gym membership, no rideshare fare — symbols on a screen and an instruction to find an equivalent form. That makes them fast when you know the moves and unbounded when you do not, which is a large part of why they cluster in the harder second module: the routing algorithm sends stronger performers to a module that leans on exactly the questions where knowing one pattern replaces two minutes of flailing.

The instruction is almost never "simplify." It is "which of the following is equivalent to the given expression," or "which equivalent form displays the minimum as a constant," or "the expression is equivalent to ax + b, what is a + b." Those look like three tasks and are one: what am I allowed to do to this expression that does not change what it is worth?

So state the definition once, because it is also a method. Two expressions are equivalent when they produce the same value at every input where both are defined. Not "look similar," not "reduce to something shorter." If two expressions disagree at even one legal input, they are not equivalent — and a single substitution can prove it in fifteen seconds. Hold onto that; it comes back as the most valuable habit in the lesson.

Everything in this lesson is a rearrangement of four things: expanding, factoring, exponent rules, and fractions. If any one of the four is shaky, the rest of the lesson reads as a list of tricks to memorise rather than a small set of moves. This block builds all four from nothing, in plain language. A reader who already has them should skip to the next block — nothing later depends on having read this one.

Expanding. Every term in the first bracket multiplies every term in the second, and nothing is skipped: (2x − 3)(x + 5) = 2x·x + 2x·5 + (−3)·x + (−3)·5 = 2x² + 10x − 3x − 15 = 2x² + 7x − 15. The only thing to be careful about is that the minus sign belongs to the 3 and travels with it into both of its products. Losing that sign is the single most common arithmetic error in this entire domain.

Factoring is expanding run backwards, and four patterns cover essentially everything the Digital SAT uses. Pattern one, common factor, and always check for it first: 6x³ − 15x² = 3x²(2x − 5). Pattern two, difference of squares: a² − b² = (a − b)(a + b), so x² − 49 = (x − 7)(x + 7) and 4x² − 9 = (2x − 3)(2x + 3). A sum of squares, a² + b², does not factor over the real numbers — there is no pattern to remember because there is no pattern.

Pattern three, a trinomial with a leading coefficient of 1: x² + bx + c wants two numbers whose product is c and whose sum is b. For x² − 7x + 12 the pair is −3 and −4, so it factors as (x − 3)(x − 4). Pattern four, a trinomial with a leading coefficient other than 1 — the ac method, which is grouping. For 6x² + 5x − 4: multiply a by c to get 6 · (−4) = −24, find the pair with product −24 and sum 5, which is 8 and −3, split the middle term into those pieces, then factor in pairs. 6x² + 8x − 3x − 4 = 2x(3x + 4) − 1(3x + 4) = (2x − 1)(3x + 4). The shared bracket appearing in both halves is the confirmation that the split was right; if it does not appear, the pair was wrong.

Exponents. x⁵ means five copies of x multiplied together, and that one sentence generates every rule. Multiplying like bases piles the copies up, so x^a · x^b = x^(a+b). Dividing cancels copies off, so x^a / x^b = x^(a−b). A power of a power makes b groups of a copies, so (x^a)^b = x^(ab). An exponent on a product reaches every factor: (xy)^a = x^a · y^a, which is why the 2 in (2x³)⁴ gets raised too.

The extensions follow from the same rules rather than being new facts. x⁰ = 1, because x^a / x^a is both 1 and x⁰. x^(−a) = 1/x^a, because x^a · x^(−a) = x⁰ = 1. x^(1/n) is the nth root of x, because raising it to the n gives x¹. And x^(m/n) = (ⁿ√x)^m. There is one thing with no rule at all: x^a + x^b does not simplify. Exponent rules govern multiplication and division only. There is nothing to do to x³ + x⁵ except factor it as x³(1 + x²).

Fractions of expressions. A fraction bar is a division, and you may cancel only what is a factor of the entire numerator and the entire denominator. The reason is not a convention: division distributes over multiplication and does not distribute over addition. (6·5)/6 = 5, but (6 + 5)/6 is not 5. Everything that goes wrong with rational expressions on this test is a version of that one sentence being violated.

The check, which you should build now and never stop using. Substitute a number into the original and into your candidate answer, and see whether they agree. Choose x = 2 or x = 3 — never 0 and never 1, because those values make coefficients and exponents invisible, and never a value that makes a denominator zero. It costs about fifteen seconds and it catches sign errors that you have already convinced yourself are correct, which is the only kind that actually costs points.

Standard form, f(x) = ax² + bx + c. The constant c is f(0), the y-intercept. The sign of a says which way the parabola opens: up when a is positive, down when a is negative. That is all standard form gives away for free.

Factored form, f(x) = a(x − r)(x − s). Here r and s are the zeros — where the graph meets the x-axis — because a product equals zero exactly when one of its factors does. Nothing about the vertex is visible without work.

Vertex form, f(x) = a(x − h)² + k. The vertex is (h, k), the axis of symmetry is the line x = h, and k is the minimum value when a > 0 or the maximum when a < 0. The reason is one line long: (x − h)² is never negative, so it is smallest when it is zero, which happens at x = h, and there f(x) = k. Note the sign convention that costs people points: in a(x + 4)² − 7 the value of h is −4, not 4, because x + 4 is x − (−4).

The SAT question built directly on this is phrased "which of the following equivalent forms of the given equation displays the [minimum / the zeros / the y-intercept] as a constant or coefficient?" It is not asking you to compute anything. All four choices are usually genuinely equivalent to the original — that is what makes it hard — and three of them display the wrong feature. Read which feature is asked for, name the form that displays it, and pick that shape. Doing algebra on this item type is the weak path; recognising the shape is the strong one.

One shortcut worth owning outright: the vertex's x-coordinate is always −b/(2a), read straight off standard form with no rearrangement. If a question asks only where the minimum occurs, that is a five-second answer and completing the square is the slow road. If it asks for the minimum value, either evaluate f(−b/(2a)) or complete the square — both work, and the substitution is usually faster.

Four steps, in this order. One: factor a out of the x² and x terms only, leaving the constant outside the parentheses. Two: halve the coefficient of x that is now inside, square it, and add and subtract that number inside the parentheses. Three: fold the first three terms inside into a square, and carry the subtracted number out — multiplied by a. Four: check by expanding back, or by substituting a number.

With a = 1 it is short enough to do in your head: x² + 10x + 3 → (x² + 10x + 25 − 25) + 3 → (x + 5)² − 25 + 3 → (x + 5)² − 22. Expanding back gives x² + 10x + 25 − 22 = x² + 10x + 3. ✓

Step three is where the points go when a ≠ 1. The number you subtracted is sitting inside the parentheses, so it gets multiplied by a on its way out. In 3(x² + 8x + 16 − 16) + 41, the −16 leaves as −48, not as −16 — and the difference between those two, 25 versus −7, is a wrong answer the test has written into the choices on purpose.

One more thing about the phrasing. A completing-the-square question frequently does not ask for the form at all. It asks for h, or for k, or for h + k, or for the minimum value, or for the x at which the minimum occurs. Producing a correct vertex form and then answering the wrong letter is a full-credit loss that feels, at the moment you bubble it, exactly like a win.

Factor the top and the bottom completely before touching anything, then cancel only whole factors that appear on both. If nothing factors, nothing cancels: the expression is already in lowest terms no matter how much it looks like it ought to shrink. "It looks like it should simplify" is the feeling that produces cancelling across a sum, which is the most punished error in this domain.

Cancelling changes the domain, and the test knows it. (x² − 9)/(x − 3) equals x + 3 everywhere except at x = 3, where the original is undefined and x + 3 is 6. That is why these items carry the phrase "for all x ≠ 3" or "where both expressions are defined." The wrapper is not legal boilerplate — it is the test protecting the equivalence claim, and you should read it as free information. It names a zero of the denominator, which is very often the factor that cancels, which tells you what to look for on top before you start.

When the numerator does not divide evenly, the answer takes the shape quotient + remainder/divisor. Polynomial long division works and is worth being able to do. On a multiple-choice item, though, multiplying back is faster and safer: given (6x² + 5x − 1)/(2x − 1) = 3x + 4 + A/(2x − 1), multiply through by (2x − 1) to get 6x² + 5x − 1 = (2x − 1)(3x + 4) + A, then substitute any convenient x. At x = 1: 10 = (1)(7) + A, so A = 3. No division performed, no sign slips available.

Adding or subtracting rational expressions needs a common denominator, and the SAT's version is usually two fractions whose denominators are related rather than unrelated. Watch specifically for (3 − x) and (x − 3): they differ by a factor of −1, and pulling that −1 out is frequently the entire question. The same applies to (a − b) and (b − a) anywhere they appear together.

If you can already factor, expand, and complete the square without hesitating and are still dropping one or two Advanced Math questions per test, the remaining errors are almost never conceptual. They are five hard variants and four execution slips. Nothing below is a new technique — that is the point of the list.

Hard variant one — the form question, where every choice is genuinely equivalent. "Which of the following equivalent forms displays the minimum value of the function as a constant or coefficient?" All four expand back to the same thing, so verifying equivalence tells you nothing. The decision is entirely about which feature each shape displays: vertex form for a maximum or minimum, factored form for zeros or x-intercepts, standard form for the y-intercept. Read the feature asked for first, before you read a single choice, and the item becomes a shape-match rather than an algebra problem.

Hard variant two — identities with unknown constants. "(2x + 5)(3x − k) = 6x² − x − 40 for all values of x." The words "for all values of x" convert this from an equation to solve into an identity to match: expand the left, then set the coefficient of x² equal to the coefficient of x², the coefficient of x equal to the coefficient of x, and the constant equal to the constant. Two independent equations usually fall out, and if only one unknown exists, one of them is a free check on the other. The alternative — substitute two convenient values of x and solve the resulting system — is faster when the expansion is ugly, and it is exactly as valid.

Hard variant three — the excluded-value wrapper as a hint. "For all x ≠ 3" or "where both expressions are defined" is not decoration and not a caveat; it names a zero of the denominator. Nine times out of ten that same binomial is a factor of the numerator, which means the question has told you what to factor before you have started. Strong students read the restriction as legal fine print and skip it. It is free information.

Hard variant four — radicals, and the absolute-value hazard. Convert radicals to fractional exponents on sight: √x = x^(1/2), ∛(x⁴) = x^(4/3), 1/√x = x^(−1/2). Once everything is an exponent, one rule set covers the whole expression. The hazard: √(x²) = |x|, not x, because the radical symbol denotes the non-negative root. The SAT usually defuses this by stipulating x > 0 somewhere in the stem — which means the moment you see an item that does not stipulate it, the sign is very likely the point of the question. Read the stem for that condition before you start simplifying, not after.

Hard variant five — division with a remainder. When the numerator does not divide evenly, the answer has the form quotient + remainder/divisor, and the question usually asks for one constant inside it. Do not long-divide unless you have to. Multiply both sides by the divisor to clear the fraction, then substitute a value of x that kills the messy term — for (…)/(2x − 1), substituting x = 1/2 makes the quotient term vanish and hands you the remainder in one line. That trick generalises: the value that zeroes the divisor is always the fastest substitution.

Execution slips, in the order they cost points. First, the distributed minus sign that reaches only the first term: −(x² − 3x + 2) becoming −x² − 3x + 2. Write the bracket out before you distribute; on a screen with no scratch discipline this is the most expensive habit on the section. Second, the sign of h in vertex form — a(x + 4)² has h = −4 — and the sign of the constant that leaves the parentheses when a ≠ 1. Third, answering the wrong letter: circle the quantity actually requested in the stem before you begin, because after two minutes of correct algebra the pull toward handing in the number you just computed is strong. Fourth, over-simplifying past the answer, or under-simplifying short of it: the choices define the target form, so look at them before deciding how far to go.

The habit that converts, and the reason this skill point is the cheapest place on the test to buy points: substitute a number. On any "which of the following is equivalent" item, pick x = 2 or 3 — never 0, never 1, never an excluded value — evaluate the original once, then evaluate choices until one matches. If two match, try a second value. This is not a fallback for when the algebra fails; on ugly items it is the faster path, and it is immune to every sign error in the trap list above. The Bluebook app has the Desmos graphing calculator built into both Math modules, which is College Board's own tooling rather than a workaround, so on a form question you can graph the original and a candidate and see whether the curves coincide.

Time. Each Math module is 22 questions in 35 minutes — about 95 seconds per question, which is arithmetic on College Board's published module structure rather than advice from anyone. The recommendation is Meridian's: equivalent-expression items should run well under that average, because the substitution check makes them verifiable rather than merely attempted, and the time they save is what funds the geometry and data questions later in the module that genuinely need two minutes.