SAT Math — equivalent expressions: factoring, expanding, exponents

Most of what the Advanced Math domain asks is a matter of rewriting: the same quantity, written a different way, so that whatever is wanted becomes visible. Factoring, expanding, cancelling and the exponent rules are all the same skill from different angles, and the errors they produce are remarkably consistent from one learner to the next. This set is built around those errors rather than around the topics.

Four questions are factoring. Two have a leading coefficient of one, where the whole task is finding two numbers that multiply to the constant and add to the middle coefficient — and the wrong options are built so that each satisfies exactly one of those two conditions, so a learner who checks only one of them cannot get through. One has a leading coefficient that is not one, where the same two constants placed in opposite brackets give a different middle term, which is the single commonest slip in the topic. One is a perfect square with a negative middle term, where the sign of the constant decides whether the two signs agree and the sign of the middle term then decides which they are.

Two questions are the special forms recognised on sight: a difference of two squares, whose middle terms cancel, and a perfect square, where the middle coefficient is twice the square root of the constant. That test is worth more than the pattern itself, because it tells you when the pattern does not apply.

Two are exponents. One puts the power-of-a-power rule and the product rule in the same expression, so that the two rules pull in opposite directions and getting one of them backwards produces a specific wrong answer for each. One uses a fractional exponent applied to a bracket containing a coefficient as well as a variable, where the tempting wrong answer takes the root of the variable and leaves the number alone.

Two simplify a fraction by factoring and cancelling, and both carry a wrong option that cancels a term out of a sum instead of a factor out of a product — the error that makes algebra look arbitrary to a learner who has not been shown why it fails. And two expand a product, one of them squaring a bracket, where the wrong option that drops the middle term is the most familiar mistake in all of school algebra.

Every explanation names the mistake behind each wrong option and checks the answer by substituting a concrete number into both forms — a habit that settles any of these questions in seconds when the algebra is not obvious. Nothing here is drawn from any official publication.

  • Factor a quadratic by finding two numbers that satisfy both the product and the sum conditions, not just one
  • Handle a leading coefficient other than one, where the placing of the constants changes the middle term
  • Recognise a difference of two squares and a perfect square, and test the latter against its middle coefficient
  • Apply the product and power-of-a-power rules in the right places when both appear in one expression
  • Apply a fractional exponent to every factor inside a bracket, the coefficient included
  • Cancel only whole factors from a fraction, and say why cancelling a term out of a sum fails
  • Check any claimed equivalence by substituting a single number into both expressions

Written for this catalogue, with no source document. The content is the equivalent-expressions material named in the College Board's own public description of the Advanced Math domain — factoring, expanding, exponent rules and rational expressions; every question, option and explanation here is original, and nothing is reproduced from any official publication or released test. Note that the digital test also asks some questions in a student-produced response format, which a multiple-choice bank cannot reproduce. Zestly is an independent study tool. It is not affiliated with the College Board, which owns the SAT, and it is not an exam centre.

Sample question

Which expression is equivalent to $x^2 + 7x + 10$?

See the answer

$(x + 2)(x + 5)$

Factoring needs two numbers whose product is $10$ and whose sum is $7$, and $2$ and $5$ are the only pair doing both. Substituting $x = 3$ gives $40$ from the original and $5 \cdot 8 = 40$ from the factors. The pair $1$ and $10$ multiplies correctly but adds to $11$, producing the wrong middle term. The pair $3$ and $4$ adds correctly but multiplies to $12$, which is the same mistake mirrored: middle term right, constant wrong. And $(x - 2)(x - 5)$ has the right numbers with both signs flipped, which makes the middle term negative.

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