Digital SAT Math · Advanced Math

Equivalent expressions (nonlinear)

Digital SAT Math · Advanced Math

An expression has many faces and all of them are true at once: $x^2 - 9$ , $(x+3)(x-3)$ and $(x-1)^2 + 2x - 10$ are the same object wearing different clothes. The test never asks you to rewrite for its own sake — it asks for something specific, and exactly one face of the expression has that something written on it. So the skill is not factoring, and it is not expanding; it is deciding which face to turn the expression toward before you touch it. Do that first and most of these items collapse into two lines of algebra.

On the test


Domain	Advanced Math (score report)
What it looks like	A polynomial, a rational expression, or a radical, plus four rewritten versions of it — or an equation said to hold for all x with letters standing in for coefficients
Often asked	“Which expression is equivalent to…?”, “…is equivalent to $ax + b$ , where $a$ and $b$ are constants”, “…displays the minimum value as a constant”, “What is the value of $k$ ?”
Format	Multiple choice and student-produced response — the live digital sets on this skill run heavily to grid-ins
Calculator	Desmos can confirm an equivalence in seconds and will also confirm a wrong one; see the box below

Recognition cues: equivalent to, for all values of x, where a and b are constants, which of the following displays…, a stated restriction like $x \neq 3$ , and choices that are all rearrangements of the same handful of numbers.

What this page owns

Rewriting shows up everywhere in Advanced Math, so the boundary matters. Here the rewrite is the answer: nothing is solved, no value of $x$ is ever found, and the question ends the moment the expression is in the right form. When the rewrite is a step toward a root, that is nonlinear equations; when the point of the new form is what it says about a graph, that is quadratic functions. The factoring muscle is the same in all three places — this page is where it gets built.

Pattern recognition

Six shapes cover essentially every item:

Expand — a product of binomials, or a difference of two squared binomials, with the choices differing only in the middle term.
Factor — a common factor, a trinomial, or the difference-of-squares pattern $a^2 - b^2$ .
Simplify a rational expression — factor top and bottom, cancel a shared factor, keep the restriction.
Combine rational expressions — one denominator, then add or subtract the rescaled numerators.
Match coefficients — an identity true for all x, with a letter in place of a coefficient or constant.
Rewrite exponents and radicals — a root turned into a power, or a power distributed over a product.

Method

Name the target form. Say it out loud in five words: “two factors”, “one fraction”, “squared plus a constant”, “a number for $k$ ”. That sentence is the whole plan.
Choose the rewrite that reaches it. Factoring exposes roots and cancels; expanding exposes coefficients and lets you compare; a common denominator turns two fractions into one; completing the square exposes a maximum or minimum. One of the four is right and the other three are work you will throw away.
Rewrite in full, with every sign carried. Distribute a negative to every term inside the parentheses. Square a binomial as $(a+b)(a+b)$ — never term by term. Cancel only whole factors, never pieces of a sum.
Match, don’t solve. If the stem says the two sides agree for all $x$ , set the $x^2$ coefficients equal, then the $x$ coefficients, then the constants. Three little equations, no algebra on $x$ at all.
Check with one number. Pick a value that is easy but not degenerate — $x = 2$ usually, never $x = 0$ or $1$ when constants might hide there — and evaluate the original and your rewrite. Equal? You are done. This costs eight seconds and catches every sign slip in step 3.

The stem asks for	The rewrite that gets there
“equivalent to $ax+b$ ”, a fraction that must simplify	factor, then cancel
“displays the minimum/maximum as a constant”	complete the square
“for all values of x”, a letter for a coefficient	expand one side, match term by term
a single fraction from two	common denominator, rescale both numerators
four factored choices, one messy stem	expand the choices
a root, or a power of a product	rational exponents, then one exponent rule per step

Three patterns are worth memorizing cold, because the test builds distractors out of confusing them with each other:

(a+b)^2 = a^2 + 2ab + b^2 \qquad (a-b)^2 = a^2 - 2ab + b^2 \qquad a^2 - b^2 = (a+b)(a-b)

The middle term is the whole point of the first two, and its absence is the whole point of the third. $(x+5)^2$ is not $x^2 + 25$ ; the missing $10x$ is the single most punished slip on this skill. Meanwhile $x^2 + 25$ — a sum of squares — does not factor at all over the real numbers, so a choice offering $(x+5)(x-5)$ for it is wrong twice over. When you meet two squares with a minus between them, the factoring is already done: take the square root of each piece, write one sum and one difference. That includes coefficients — $9x^2 - 64$ is $(3x)^2 - 8^2$ , so the factors start with $3x$ , not $9x$ .

Cancelling is a factor privilege. In $\dfrac{x^2-9}{x^2+7x+12}$ you may cancel $(x+3)$ only after both top and bottom are written as products: $\dfrac{(x-3)(x+3)}{(x+3)(x+4)} = \dfrac{x-3}{x+4}$ . What you may never do is strike out matching terms — the $x^2$ on top against the $x^2$ below, or the 9 against the 12. A term is part of a sum; deleting it changes the value. The tell that you have done this is an answer that no longer contains the variable in a place where it obviously should. And the cancelled factor leaves a trace: the simplified expression is equal to the original except where the cancelled factor was zero, which is why real items print restrictions like $x \neq -3$ . Keep them; they are how you know your rewrite is honest.

Desmos on this skill — the check, and the two ways it lies. Type the stem expression on one line and a choice on the next; if the two graphs sit exactly on top of each other, they are equivalent, and this takes about ten seconds per choice. When the expression has no natural graph — a rewrite in terms of $a$ , $b$ and $k$ — set both sides equal to the same convenient number and compare, or just substitute a number for each letter and compare values. Two cautions the free Desmos guides skip. First, removed points are invisible: $\frac{x^2-9}{x+3}$ and $x-3$ plot as the same line even though the first is undefined at $x=-3$ , so a graph check will happily bless a choice that has dropped a restriction. Second, the graph cannot tell you which form the question wanted — on a “displays the minimum as a constant” item, every choice that is genuinely equivalent graphs identically, and the equivalent ones that fail to display the minimum are still wrong answers. Use the graph to catch arithmetic, never to replace reading the target.

Worked example 1 — the middle term is the question

Stem. Expand $(2x-5)(3x+4)$ , and then expand $(2x-5)^2$ .

Step 1 — the product, one pair at a time. Every term in the first factor meets every term in the second:

(2x-5)(3x+4) = 6x^2 + 8x - 15x - 20 = 6x^2 - 7x - 20

The $x$ -coefficient, $-7$ , is built from two products, $+8$ and $-15$ . Nothing else in the expansion can produce it.

Step 2 — the square is the same operation, not a shortcut. Write it out as a product before doing anything:

(2x-5)^2 = (2x-5)(2x-5) = 4x^2 - 10x - 10x + 25 = 4x^2 - 20x + 25

Both cross terms are $-10x$ , which is exactly why the pattern reads $a^2 - 2ab + b^2$ : the middle term is the cross product counted twice.

Check. Substitute $x = 2$ . Directly: $(4-5)(6+4) = -10$ , and $6(4) - 7(2) - 20 = 24 - 14 - 20 = -10$ ✓. For the square: $(4-5)^2 = 1$ , and $4(4) - 20(2) + 25 = 16 - 40 + 25 = 1$ ✓. Answers: $6x^2 - 7x - 20$ and $4x^2 - 20x + 25$ .

Trap watch. Writing $(2x-5)^2 = 4x^2 + 25$ squares the terms separately and throws away $-20x$ ; writing $4x^2 - 25$ confuses the square with the difference-of-squares pattern, which belongs to $(2x-5)(2x+5)$ — a different product entirely. In the first expansion, losing the sign on $-15x$ gives $6x^2 + 23x - 20$ , and it looks perfectly respectable next to the right answer.

Worked example 2 — factor first, cancel second

Stem. For $x \neq 3$ and $x \neq -3$ , write $\dfrac{2x^2 + x - 15}{x^2 - 9}$ in simplest form.

Step 1 — target form. One fraction with nothing left to cancel. That means both parts must become products before anything is struck out.

Step 2 — the denominator is a difference of squares.

x^2 - 9 = (x+3)(x-3)

Step 3 — factor the numerator. Two numbers multiplying to $2 \cdot (-15) = -30$ and adding to $+1$ : $+6$ and $-5$ . Split and group:

2x^2 + x - 15 = 2x^2 + 6x - 5x - 15 = 2x(x+3) - 5(x+3) = (2x-5)(x+3)

Step 4 — now cancel, and only a whole factor.

\frac{(2x-5)(x+3)}{(x+3)(x-3)} = \frac{2x-5}{x-3}

Check. At $x = 1$ : the original is $\frac{2+1-15}{1-9} = \frac{-12}{-8} = \frac{3}{2}$ , and the answer is $\frac{2-5}{1-3} = \frac{-3}{-2} = \frac{3}{2}$ ✓. Answer: $\dfrac{2x-5}{x-3}$ , for $x \neq \pm 3$ .

Trap watch. Cancelling the $x^2$ terms before factoring produces $\frac{x-15}{-9}$ or similar nonsense — those are terms inside sums, not factors. Factoring the denominator as $(x-3)^2$ forgets that $-9$ needs one plus and one minus. And the restriction $x \neq -3$ survives the cancellation even though $(x+3)$ has vanished from the page: the original expression never had a value there, so neither does anything claiming to equal it.

Worked example 3 — an identity is three tiny equations

Stem. The equation $(5x-2)(x+k) = 5x^2 + 13x - 6$ is true for all values of $x$ , where $k$ is a constant. What is the value of $k$ ?

Step 1 — read “for all x”. Both sides are the same polynomial, so their coefficients agree term by term. This is not an equation to solve for $x$ ; $x$ is never found.

Step 2 — expand the side with the letter in it.

(5x-2)(x+k) = 5x^2 + 5kx - 2x - 2k = 5x^2 + (5k-2)x - 2k

Step 3 — match, starting with the cheapest term. The constants give $-2k = -6$ , so $k = 3$ in one step. The $x^2$ coefficients already agree ( $5 = 5$ ) and cost nothing to confirm.

Step 4 — use the third equation as the check, not as a second unknown. The $x$ -coefficients must satisfy $5k - 2 = 13$ ; with $k = 3$ , $15 - 2 = 13$ ✓. Every coefficient agrees, so the identity holds.

Check. Substitute $x = 2$ into both sides with $k = 3$ : $(10-2)(2+3) = 40$ , and $5(4) + 13(2) - 6 = 20 + 26 - 6 = 40$ ✓. Answer: $k = 3$ .

Trap watch. Matching only the middle term and stopping — $5k - 2 = 13$ also gives 3 here, but on items where the two equations disagree, the constant term is the one the test builds the trap around, and a student who checked only one of them cannot tell which answer is real. The other classic is reading the question as “solve for $x$ ” and producing a value of $x$ that satisfies the equation at one point; an identity holds at every point, so a single $x$ proves nothing.

Practice

Answer before opening the explanation. Two items are student-produced response — no choices to reverse-engineer, which is exactly the format the live digital sets on this skill lean on. Every wrong choice below is one specific, named misconception: a lost cross term, a sign that stayed behind in the parentheses, a cancelled term that was never a factor. When you miss one, log the misconception, not the item number.

12 questions — 10 multiple choice, 2 student-produced response. Every wrong choice has its own explanation.

Question 1 Warm-up

Which of the following expressions is equivalent to (x + 5)²?

Show the answer Choice D

Why it is right

A squared binomial is a product, so write it as one before expanding: (x + 5)² = (x + 5)(x + 5). Multiplying every term of the first factor by every term of the second gives x·x = x², x·5 = 5x, 5·x = 5x, and 5·5 = 25. The two cross products are identical, so they combine into 10x, and the expansion is x² + 10x + 25 — the pattern (a + b)² = a² + 2ab + b².

Why each other choice fails

Choice A: Squares the two terms separately, as if (a + b)² were a² + b². That discards both cross products; at x = 1 the original is 36 while this expression is 26.
Choice B: Counts the cross product once instead of twice. Multiplying out produces 5x from x·5 and another 5x from 5·x, so the middle term is 10x.
Choice C: Gets the middle term right but adds the two 5s to build the constant instead of multiplying them. The constant term comes from 5·5 = 25.

Question 2 Standard

Which of the following expressions is equivalent to 9x² - 64?

Show the answer Choice B

Why it is right

Both terms are perfect squares and a minus sign separates them, so the difference-of-squares pattern a² - b² = (a + b)(a - b) applies. Here 9x² = (3x)², so a = 3x, and 64 = 8², so b = 8, giving (3x + 8)(3x - 8). Expanding confirms it: 9x² - 24x + 24x - 64 = 9x² - 64. The cross terms cancel precisely because one factor is a sum and the other is a difference.

Why each other choice fails

Choice A: Uses two identical factors, which is the perfect-square pattern rather than the difference of squares. It expands to 9x² - 48x + 64, a middle term and a sign the original does not have.
Choice C: Keeps the coefficient 9 instead of its square root. Since 9x² = (3x)², each factor must begin with 3x; this product expands to 81x² - 64.
Choice D: Takes the square root of the coefficient but not of the constant. Because 64 = 8², the factors need 8; this product expands to 9x² - 4096.

Question 3 Standard

Which of the following expressions is equivalent to -3x² + 12x - 15?

Show the answer Choice C

Why it is right

Factoring out -3 divides every term by -3, and dividing by a negative flips each sign: -3x² ÷ (-3) = x², 12x ÷ (-3) = -4x, and -15 ÷ (-3) = 5. The factored form is therefore -3(x² - 4x + 5). Distributing back is the check: -3·x² = -3x², -3·(-4x) = 12x, and -3·5 = -15, which rebuilds the original expression exactly.

Why each other choice fails

Choice A: Leaves the middle term positive inside the parentheses. Distributing gives -3x² - 12x - 15, so the x-term has the wrong sign; factoring out a negative changes the sign of every remaining term.
Choice B: Fixes the x-term but not the constant: -3 times -5 is +15, so this expands to -3x² + 12x + 15 instead of ending in -15.
Choice D: Pulls out +3 while copying the inside of the correct answer, which negates every term. It expands to 3x² - 12x + 15, the opposite of the given expression.

Question 4 Standard Student-produced response

The expression (2x + 5)(3x - 4) is equivalent to 6x² + bx - 20, where b is a constant. What is the value of b?

Your answer

Show the answer 7

Why it is right

Expand the product term by term: 2x·3x = 6x², 2x·(-4) = -8x, 5·3x = 15x, and 5·(-4) = -20. The x-coefficient is built from the two cross products, -8x + 15x = 7x, so the product is 6x² + 7x - 20. Comparing that with 6x² + bx - 20 gives b = 7, and the matching constant term, -20, confirms the expansion was carried out correctly.

Answers students type instead

1: Adds the two constants, 5 + (-4) = 1, instead of forming cross products. Two constants multiplied together can only produce the constant term, never the x-term.
23: Adds the sizes of the two cross products and ignores that one is negative: 8 + 15 = 23. Only 15x is positive; the other product is -8x, so they partly cancel.
-7: Reverses the signs of both cross products, computing +8x - 15x = -7x. The -4 sits in the second factor, so it multiplies 2x to give -8x and multiplies 5 to give -20.

Question 5 Standard

Which of the following expressions is equivalent to (x - 3)(x² + 2x - 5)?

Show the answer Choice A

Why it is right

Multiply every term of the binomial by every term of the trinomial. First x·(x² + 2x - 5) = x³ + 2x² - 5x. Then -3·(x² + 2x - 5) = -3x² - 6x + 15, where the -3 changes the sign of all three terms it meets. Adding the two results and collecting like terms gives x³ + (2x² - 3x²) + (-5x - 6x) + 15 = x³ - x² - 11x + 15.

Why each other choice fails

Choice B: Mishandles the final product: -3 times -5 is +15, not -15. Two negatives multiply to a positive, so the constant term of the expansion is positive.
Choice C: Takes -3 times 2x as +6x, which turns -5x - 6x into -5x + 6x = x. The -3 multiplies every term of the trinomial, so that product is -6x.
Choice D: Combines the x² terms in the wrong order, computing 3x² - 2x² = x² instead of 2x² - 3x² = -x². The 2x² comes from x·2x and the -3x² from -3·x².

Question 6 Harder

For x ≠ -3 and x ≠ -4, which of the following expressions is equivalent to (x² - 9)/(x² + 7x + 12)?

Show the answer Choice C

Why it is right

Factor both parts before cancelling anything. The numerator is a difference of squares: x² - 9 = (x - 3)(x + 3). The denominator factors as (x + 3)(x + 4), since 3 and 4 multiply to 12 and add to 7. Now (x + 3) is a whole factor of both, so it cancels, leaving (x - 3)/(x + 4). Checking at x = 1: the original is (1 - 9)/(1 + 7 + 12) = -8/20 = -2/5, and the answer is (1 - 3)/(1 + 4) = -2/5.

Why each other choice fails

Choice A: Cancels the x² terms, which are terms inside sums rather than factors of a product. Deleting part of a sum changes the value; only whole factors may be cancelled.
Choice B: Factors x² - 9 as (x + 3)(x + 3), a perfect square. A difference of squares needs one sum and one difference, so the factor surviving the cancellation is (x - 3).
Choice D: Factors the denominator as (x + 3)(x - 4), which expands to x² - x - 12 rather than x² + 7x + 12. Both constants must be positive to produce +12 and +7.

Question 7 Harder

For x > 0, which of the following expressions is equivalent to ∛(64x⁹)?

Show the answer Choice B

Why it is right

A cube root applies to the entire product, coefficient and variable alike. For the coefficient, 4³ = 64, so the cube root of 64 is 4. For the variable, a cube root is the exponent 1/3, and (x⁹)^(1/3) = x^(9/3) = x³. The expression is therefore 4x³. Cubing the answer is the check: (4x³)³ = 4³·x⁹ = 64x⁹, the expression we started with.

Why each other choice fails

Choice A: Handles the coefficient correctly but subtracts the index from the exponent, 9 - 3 = 6, instead of dividing by it. Taking a cube root divides an exponent by 3, so x⁹ becomes x³.
Choice C: Takes the square root of 64 rather than the cube root. 8² = 64 but 8³ = 512, and the number whose cube is 64 is 4.
Choice D: Roots the variable but leaves the coefficient untouched. The radical covers the whole product, so 64 must also be reduced, to its cube root 4.

Question 8 Harder

The equation 3x² + kx + 20 = (3x + a)(x + 4) is true for all values of x, where k and a are constants. What is the value of k?

Show the answer Choice D

Why it is right

Expand the right side: (3x + a)(x + 4) = 3x² + 12x + ax + 4a = 3x² + (12 + a)x + 4a. Because the equation holds for all x, the two sides are the same polynomial, so their coefficients match term by term. The constants give 4a = 20, so a = 5. The x-coefficients then give k = 12 + a = 17. Substituting back, (3x + 5)(x + 4) = 3x² + 17x + 20, which matches the left side exactly.

Why each other choice fails

Choice A: This is a, found correctly from the constant term, but the question asks for k, the coefficient of x. The two values come from different matched terms.
Choice B: Adds the constants appearing inside the factors, a + 4 = 5 + 4 = 9. The x-coefficient is built from two products, 3x·4 and a·x, not from adding constants.
Choice C: Counts only the product 3x·4 = 12x and misses the second cross product, a·x = 5x. Both contribute to the middle term, so k = 12 + 5.

Question 9 Harder

For x ≠ 2 and x ≠ -1, which of the following expressions is equivalent to 1/(x - 2) + 3/(x + 1)?

Show the answer Choice A

Why it is right

The common denominator is the product (x - 2)(x + 1). Rescale each numerator by whatever its own denominator was missing: 1/(x - 2) becomes (x + 1) over the common denominator, and 3/(x + 1) becomes 3(x - 2) over the same denominator. Adding the numerators gives (x + 1) + (3x - 6) = 4x - 5, so the sum is (4x - 5)/((x - 2)(x + 1)). At x = 0 both the original and the answer equal 5/2.

Why each other choice fails

Choice B: Distributes the 3 over (x - 2) as 3x + 6, giving x + 1 + 3x + 6 = 4x + 7. Three times -2 is -6, so the combined numerator is 4x - 5.
Choice C: Builds the common denominator correctly but adds the original numerators, 1 + 3, without rescaling them. Each numerator must be multiplied by the factor its own denominator was missing.
Choice D: Adds straight across, numerator to numerator and denominator to denominator. Fractions are never combined that way; the denominators combine by multiplication, not addition.

Question 10 Hardest

Which of the following expressions is equivalent to x² - 10x + 18 and displays the minimum value of the expression as a constant?

Show the answer Choice B

Why it is right

Complete the square. Half of -10 is -5, and (-5)² = 25, so x² - 10x + 25 = (x - 5)². Since only 18 is present, add and subtract 25: x² - 10x + 18 = (x² - 10x + 25) - 25 + 18 = (x - 5)² - 7. The rewrite displays the minimum because (x - 5)² is never negative and equals 0 at x = 5, so the smallest value of the whole expression is -7.

Why each other choice fails

Choice A: Adds 25 to build the perfect square but never subtracts it back, so this expression is 25 too large: at x = 0 it gives 43 while x² - 10x + 18 gives 18.
Choice C: Computes 25 - 18 = 7 instead of 18 - 25 = -7. The 25 was added inside the square, so it must be subtracted outside, leaving a negative constant.
Choice D: Flips the sign inside the parentheses. Expanding (x + 5)² produces +10x, but the original expression has -10x, so the binomial must be (x - 5).

Question 11 Hardest Student-produced response

For x ≠ -7/2, the expression (6x² + 19x - 7)/(2x + 7) is equivalent to ax + b, where a and b are constants. What is the value of a + b?

Your answer

Show the answer 2

Why it is right

The target form ax + b says the division comes out even, so factor the numerator with (2x + 7) as one of its factors. Testing that guess: (2x + 7)(3x - 1) = 6x² - 2x + 21x - 7 = 6x² + 19x - 7, which matches. Cancelling the common factor (2x + 7) leaves 3x - 1, so a = 3 and b = -1. The question asks for the combination, so a + b = 3 + (-1) = 2.

Answers students type instead

3: Stops at a, the coefficient of x. The question asks for a + b, so the constant term of 3x - 1 still has to be added.
4: Adds the size of b instead of b itself, 3 + 1. The simplified expression is 3x - 1, so its constant is negative and the sum is 3 - 1.
-1: Reports b, the constant term of the simplified expression, rather than the requested sum a + b.

Question 12 Hardest

Which of the following expressions is equivalent to (2x + 3)² - (x - 1)²?

Show the answer Choice C

Why it is right

Expand each square in full: (2x + 3)² = 4x² + 12x + 9 and (x - 1)² = x² - 2x + 1. Subtracting the second square means subtracting all three of its terms: 4x² + 12x + 9 - x² + 2x - 1 = 3x² + 14x + 8. The difference-of-squares route confirms it: [(2x + 3) + (x - 1)]·[(2x + 3) - (x - 1)] = (3x + 2)(x + 4) = 3x² + 14x + 8.

Why each other choice fails

Choice A: Squares each binomial term by term, treating (2x + 3)² as 4x² + 9 and (x - 1)² as x² + 1. Both squares have middle terms, and 12x and -2x do not cancel.
Choice B: Subtracts the second middle term with the wrong sign, computing 12x - 2x rather than 12x - (-2x). Subtracting a negative adds, so the middle term is 14x.
Choice D: Applies the minus sign to the x² and x terms of the second square but not to its constant, giving 9 + 1 = 10 instead of 9 - 1 = 8.

Common mistakes

Squaring term by term — $(x+5)^2$ written as $x^2 + 25$ , or $(2x-5)^2$ as $4x^2 + 25$ . A squared binomial is a product; expand it as one and the middle term appears on its own.
Confusing the square with the difference of squares — $(a-b)^2$ is $a^2 - 2ab + b^2$ ; $a^2 - b^2$ is $(a+b)(a-b)$ . One has a middle term and two identical factors, the other has no middle term and two different factors.
Forgetting the square root of a coefficient — factoring $9x^2 - 64$ as $(9x+8)(9x-8)$ . The pattern needs $(3x)^2$ , so the factors start with $3x$ .
A negative that stays in the parentheses — pulling $-3$ out of $-3x^2 + 12x - 15$ and writing $-3(x^2 + 4x + 5)$ . Factoring out a negative flips the sign of every remaining term, not just the first.
Cancelling terms instead of factors — striking the $x^2$ on top against the $x^2$ below, or a 9 against a 12. Cancellation is only legal on whole factors of a product.
Dropping the domain restriction — a cancelled $(x+3)$ still forbids $x = -3$ . The rewrite is equal to the original everywhere the original exists, and nowhere else.
Adding fractions straight across — $\frac{1}{x-2} + \frac{3}{x+1}$ turned into $\frac{4}{2x-1}$ . Denominators combine by multiplication, and each numerator has to be rescaled by whatever its denominator was missing.
Matching coefficients but skipping the constant — solving from the $x$ -term alone and never confirming the number at the end, which is where the test hides the disagreement.
Answering the wrong piece of an identity — reporting $a$ when the question asked for $a+b$ , or $k$ when it asked for $k^2$ . Reread the last six words before entering anything.
Rooting only part of a product — $\sqrt[3]{64x^9}$ treated as $64x^3$ or $8x^3$ . The root applies to the coefficient and the variable alike, and index 3 means divide the exponent by 3, not subtract 3.
Rewriting past the target — factoring an expression the question wanted expanded, or fully factoring when it asked which form displays the minimum. A correct rewrite that is not the requested one earns nothing.

FAQ

Can I just plug in a number and test the choices? Yes, and on multiple choice it is often the fastest safe route: substitute a single value into the stem and into each choice, and discard everything that disagrees. Two rules make it reliable. Avoid $x = 0$ and $x = 1$ , which collapse too many different expressions onto the same value, and avoid any value that makes a denominator zero. If two choices survive, run a second value — that resolves it. On grid-ins there is nothing to test, so the algebra has to be right.

How do I know whether to factor or expand? Look at the choices, not the stem. If the choices are factored and the stem is not, expand the choices until one matches. If the choices are expanded, factor the stem. The test writes the choices in the destination form, so they tell you which direction to travel.

What if I factor correctly and my answer still isn’t there? Check whether the question asked for a complete factorization or a particular form, and check for a sign pulled out front: $(3-x)$ and $(x-3)$ differ by a factor of $-1$ , and a choice may carry that $-1$ somewhere else. Expanding your answer and expanding the closest choice settles it in ten seconds.

Do I need to know completing the square for this skill? You need the version used here, which is mechanical: halve the $x$ -coefficient, square it, add and subtract it. For $x^2 - 10x + 18$ , half of $-10$ is $-5$ , squared is 25, so $x^2 - 10x + 25 - 25 + 18 = (x-5)^2 - 7$ . The heavier version — with a leading coefficient other than 1, aimed at a vertex — lives on the quadratic functions page.

Why do the questions keep saying “where a and b are constants”? It is the test’s way of saying: the answer is a number, and the letters are placeholders in a form you must produce. It also signals an identity — the two sides agree for every $x$ — which is your cue to match coefficients rather than to solve.

Is a sum of squares ever factorable? Not with real numbers. $x^2 + 16$ has no real factorization, and a choice offering $(x+4)(x-4)$ for it is testing exactly this. Only a difference of squares splits.

Grid-in format questions? Enter the exact value. Negatives are allowed and common on this skill; a fraction such as 7/2 should be entered as a fraction rather than rounded. If the question asks for a combination like $a+b$ , enter the combination — the individual values are the most frequent wrong entry on identity items.

Coming next in this domain: polynomial operations, rational equations and expressions, radicals and rational exponents.

Editorial team · last reviewed 2026-08-06

Equivalent expressions (nonlinear)

On the test

What this page owns

Pattern recognition

Method

Worked example 1 — the middle term is the question

Worked example 2 — factor first, cancel second

Worked example 3 — an identity is three tiny equations

Practice

Question 1 Warm-up

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Question 2 Standard

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Question 3 Standard

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Question 4 Standard Student-produced response

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Question 5 Standard

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Question 6 Harder

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Question 7 Harder

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Question 8 Harder

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Question 9 Harder

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Question 10 Hardest

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Question 11 Hardest Student-produced response

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Question 12 Hardest

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Common mistakes

FAQ

Related skills