NY Regents Algebra I — August 2025: Full Walkthrough of All 35 Questions

Answer: (3) (10x − 4)(10x + 4)

This is a difference of squares: a² − b² = (a − b)(a + b).

Rewrite: 100x² = (10x)² and 16 = 4², so 100x² − 16 = (10x)² − 4² = (10x − 4)(10x + 4).

Choice (1) uses 50 instead of 10 for the coefficient — a common slip since 50² = 2500 ≠ 100. Choices (2) and (4) repeat the same sign, which would expand to a perfect square trinomial, not a difference of squares.

Answer: (3) 0.25x + 0.10(x + 2) = 2.30

Define the variables: x = number of quarters. Since she has two more dimes than quarters, the number of dimes is x + 2.

Set up value equation: (value of quarters) + (value of dimes) = total value, so 0.25x + 0.10(x + 2) = 2.30.

Choice (4) swaps the sign, which would mean fewer dimes than quarters. Choice (2) attaches the "+2" to the quarters instead of the dimes.

Answer: (2) −2

Substitute x = −3:

Watch order of operations: square −3 first to get 9, then multiply by −2. Squaring before applying the negative sign is the step most students rush through.

Answer: (2) 10 and −2

Factor: find two numbers that multiply to −20 and add to −8 — that’s −10 and 2.

The zeros are x = 10 and x = −2.

Answer: (1) (−2, 15.5)

Substitute x = −2:

This matches (−2, 15.5). Choice (4) has the same y-value but the wrong sign on x — testing it gives 3(4) − 0.5 + 3 = 14.5, not 15.5, since the middle term’s sign flips with x.

Answer: (3) 3 and 5

Set the functions equal and solve:

Both values make the two functions intersect, so the graphs meet at x = 3 and x = 5.

Answer: (4) 4x⁵ − 8x² + 5

Standard form means the terms are written in descending order of degree. Constant term is the term with no variable.

Check each choice: (1) has constant term 10, not 5. (2) is written with 2x² before 6x⁴ — degrees are out of order (ascending, not descending). (3) has constant term 1. (4) is in descending order (x⁵ then x² then constant) and ends in 5. ✓

Answer: (4) addition property of equality

What happened: the student added 8 to both sides of the equation — the left side went from x² + 6x to x² + 6x + 8, and the right side went from −8 to 0.

Adding the same quantity to both sides of an equation is exactly what the addition property of equality allows — it keeps the equation balanced. Associative and commutative properties only rearrange or regroup terms; they don’t add anything new to an equation.

Answer: (4) j(x)

Test for a constant first difference — that’s the signature of a linear function.

f(x) has a constant second difference (it’s quadratic: 6,1,−2,−3,−2,1 — a parabola shape). g(x) decreases then increases symmetrically, like |x| — absolute value, not linear. h(x) doubles each step (−1,−2,−4,−8,−16,−32) — exponential.

Answer: (2) {0, 8}

Move everything to one side — do NOT divide both sides by x, since that discards the x = 0 solution.

The solution set is {0, 8}. Choice (1) is the classic trap from dividing both sides by x too early, which silently loses the root x = 0.

Answer: (2) −3.75

Average rate of change = (change in output) / (change in input), using the endpoints of the interval t = 1 and t = 3.

The negative sign makes sense — radioactivity decays over time, so the rate of change is negative.

Answer: (1) 0.984 and strong

On a TI-84: enter day numbers in L1, minutes in L2, turn on DiagnosticOn, then run STAT → CALC → 4: LinReg(ax+b).

Calculator output: r ≈ 0.984. Since |r| is very close to 1, the linear fit is strong.

Answer: (4) k(x) = (x + 3)²

Horizontal shift rule: f(x + h) shifts left by h (opposite of what most students expect); f(x − h) shifts right.

Shifting f(x) = x² left 3 units replaces x with x + 3, giving k(x) = (x + 3)². Choices (1) and (2) are vertical shifts, not horizontal.

Answer: (2) The median and the mode are the same.

Mean: add up (number of pets × frequency) and divide by 20.

Median: with 20 ordered values, average the 10th and 11th. Counting cumulatively: 1 zero, then 4 ones bringing us to position 5, then 6 twos filling positions 6 through 11. Both the 10th and 11th values are 2, so the median is 2.

Mode: the most frequent value is 2 pets (6 students — the tallest column), so the mode is 2.

Median (2) and mode (2) match, while the mean (2.5) does not. Answer: (2).

Answer: (1) y ≥ −5

Range describes the set of possible y-values (output), which immediately eliminates choices (3) and (4) — those describe x-values.

The absolute value |x + 2| can never be negative; its smallest possible value is 0, achieved at x = −2. So the minimum of f(x) = |x+2| − 5 is 0 − 5 = −5, and f(x) can grow without bound above that. Range: y ≥ −5.

Answer: (3) (ab)^x = a^xb^x

Exponent rule: a power distributes over a product: (ab)ⁿ = aⁿbⁿ. This is always true, no matter what a, b, or x are.

Choice (1) confuses the product rule (which adds exponents when bases match: a³ · a^x = a^3+x) with multiplying them. Choice (2) confuses the power rule (which multiplies exponents: (a⁴)^x = a^4x) with adding them. Choice (4) has mismatched bases, so there is no valid combination rule.

Answer: (1) 2A / (b₁ + b₂)

Isolate h by undoing the operations attached to it:

Choice (3) inverts the correct fraction; choice (2) mistakenly multiplies instead of dividing.

Answer: (1) f(x) and g(x)

Compute each y-intercept by evaluating at x = 0 (or reading the table at x = 0):

f(0) = g(0) = 5, while h(0) = −3 is different. So f(x) and g(x) share the same y-intercept.

Answer: (3) 2x² + 8x + 58

Expand each square then add:

A common error is forgetting the middle term entirely and just adding x² + 49 and x² + 9, which gives the wrong choice (1).

Answer: (1) 12√5

Multiply coefficients and radicals separately:

The key step often missed is simplifying √20 down to 2√5 before finalizing the answer.

Answer: (3) −x³ + 5x − 12

Rewrite carefully: “A is subtracted from B” means B − A. Here B = 5x³ + 3x − 4, A = 6x³ − 2x + 8.

The most common error is forgetting to flip all three signs inside the second polynomial when distributing the subtraction.

Answer: (4) I, II, and III

Function rule: every input maps to exactly one output (no input repeats with two different outputs).

Relation I: each of the four x-values (0, 1, 2, 3) is distinct, so it’s a function. Relation II: the arrow diagram sends each of 3, 4, 5, 6 to exactly one output, so it’s a function (in this case, the identity mapping). Relation III’s graph passes the vertical line test (no vertical line crosses it more than once), so it is a function too. All three qualify.

Answer: (1) x = y − 1

Substitution starts by isolating one variable in the simpler equation. Here, x − y = −1 is the simpler (linear, one-step) equation.

Choice (2) accidentally isolates y incorrectly — adding y and subtracting 1 from x − y = −1 gives x = y − 1, not y = x − 1. Choices (3) and (4) are linear combinations that don’t isolate a single variable.

Answer: (4) (100 m/9.58 sec) · (60 sec/1 min) · (1 km/1000 m) · (60 min/1 hr)

Start with the rate in meters per second: 100 m / 9.58 sec. We need to convert this to kilometers per hour.

Chain the conversions so units cancel diagonally: seconds → minutes → hours (multiply by 60 sec/1 min, then 60 min/1 hr, both flipped to cancel seconds and minutes from the denominator), and meters → kilometers (multiply by 1 km/1000 m).

Choice (3) has the seconds-to-minutes and minutes-to-hours ratios upside down, which would shrink instead of grow the value as needed to convert small units to larger ones in the denominator.

Answer: x = 10.5

Clearing the fraction first (multiplying by 6) avoids working with fractions throughout the rest of the problem.

Answer: Irrational.

Reasoning: both 3√2 and √5 are irrational numbers on their own (neither 2 nor 5 is a perfect square, so their square roots never terminate or repeat). They are also “unlike” radicals — there is no common factor that lets them combine or cancel into something rational.

Contradiction argument: suppose 3√2 + √5 = r for some rational number r. Then √5 = r − 3√2. Squaring both sides gives 5 = r² − 6r√2 + 18, which forces 6r√2 to equal the rational quantity r² + 13. That’s only possible if r = 0, but then 3√2 = −√5, which is impossible since one side is positive and the other negative. So no rational r works — the sum is irrational.

Answer: V-shaped graph with vertex at (2, 0), opening upward, passing through (−4, 6) and (4, 2).

Strategy: the parent function |x| is a V with vertex at the origin. Replacing x with x − 2 shifts the whole V right by 2, so the new vertex is (2, 0). The two arms have slopes ±1.

Sample points across the domain:

Plot these five points — (−4,6), (−2,4), (0,2), (2,0), (4,2) — and connect them with two straight rays meeting at the vertex (2, 0).

Answer: Adults — Brand A 40, Brand B 59, Total 99. Teenagers — Brand A 32, Brand B 49, Total 81. Column totals: Brand A 72, Brand B 108, grand total 180.

Step 1 — total Brand A: 40% of 180.

Step 2 — split Brand A between adults and teenagers. Teenagers who own Brand A = 32, so:

Step 3 — find adult and teenager totals. Adults own 40 (A) + 59 (B):

Completed table:

Answer: 10,935

Apply a_n = a₁ · rⁿ⁻¹ with a₁ = 5, r = 3, n = 8:

Remember the exponent is n − 1, not n — a very common off-by-one error on this formula.

Answer: (x + 7)² = 77 — i.e. p = −7 and q = 77.

Written as (x − p)² = q, this is (x − (−7))² = 77, so p = −7 and q = 77.

Answer: Vertex (0, 4). Axis of symmetry: x = 0.

Identify the vertex directly: the function is already in vertex form f(x) = a(x − h)² + k with h = 0 and k = 4, so the vertex is (0, 4). Since a = −⅓ is negative, the parabola opens downward.

Axis of symmetry is the vertical line through the vertex: x = 0 (the y-axis).

Sample points to plot the graph (using symmetric x-values on either side of the vertex):

To sketch: plot (−6,−8), (−3,1), (0,4) as the peak, (3,1), and (6,−8), then connect them with a smooth downward-opening curve that is symmetric about the y-axis. The parabola is fairly wide and flat near the vertex because |a| = ⅓ is small, and it falls steeply as |x| grows.

Answer: Inequality 18 + 7.50x ≤ 78. Maximum total rental time: 9 hours (1 first hour + 8 additional hours).

Set up the inequality: the flat $18 covers the first hour, and $7.50 is charged per additional hour x. Total cost must be at most $78:

Solve for x:

x represents additional hours beyond the first, so the maximum x = 8. The total rental time is 1 + 8 = 9 hours.

Answer: any point in the overlap region works — e.g. (0, 0).

Rewrite each in slope-intercept form:

Describing the graph: the first boundary is a solid line with y-intercept −3 and slope ½, shaded above (since y ≥). The second boundary is a dashed line with y-intercept 5 and slope 2, shaded below (since y <). The solution region is the overlap of “above the shallow solid line” and “below the steep dashed line” — a wedge-shaped region that opens toward the lower-right, bounded on the left where the two lines cross.

Test the point (0, 0):

Both are satisfied, so (0, 0) lies in the solution region common to both inequalities.

Answer: x = 3 + √6 or x = 3 − √6

Identify: a = 1, b = −6, c = 3.

Quadratic formula: x = (−b ± √(b² − 4ac)) / (2a).

Dividing every term in the numerator by 2 (not just part of it) is the step students most often skip, leaving an unsimplified fraction like (6 ± 2√6)/2 instead of the fully reduced 3 ± √6.

Answer: (a) System: x + y = 25 and 2.25x + 1.50y = 45. (b) Cameron sold 10 hot dogs and 15 sodas. (c) Maximum of 6 hot dogs.

(a) Set up the system. One equation counts items, the other counts dollars:

(b) Solve by substitution. From the first equation, y = 25 − x. Substitute into the second:

So Cameron sold 10 hot dogs and 15 sodas. Check: 10 + 15 = 25 items ✓, and 2.25(10) + 1.50(15) = 22.50 + 22.50 = 45 ✓.

(c) New scenario, different customer, $20 budget, exactly 4 sodas. First find how much the 4 sodas cost, then see how many hot dogs the remaining money buys:

Since you can’t buy a fraction of a hot dog, round down to the nearest whole number: maximum 6 hot dogs. Check: 6($2.25) + 4($1.50) = $13.50 + $6.00 = $19.50 ≤ $20 ✓ (7 hot dogs would cost $15.75 + $6.00 = $21.75, over budget).