How This Exam Is Built
The Regents Examination in Algebra II (Next Generation) was first given in June 2026. It has 36 questions worth a total of 82 raw credits, and you have 3 hours to finish.
- Part I: 24 multiple-choice questions, 2 credits each, for 48 credits. There is no partial credit.
- Part II: 8 constructed-response questions, 2 credits each, for 16 credits.
- Part III: 3 constructed-response questions, 4 credits each, for 12 credits.
- Part IV: 1 constructed-response question worth 6 credits.
In Parts II, III, and IV, show the necessary steps, including formula substitutions, graphs, and charts. When a question tells you to solve algebraically, a graph or a list of guesses alone will not earn full credit. Write in pen, except graphs and drawings, which are done in pencil.
A graphing calculator is required for the whole exam, and calculators with symbol manipulation are not allowed. A straightedge must be available. The reference sheet in the exam booklet lists the quadratic formula, the sequence and series formulas, the sum and difference of cubes, the trigonometric identities, the probability rules, and the exponential growth and decay formulas, so learn how to use them rather than memorizing them.
Your raw score converts to a scale score from 0 to 100. A 65 is passing and an 85 is mastery. On the August 2026 chart, 30 raw credits converted to 65 and 66 converted to 85, and on the June 2026 chart the numbers were 30 and 64. The official blueprint puts 38 to 45 percent of the credits in Functions, 30 to 39 percent in Algebra, 14 to 22 percent in Statistics and Probability, and 4 to 8 percent in Number and Quantity. The sections below follow those categories.
Number and Quantity
Rational Exponents and Radicals
A rational exponent combines a power and a root: the denominator is the root and the numerator is the power, so \(x^{\frac{m}{n}} = \sqrt[n]{x^{m}}\). A negative exponent means a reciprocal. All the usual exponent rules still apply: multiply exponents when you raise a power to a power, and add them when you multiply powers with the same base.
Worked example. Simplify \(\left(8x^{6}\right)^{\frac{2}{3}}\) for \(x > 0\). The number: \(8^{\frac{2}{3}} = \left(\sqrt[3]{8}\right)^{2} = 2^{2} = 4\). The variable: \(\left(x^{6}\right)^{\frac{2}{3}} = x^{4}\). The result is \(4x^{4}\).
Common mistake. Multiplying the coefficient by the exponent, as in \(8 \cdot \frac{2}{3}\). The exponent applies to every factor inside the parentheses, so the number must be raised to the power too.
Practice next. Practice Test 1, Question 1, and Sample Question 9.
Complex Numbers
The imaginary unit \(i\) is defined by \(i^{2} = -1\), so \(\sqrt{-k} = i\sqrt{k}\) for \(k > 0\). Add and subtract complex numbers by combining real parts and imaginary parts. Multiply like binomials, then replace \(i^{2}\) with \(-1\). Powers of \(i\) repeat in the cycle \(i, -1, -i, 1\).
Worked example. \((1 + 3i)(2 - i) = 2 - i + 6i - 3i^{2} = 2 + 5i + 3 = 5 + 5i\).
Common mistake. Leaving \(i^{2}\) in the answer or replacing it with \(+1\). Every \(i^{2}\) becomes \(-1\), which changes the real part of the product.
Practice next. Practice Test 1, Questions 5 and 25, and Sample Question 1.
Algebra
Polynomials: Factoring and the Remainder Theorem
Factor completely: look for a greatest common factor, then grouping, a difference of squares, a sum or difference of cubes, or a quadratic-form trinomial. The remainder theorem says that the remainder when \(p(x)\) is divided by \(x - a\) is \(p(a)\), and the factor theorem says \(x - a\) is a factor exactly when \(p(a) = 0\). Zeros of a polynomial are the values that make a factor zero.
Worked example. Factor \(x^{3} - 2x^{2} - 16x + 32\). Group: \(x^{2}(x - 2) - 16(x - 2) = (x - 2)(x^{2} - 16) = (x - 2)(x - 4)(x + 4)\). The zeros are 2, 4, and \(-4\). As a check with the remainder theorem, \(p(4) = 64 - 32 - 64 + 32 = 0\).
Common mistake. Stopping too early. In the example, \((x - 2)(x^{2} - 16)\) is not complete, because \(x^{2} - 16\) is a difference of squares. Another common slip is evaluating \(p(3)\) to test the factor \(x + 3\); use \(p(-3)\).
Practice next. Practice Test 1, Questions 2, 7, 16, and 24, and Sample Question 10.
Rational and Radical Equations
To solve a rational equation, note the values that make any denominator zero, then multiply every term by the least common denominator. To solve a radical equation, isolate the radical and raise both sides to the matching power. Both methods can produce extraneous solutions, so check every answer in the original equation.
Worked example. Solve \(\sqrt{x + 10} = x - 2\). Square both sides: \(x + 10 = x^{2} - 4x + 4\), so \(x^{2} - 5x - 6 = 0\) and \(x = 6\) or \(x = -1\). Check: \(\sqrt{16} = 4 = 6 - 2\), so 6 works. For \(-1\), \(\sqrt{9} = 3\) but \(-1 - 2 = -3\), so \(-1\) is extraneous. The solution is \(x = 6\).
Common mistake. Skipping the check. In a rational equation such as \(\frac{x}{x - 4} = \frac{4}{x - 4} + 2\), the algebra gives \(x = 4\), but 4 makes the denominators zero, so the equation has no solution at all.
Practice next. Practice Test 1, Questions 9, 26, 28, and 33.
Quadratics, Complex Solutions, and Systems
Solve quadratics by factoring, completing the square, the quadratic formula, or graphing. The discriminant \(b^{2} - 4ac\) tells you what kind of solutions to expect: a negative discriminant means two complex conjugate solutions. For a system with a line and a parabola or circle, substitute the linear equation into the other one and solve the quadratic that results.
Worked example. Solve \(x^{2} - 6x + 13 = 0\). The discriminant is \(36 - 52 = -16\), so the solutions are non-real: \(x = \frac{6 \pm \sqrt{-16}}{2} = \frac{6 \pm 4i}{2} = 3 \pm 2i\). For the system \(y = x^{2} + 1\) and \(y = x + 3\), set \(x^{2} + 1 = x + 3\), so \(x^{2} - x - 2 = 0\) and \(x = 2\) or \(x = -1\), giving the points \((2, 5)\) and \((-1, 2)\).
Common mistake. Dividing only part of the numerator by \(2a\), as in \(6 \pm \frac{4i}{2}\). Simplify the whole fraction, or split it as \(\frac{6}{2} \pm \frac{4i}{2}\).
Practice next. Practice Test 1, Questions 12, 20, 22, and 27.
Functions
Exponential and Logarithmic Functions
In \(y = a(b)^{t}\), \(a\) is the starting amount and \(b\) is the growth factor per time period. Growth means \(b > 1\) with rate \(b - 1\); decay means \(0 < b < 1\) with rate \(1 - b\). The reference sheet gives \(A = P\left(1 + \frac{r}{n}\right)^{nt}\), \(A = Pe^{rt}\), and the half-life model \(A = A_0\left(\frac{1}{2}\right)^{\frac{t}{h}}\). Logarithms undo exponents, so they are the tool for solving for a variable in an exponent.
Worked example. When will $2000 growing at 6% per year, compounded annually, reach $5000? Solve \(2000(1.06)^{t} = 5000\): \((1.06)^{t} = 2.5\), so \(t = \frac{\ln 2.5}{\ln 1.06} \approx 15.7\) years.
Common mistake. Dividing an annual rate by 12 to get a monthly rate in a growth model. Use exponent rules instead: the monthly factor for \(b^{t}\) is \(b^{\frac{1}{12}}\).
Practice next. Practice Test 1, Questions 6, 14, 19, 29, and 36, and Sample Questions 2 and 8.
Sequences and Series
An arithmetic sequence adds a common difference, \(a_n = a_1 + (n - 1)d\), and a geometric sequence multiplies by a common ratio, \(a_n = a_1 r^{n - 1}\). A recursive formula gives the first term and the rule from one term to the next. A finite geometric series adds the terms: \(S_n = \frac{a_1 - a_1 r^{n}}{1 - r}\), and sigma notation tells you where to start and stop.
Worked example. Find \(2 + 6 + 18 + \cdots\) for 7 terms. Here \(a_1 = 2\) and \(r = 3\), so \(S_7 = \frac{2 - 2(3)^{7}}{1 - 3} = \frac{2 - 4374}{-2} = 2186\).
Common mistake. Miscounting terms. \(\sum_{k=1}^{6}\) has 6 terms, but \(\sum_{k=0}^{6}\) has 7.
Practice next. Practice Test 1, Questions 3, 18, and 32, and Sample Question 3.
Transformations, Inverses, and Key Features
The graph of \(f(x - h) + k\) is the graph of \(f\) shifted \(h\) right and \(k\) up. Even functions are symmetric about the \(y\)-axis and odd functions about the origin. To find an inverse, swap \(x\) and \(y\) and solve for \(y\). Know how to read zeros, intercepts, end behavior, and intervals of increase from a graph or an equation, and how to find an average rate of change, \(\frac{f(b) - f(a)}{b - a}\).
Worked example. The inverse of \(f(x) = \frac{x + 4}{3}\): write \(x = \frac{y + 4}{3}\), so \(3x = y + 4\) and \(f^{-1}(x) = 3x - 4\). Check: \(f(2) = 2\) and \(f^{-1}(2) = 2\); \(f(5) = 3\) and \(f^{-1}(3) = 5\).
Common mistake. Reading a horizontal shift backward: \(f(x + 3)\) moves the graph 3 units left, not right.
Practice next. Practice Test 1, Questions 10, 13, and 21, and Sample Questions 4 and 5.
Trigonometric Functions
On the unit circle, an angle \(\theta\) in standard position ends at the point \((\cos\theta, \sin\theta)\). Learn the values at multiples of \(\frac{\pi}{6}\) and \(\frac{\pi}{4}\), and use the reference angle and the quadrant to get the sign. The Pythagorean identity, \(\sin^{2}\theta + \cos^{2}\theta = 1\), lets you find one value from another. In \(y = A\sin(B(x - h)) + k\), the amplitude is \(|A|\), the period is \(\frac{2\pi}{B}\), the horizontal shift is \(h\), and the midline is \(y = k\).
Worked example. If \(\sin\theta = -0.6\) and \(\theta\) is in Quadrant III, then \(\cos^{2}\theta = 1 - 0.36 = 0.64\), and cosine is negative in Quadrant III, so \(\cos\theta = -0.8\) and \(\tan\theta = \frac{-0.6}{-0.8} = 0.75\). For a model: daily temperatures that range from 60 to 84 degrees over a 24-hour cycle have midline \(\frac{60 + 84}{2} = 72\), amplitude 12, and \(B = \frac{2\pi}{24} = \frac{\pi}{12}\).
Common mistake. Using \(B\) as the period. The period is \(\frac{2\pi}{B}\), so a larger \(B\) means a shorter period.
Practice next. Practice Test 1, Questions 8, 15, 30, and 34.
Statistics and Probability
Normal Distributions and Regression
When data are roughly symmetric and mound-shaped, a normal model fits. Use the calculator's normal distribution feature with the mean and standard deviation to find the percent of data in an interval, and multiply by the population size to estimate a count. For bivariate data, run the regression the question names (linear, quadratic, exponential, or power), round as directed, and use the rounded equation to predict.
Worked example. Test scores with a mean of 500 and standard deviation 100: the percent between 450 and 650 is about 62.5%, from the area under the normal curve between those two values.
Common mistake. Using the 68, 95, and 99.7 percent rule for values that are not whole numbers of standard deviations from the mean. 650 is 1.5 standard deviations above 500, so use the calculator.
Practice next. Practice Test 1, Questions 11, 31, and 35, and Sample Question 6.
Study Design and Simulation
A survey asks questions, an observational study records data without imposing a treatment, and an experiment randomly assigns treatments, which is what allows a cause-and-effect conclusion. A simulation shows how much a sample statistic varies by chance. In this course, a result within 2 standard deviations of the simulation's mean (the middle 95%) is considered likely, and a result outside that interval is unlikely.
Worked example. A simulation of sample proportions has a mean of 0.55 and a standard deviation of 0.03. The likely range is \(0.55 \pm 0.06\), or 0.49 to 0.61. A sample proportion of 0.47 would be unusual, so it casts doubt on a true proportion of 0.55.
Common mistake. Concluding cause and effect from an observational study. Only random assignment of a treatment supports that kind of claim.
Practice next. Practice Test 1, Questions 4 and 23.
Probability
Use the addition rule, \(P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)\), and conditional probability, \(P(A \mid B) = \frac{P(A \text{ and } B)}{P(B)}\). Events are independent when \(P(A \text{ and } B) = P(A) \cdot P(B)\), or equivalently when \(P(A \mid B) = P(A)\). In a two-way table, a conditional probability uses only the row or column of the given condition.
Worked example. In a group of 80 students, 30 play an instrument and 20 of those also sing, while 25 sing in all. The probability that a randomly chosen singer plays an instrument is \(\frac{20}{25} = 0.8\), while \(P(\text{instrument}) = \frac{30}{80} = 0.375\). Because \(0.8 \ne 0.375\), the events are not independent.
Common mistake. Swapping the condition: \(P(A \mid B)\) and \(P(B \mid A)\) usually have different denominators and different values.
Practice next. Practice Test 1, Question 17, and Sample Question 7.