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Algebra II

Common Core · 4 units · high school

Algebra II: every unit as Common Core publishes it, with practice questions per unit and a countdown to the exam.

Set up Algebra II → One set per unit, 4 of them pre-written and answer-checked. Free, no account needed.

Units

Unit 1: Polynomial, Rational, and Radical Relationships

Complex numbers, polynomial identities and arithmetic, the remainder theorem, zeros and graphs of polynomials, rational expressions, radical and rational equations, systems.

3 of 14 questions in this unit, with the worked answer

Simplify $(3+2i)(1-4i)$ and write the result in the form a+bi.

Answer. 11-10i

Distribute: $3(1)+3(-4i)+2i(1)+2i(-4i)=3-12i+2i-8i^2$. Since $i^2=-1$, $-8i^2=8$, giving $11-10i$.

Solve $x^2+4x+13=0$ over the complex numbers.

Answer. x = -2+3i or x = -2-3i

By the quadratic formula, the discriminant is $16-52=-36$, so $\sqrt{-36}=6i$. Then $x=\dfrac{-4\pm6i}{2}=-2\pm3i$.

Use synthetic division to divide $x^3-4x^2+x+6$ by $x-3$, and use the Remainder Theorem to state the remainder.

Answer. Quotient x^2-x-2, remainder 0

Synthetic division with c=3 on coefficients 1,-4,1,6 gives 1, -1, -2, and a final 0. The remainder 0 equals p(3), confirming x-3 is a factor.

All 14 questions for Unit 1, with the working →

What people get wrong

  1. Squaring both sides of a radical equation and stopping there — this can introduce extraneous roots. Always substitute each candidate solution back into the original (unsquared) equation.
  2. Treating i like an ordinary variable when multiplying complex numbers and forgetting that $i^2=-1$. Simplify every $i^2$ term you produce before combining like terms.
  3. Setting up synthetic division with the wrong sign — dividing by $(x-c)$ uses c, but dividing by $(x+c)$ requires using $-c$. Rewrite the divisor in $(x-c)$ form first.
  4. Canceling a common factor in a rational expression without noting that the canceled factor's zero is still excluded from the domain. State the restriction before or as you simplify.

Unit 2: Trigonometric Functions

Radian measure, the unit circle, trigonometric functions of any angle, graphing sine and cosine, modelling periodic phenomena, the Pythagorean identity.

3 of 13 questions in this unit, with the worked answer

Convert $\dfrac{5\pi}{6}$ radians to degrees.

Answer. 150°

Multiply by $\dfrac{180}{\pi}$: $\dfrac{5\pi}{6}\times\dfrac{180}{\pi} = \dfrac{5\times180}{6} = 150°$.

Convert 315° to radians, expressed as a fraction of $\pi$.

Answer. $\dfrac{7\pi}{4}$

Multiply by $\dfrac{\pi}{180}$: $315\times\dfrac{\pi}{180} = \dfrac{315\pi}{180} = \dfrac{7\pi}{4}$.

Find a positive angle coterminal with $-50°$.

Answer. 310°

Add 360° to get a coterminal angle in the standard range: $-50°+360° = 310°$.

All 13 questions for Unit 2, with the working →

What people get wrong

  1. Leaving the calculator in the wrong angle mode. If the problem is in radians and the calculator is set to degrees (or vice versa), every trig value comes out wrong even though the keystrokes are correct — always check the mode before computing.
  2. Reporting the reference angle as if it were the final answer. The reference angle only gives the magnitude of the trig value; you must still attach the correct sign using the quadrant (ASTC) before the answer is complete.
  3. Using the identity $\sin^2\theta+\cos^2\theta=1$ and forgetting the ± choice. Solving for $\sin\theta$ or $\cos\theta$ from the identity gives a square root, which is only correct once you've picked the sign that matches the given quadrant.
  4. Confusing amplitude with the maximum value, or midline with zero. Amplitude is half the distance between max and min, and midline is the average of max and min — read both off the graph or context before writing the equation, don't assume midline is always 0.

Unit 3: Modeling with Functions

Creating equations, interpreting and analysing functions, building functions from context, inverses, logarithms and exponential equations, comparing models.

3 of 14 questions in this unit, with the worked answer

A moving company charges a flat fee of $80 plus $35 per hour of labor. Write a function C(h) for the total cost of h hours of work, and state a reasonable domain for this context.

Answer. C(h) = 80 + 35h, with domain h ≥ 0.

The flat fee is the constant term and the hourly rate is the constant rate of change, giving a linear model in slope-intercept form. Since negative hours don't make sense, the domain is restricted to non-negative real numbers.

A ball is thrown upward so that its height in feet after t seconds is modeled by h(t) = -16t^2 + 96t + 4. Find the maximum height the ball reaches and the time at which it occurs.

Answer. Maximum height is 148 feet at t = 3 seconds.

The vertex of a downward parabola gives the maximum, located at $t=-\dfrac{b}{2a}=-\dfrac{96}{2(-16)}=3$. Substituting back gives $h(3)=-16(9)+96(3)+4=148$.

For f(x) = x^2 - 3x + 2, find the average rate of change of f over the interval [1,4].

Answer. 2

Average rate of change is $\dfrac{f(4)-f(1)}{4-1}$. Since $f(4)=6$ and $f(1)=0$, this equals $\dfrac{6-0}{3}=2$.

All 14 questions for Unit 3, with the working →

What people get wrong

  1. Writing a domain that matches the algebra but not the real situation (e.g., allowing negative time or negative quantity). Instead, always restrict the domain to values that make sense in context, such as $t\ge0$.
  2. Making an arithmetic slip while isolating y when finding an inverse, then not catching it. Instead, verify every inverse by composing $f(f^{-1}(x))$ and confirming it simplifies to x.
  3. Treating $\log(a+b)$ as $\log a+\log b$. The addition rule for logs only applies to multiplication inside the log, i.e., $\log(ab)=\log a+\log b$; there is no rule for the log of a sum.
  4. Accepting every algebraic solution to a log equation without checking that each log's argument is positive. A root that makes an argument zero or negative must be rejected as extraneous.

Unit 4: Inferences and Conclusions from Data

Normal distributions, sampling and study design, estimating population parameters, margin of error, simulation, comparing treatments, evaluating reports.

3 of 14 questions in this unit, with the worked answer

Scores on a standardized test are normally distributed with mean 500 and standard deviation 100. What percentage of test-takers scored between 400 and 600?

Answer. About 68%

400 and 600 are exactly 1 standard deviation below and above the mean of 500. The empirical rule states that about 68% of normally distributed data falls within 1 standard deviation of the mean.

Using the same test (mean 500, standard deviation 100), what is the z-score for a student who scored 650?

Answer. z = 1.5

$z=\dfrac{x-\mu}{\sigma}=\dfrac{650-500}{100}=1.5$. This tells you the score is 1.5 standard deviations above the mean.

About what percentage of test-takers scored above 700 on the same test (mean 500, SD 100)?

Answer. About 2.5%

700 is exactly 2 standard deviations above the mean. By the empirical rule, 95% of scores fall within 2 SDs, leaving 5% in the two tails combined; by symmetry, 2.5% lies above 700.

All 14 questions for Unit 4, with the working →

What people get wrong

  1. Applying the empirical rule to find a percentage at a boundary that isn't exactly 1, 2, or 3 standard deviations from the mean — instead convert to a z-score and use the standard normal table or software for exact boundaries.
  2. Concluding that an observational study 'proves' one variable causes another — instead remember that only a randomized controlled experiment supports a causal claim; an observational study can only show association.
  3. Treating margin of error as if it fixes bias — instead recognize that MOE only measures variability from random sampling; a biased sampling method (e.g., a self-selected online poll) produces a wrong estimate no matter how small the MOE is.
  4. Confusing random sampling with random assignment — instead ask separately 'were subjects chosen randomly from the population?' (affects generalizability) and 'were subjects randomly placed into groups?' (affects causal claims).

How it works

Choosing this curriculum creates a course in Chalkset with one study set per unit and the exam date set to the start of the next testing window. Each unit is drilled card by card, and the review queue brings back what you miss. Pre-written units come with worked explanations and an independently checked answer key; the rest are written by the agent when you open them.

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