What does the Remainder Theorem say?

When a polynomial $P(x)$ is divided by $x - a$, the remainder equals $P(a)$. So you can find a remainder by substituting $a$ into the polynomial, with no division. The Factor Theorem is the zero case: $x - a$ is a factor of $P(x)$ if and only if $P(a) = 0$, meaning $a$ is a zero. Watch the sign: the factor $x + 2$ corresponds to $a = -2$.

How does multiplicity affect a polynomial graph?

At a zero with odd multiplicity (1, 3, ...), the graph crosses the x-axis. At a zero with even multiplicity (2, 4, ...), the graph touches the x-axis and turns around without crossing. The multiplicity is the exponent on that zero's factor, so $(x - 3)^2$ gives a touch-and-turn at $x = 3$ while $(x - 3)$ gives a crossing.

Why do rational and radical equations need a solution check?

Clearing denominators (rational equations) or squaring both sides (radical equations) can introduce extraneous solutions that satisfy the transformed equation but not the original. A rational solution that makes a denominator zero is rejected, and a radical solution that forces a square root to equal a negative is rejected. State restrictions first, then test every candidate in the original equation.

How do you convert a radical to a rational exponent?

Use $\sqrt[n]{x^m} = x^{m/n}$: the root index $n$ is the denominator and the power $m$ is the numerator. So a cube root of $x$ squared is $x^{2/3}$. Once in exponent form, the ordinary laws of exponents (add when multiplying, subtract when dividing, multiply when raising to a power) apply directly.

What does the discriminant tell you about a quadratic's roots?

The discriminant $b^2 - 4ac$ classifies the roots. Positive gives two distinct real roots; zero gives one repeated real root; negative gives two complex conjugate roots of the form $a \pm bi$. When it is negative, the quadratic formula still works, with the negative under the radical rewritten using $i = \sqrt{-1}$.

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NY Regents Algebra II: a complete guide to polynomials and rationals on the exam

A deep-dive NY Regents Algebra II guide to the polynomials-and-rationals strand. Covers polynomial division and the Remainder and Factor Theorems, zeros, multiplicity and end behavior, rational expressions and equations with extraneous solutions, radicals and rational exponents, and complex numbers with the discriminant, plus the credit-based exam technique the Regents rewards.

Generated by Claude Opus 4.816 min readA-APR, A-REI, N-RN, N-CNUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section

What this strand demands
Polynomial division and the theorems
Zeros, multiplicity, and end behavior
Rational expressions and equations
Radicals and rational exponents
Complex numbers and the discriminant
How this strand is examined
Check your knowledge

What this strand demands

The polynomials-and-rationals strand opens NY Regents Algebra II and supplies many of its constructed-response questions. It rewards fluent polynomial division and the Remainder and Factor Theorems, accurate graph reasoning from zeros and end behavior, careful rational-expression work with domain restrictions, confident handling of radicals and rational exponents, and operations with complex numbers. This guide ties together the dot-point pages, each with its own practice: polynomial arithmetic and the Remainder Theorem, polynomial zeros and end behavior, rational expressions and equations, radicals and rational exponents, and complex numbers and quadratics.

Polynomial division and the theorems

Divide polynomials by long division or, for a linear $x - a$ , synthetic division. The Remainder Theorem says the remainder on dividing by $x - a$ is $P(a)$ , and the Factor Theorem says $x - a$ is a factor exactly when $P(a) = 0$ . The workflow: find a value where $P(a) = 0$ , confirm the factor, divide to reduce the degree, and factor the rest.

Zeros, multiplicity, and end behavior

Zeros come straight from factored form. Odd multiplicity crosses the x-axis; even multiplicity touches and turns. End behavior is set by the leading term: even degree sends both ends the same way (sign of the leading coefficient), odd degree sends them opposite ways. A sketch uses local behavior (multiplicity) and global behavior (leading term) together.

Rational expressions and equations

Simplify by factoring and cancelling common factors (never terms), keeping the domain restrictions. Multiply across, divide by the reciprocal, add or subtract over a common denominator. Solve equations by clearing denominators, then reject any candidate that makes an original denominator zero.

Radicals and rational exponents

A radical is a fractional exponent: $\sqrt[n]{x^m} = x^{m/n}$ , index on the bottom. The exponent laws then apply to fractions. Solve radical equations by isolating the radical and raising to the matching power, then checking, because squaring can introduce extraneous solutions and the principal root is never negative.

Complex numbers and the discriminant

The imaginary unit is $i = \sqrt{-1}$ , so $i^2 = -1$ . Add and subtract by parts; multiply like binomials and replace $i^2$ with $-1$ . The discriminant $b^2 - 4ac$ classifies roots: positive (two real), zero (one repeated), negative (two complex conjugates). The quadratic formula handles a negative discriminant by rewriting the radical with $i$ .

How this strand is examined

Part I (2 credits). A remainder by the theorem, a factor test, multiplicity behavior, a rational simplification, a complex product, or a radical-to-exponent conversion.
Part II (2 credits). A factor-theorem justification, an end-behavior description, or a discriminant classification. Show the key reasoning.
Part III and IV (4 to 6 credits). A full factoring or zeros problem, a rational equation with restrictions and a check, or a radical equation with an extraneous-solution check. Show every step.

Check your knowledge

Work these as you would for credit.

Find the remainder when $P(x) = x^3 + 2x - 7$ is divided by $x - 2$ . (2 credits)
Is $x + 1$ a factor of $P(x) = x^3 + 1$ ? Justify. (2 credits)
State the behavior of $f(x) = (x - 1)^2(x + 4)$ at each x-intercept. (2 credits)
Describe the end behavior of $g(x) = -x^5 + 2x$ . (2 credits)
Simplify $\frac{x^2 - 16}{x^2 + 8x + 16}$ with restrictions. (2 credits)
Solve $\frac{3}{x} + \frac{1}{2} = \frac{5}{x}$ , checking restrictions. (4 credits)
Solve $\sqrt{3x + 1} = x - 1$ , checking for extraneous solutions. (4 credits)
Multiply $(2 - 3i)(1 + i)$ and write in $a + bi$ form. (2 credits)

Solutions

Step 1: Apply the Remainder Theorem (Q1)

The Remainder Theorem says that when you divide $P(x)$ by $x - a$ , the remainder equals $P(a)$ . The divisor here is $x - 2$ , so $a = 2$ , not $-2$ . Substitute $x = 2$ directly into the polynomial to find the remainder without performing any long division.

P(2) = 2^3 + 2(2) - 7 = 8 + 4 - 7 = 5

Final answer: The remainder is 5.

Step 2: Use the Factor Theorem to test the factor (Q2)

The Factor Theorem is the zero case of the Remainder Theorem: $x - a$ is a factor of $P(x)$ if and only if $P(a) = 0$ . The candidate factor is $x + 1$ , which equals $x - (-1)$ , so $a = -1$ . Evaluate $P(-1)$ and check whether the result is zero.

P(-1) = (-1)^3 + 1 = -1 + 1 = 0

Because $P(-1) = 0$ , the Factor Theorem guarantees that $x + 1$ is a factor of $P(x)$ .

Final answer: Yes, $x + 1$ is a factor.

Step 3: Read multiplicity from each factor to determine crossing or touching (Q3)

The behavior at each x-intercept is controlled by the multiplicity of the corresponding factor. Even multiplicity means the graph touches the x-axis and turns back without crossing; odd multiplicity means it crosses straight through. In $f(x) = (x - 1)^2(x + 4)$ , the factor $(x - 1)$ appears to the power 2 (even) and the factor $(x + 4)$ appears to the power 1 (odd).

At $x = 1$ (multiplicity 2): the graph touches the x-axis and turns around.

At $x = -4$ (multiplicity 1): the graph crosses the x-axis.

Final answer: Touch and turn at $x = 1$ ; crosses at $x = -4$ .

Step 4: Identify the leading term to determine end behavior (Q4)

End behavior depends entirely on the leading term, because for large $|x|$ all lower-degree terms become negligible. For $g(x) = -x^5 + 2x$ , the leading term is $-x^5$ : odd degree and negative leading coefficient. Odd degree means the two ends go in opposite directions; the negative coefficient flips the usual odd-degree pattern, so the right end falls and the left end rises.

As $x \to \infty$ , $g \to -\infty$ ; as $x \to -\infty$ , $g \to +\infty$ .

Final answer: Right end falls, left end rises.

Step 5: Factor both expressions and cancel, then state the restriction (Q5)

To simplify a rational expression, factor numerator and denominator completely and cancel any common factor. The restriction must come from the original denominator, before canceling, because the simplified form hides the excluded value. Factor $x^2 - 16 = (x - 4)(x + 4)$ and $x^2 + 8x + 16 = (x + 4)^2$ , then cancel one copy of $(x + 4)$ .

\frac{(x - 4)(x + 4)}{(x + 4)^2} = \frac{x - 4}{x + 4}, \quad x \neq -4

Final answer: $\dfrac{x - 4}{x + 4}$ , with the restriction $x \neq -4$ .

Step 6: State restrictions first, then clear denominators and check (Q6)

For a rational equation, identify the values that make any denominator zero before solving, because those values must be rejected even if they satisfy the cleared equation. The denominators are $x$ and $2$ , so the only restriction is $x \neq 0$ . Multiply every term by the LCD, which is $2x$ , to clear all fractions, then solve the resulting linear equation.

\frac{3}{x} + \frac{1}{2} = \frac{5}{x}

Multiplying through by $2x$ : $6 + x = 10$ , so $x = 4$ .

Check: $x = 4 \neq 0$ , so the restriction is satisfied. Substituting back: $\frac{3}{4} + \frac{1}{2} = \frac{3}{4} + \frac{2}{4} = \frac{5}{4} = \frac{5}{4}$ . Valid.

Final answer: $x = 4$ .

Step 7: Isolate, square both sides, and reject extraneous solutions (Q7)

When you square both sides of a radical equation, you can introduce solutions that satisfy the squared equation but not the original. This happens because squaring removes the sign: $\sqrt{u} = v$ requires $v \geq 0$ , but $v^2 = u$ does not. Solve the resulting quadratic, then check each candidate in the original equation.

The radical is already isolated. Square both sides:

3x + 1 = (x - 1)^2 = x^2 - 2x + 1

Rearranging: $x^2 - 5x = 0$ , so $x(x - 5) = 0$ , giving candidates $x = 0$ and $x = 5$ .

Check $x = 5$ : $\sqrt{3(5) + 1} = \sqrt{16} = 4$ and $5 - 1 = 4$ . Valid.

Check $x = 0$ : $\sqrt{3(0) + 1} = \sqrt{1} = 1$ and $0 - 1 = -1$ . Since $1 \neq -1$ , this is extraneous.

Final answer: $x = 5$ .

Step 8: Multiply like binomials and replace $i^2 = -1$ (Q8)

Complex number multiplication follows the same distributive rule as binomial multiplication (FOIL). The only difference from real algebra is that $i^2 = -1$ , so any $i^2$ term is replaced with $-1$ and combined with the real parts. Expand $(2 - 3i)(1 + i)$ fully before collecting terms.

(2 - 3i)(1 + i) = 2 + 2i - 3i - 3i^2 = 2 - i - 3(-1) = 2 - i + 3 = 5 - i

Final answer: $5 - i$ .

Sources & how we know this

Educator Guide to the Regents Examination in Algebra II — NYSED (2025)