Divide polynomials using long division and synthetic division; apply the Remainder Theorem (the remainder when dividing by x minus a equals the value of the polynomial at a) and the Factor Theorem to test for factors and find zeros.

What this topic is asking

The Regents Algebra II exam (the Arithmetic with Polynomials and Rational Expressions, A-APR, cluster) wants you to divide polynomials (by long division and synthetic division) and to use the Remainder Theorem and Factor Theorem: the remainder on dividing $P(x)$ by $x - a$ is just $P(a)$, and $x - a$ is a factor exactly when $P(a) = 0$. These connect division, factoring, and zeros into one idea.

Dividing polynomials

You can divide one polynomial by another with long division, just like numbers: divide the leading terms, multiply back, subtract, and bring down. When the divisor is a linear $x - a$, synthetic division is a faster shortcut using only the coefficients.

For $P(x) = x^3 - 4x^2 + 2x + 5$ divided by $x - 3$, synthetic division uses $a = 3$ and the coefficients $1, -4, 2, 5$, producing a quotient and a remainder. The remainder is the single most useful output, because of the next theorem.

The Remainder Theorem

This is a powerful shortcut: to find the remainder, you do not need to divide at all, just evaluate $P(a)$. For a multiple-choice question asking for a remainder, substitution is almost always faster than long division and far less error-prone.

The Factor Theorem

The Factor Theorem is the Remainder Theorem when the remainder is zero.

So testing whether $x - a$ divides evenly reduces to checking whether $P(a) = 0$. This is how you confirm a candidate factor and, once confirmed, peel it off to factor the rest of the polynomial.

Putting it together

The three ideas form a workflow: use the Remainder Theorem (or a rational-root candidate) to find a value where $P(a) = 0$, apply the Factor Theorem to confirm $x - a$ is a factor, then divide to reduce the polynomial's degree and factor what remains. A clarifying point that prevents the most common error is the sign of $a$: the factor $x + 2$ corresponds to $a = -2$, because $x + 2 = x - (-2)$. Evaluating at the wrong sign is the single biggest slip on Factor Theorem questions, so always read off $a$ as the value that makes the factor zero.

A second useful idea is that the degree of a polynomial fixes how many zeros it has, counting multiplicity and complex roots. A cubic such as $P(x) = x^3 - 4x^2 + x + 6$ has exactly three zeros; once you find one with the Factor Theorem and divide it out, the remaining quadratic gives the other two. This is why the workflow always terminates: each confirmed factor lowers the degree by one until a quadratic (or simpler) is left, which you can finish by factoring or the quadratic formula. Knowing the target number of zeros also tells you when you are done, so you do not stop short or hunt for extra roots that do not exist.

Try this

Q1. Find the remainder when $P(x) = x^2 + x - 6$ is divided by $x - 2$. [1 credit]

• Cue. $P(2) = 4 + 2 - 6 = 0$.

Q2. Is $x - 1$ a factor of $P(x) = x^3 - 1$? [2 credits]

• Cue. $P(1) = 1 - 1 = 0$, so yes.