Polynomial Division & the Remainder Theorem

You’ll be able to

Divide one polynomial by another and write the result as quotient plus remainder over divisor
Use the Remainder Theorem to evaluate a polynomial and to test for factors
Use a known factor to reduce a polynomial and find its remaining zeros

Division with a remainder, one degree up

Integer division writes 17 ÷ 5 as 3 remainder 2, or 17 = 5·3 + 2, where the remainder is smaller than the divisor. Polynomial division does the same with "smaller" meaning lower degree: dividing p(x) by d(x) gives p(x) = d(x)·q(x) + r(x), where the degree of r is strictly less than the degree of d. Dividing by a linear factor therefore always leaves a constant remainder.

Division algorithm for polynomials

p(x) = d(x)·q(x) + r(x), with deg r < deg d — equivalently p(x)/d(x) = q(x) + r(x)/d(x)

The second form is the one used to find slant asymptotes and to rewrite rational functions.

Worked example

Divide p(x) = 2x³ − 3x² + 0x − 4 by (x − 2).

1.Set up synthetic division with the zero of the divisor, x = 2, and the coefficients 2, −3, 0, −4.
2.Bring down 2. Multiply by 2 to get 4; add to −3 to get 1.
3.Multiply 1 by 2 to get 2; add to 0 to get 2.
4.Multiply 2 by 2 to get 4; add to −4 to get 0.
5.The bottom row 2, 1, 2 gives the quotient 2x² + x + 2, and the last entry 0 is the remainder.

Answer: p(x) = (x − 2)(2x² + x + 2) + 0. The remainder is 0, so (x − 2) is a factor of p and x = 2 is a zero.

The Remainder Theorem, and the shortcut hiding in it

Substituting x = a into p(x) = (x − a)q(x) + r kills the first term entirely, leaving p(a) = r. So the remainder on division by (x − a) is just p(a). Two consequences follow immediately. First, synthetic division evaluates a polynomial — often faster than substitution. Second, and more useful: (x − a) is a factor of p exactly when p(a) = 0. That is the Factor Theorem, and it is why hunting for zeros and hunting for factors are the same activity.

Remainder and Factor Theorems

p(x) ÷ (x − a) leaves remainder p(a) · and (x − a) is a factor ⟺ p(a) = 0

Watch the sign: dividing by (x + 3) means a = −3, so the relevant value is p(−3).

Checkpoint

When p(x) is divided by (x + 3), the remainder is 7. What is p(−3)?

Using one zero to find the rest

A cubic is hard to factor by inspection; a quadratic is not. So the standard strategy is: find one zero (by the Rational Root Theorem, a graph, or a hint in the problem), divide it out, and factor what is left. Each division drops the degree by one, which turns an intractable problem into a familiar one. The Rational Root Theorem narrows the search: any rational zero p/q of a polynomial with integer coefficients has p dividing the constant term and q dividing the leading coefficient.

Worked example

Find all zeros of p(x) = x³ − 4x² + x + 6.

1.Rational candidates divide the constant 6 over the leading coefficient 1: ±1, ±2, ±3, ±6.
2.Test x = −1: (−1) − 4(1) + (−1) + 6 = −1 − 4 − 1 + 6 = 0. So (x + 1) is a factor.
3.Synthetic division by −1 on coefficients 1, −4, 1, 6 gives 1, −5, 6 with remainder 0.
4.So p(x) = (x + 1)(x² − 5x + 6).
5.Factor the quadratic: x² − 5x + 6 = (x − 2)(x − 3).

Answer: p(x) = (x + 1)(x − 2)(x − 3), so the zeros are x = −1, 2 and 3. Each has multiplicity 1, so the graph crosses the x-axis at all three.

Tip

Synthetic division needs every coefficient, including zeros for missing powers. Dividing 2x³ − 3x² − 4 requires the row 2, −3, 0, −4. Omitting the placeholder is the single most common synthetic-division error.

Checkpoint

A cubic p has p(2) = 0. Dividing p by (x − 2) gives a quotient with no real zeros. How many real zeros does p have?

Answer the 2 checkpoints as you read.