Dividing polynomials works like dividing numbers: a quotient with a remainder that is smaller than the divisor.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The Remainder Theorem
Dividing f(x) by (x − a) leaves a remainder of exactly f(a). Evaluating is faster than dividing.
The Factor Theorem
If f(a) = 0 the remainder is zero, so (x − a) is a factor. Roots and factors are two views of one fact.
The quotient shape
Dividing a degree-n polynomial by a linear factor leaves degree n − 1. Each division shrinks the problem.
Why it matters for rationals
When the numerator degree exceeds the denominator, division rewrites the function as a polynomial plus a proper fraction. That form reveals the end behaviour.
Slant asymptotes
If that polynomial part is linear, its line is a slant asymptote, and the fractional part fades to nothing far out.
Long division, with polynomials
Divide leading terms, multiply back, subtract, bring down. The procedure mirrors numerical long division, and including zero coefficients for missing powers is what keeps the columns aligned.
The remainder theorem
Dividing f(x) by x − a leaves remainder f(a). So evaluating a polynomial and dividing it by a linear factor are the same computation, which is a surprisingly useful equivalence.
A zero remainder means a factor
If f(a) = 0 then x − a divides f exactly. Roots and factors are the same information, so each root found reduces the degree of the problem that remains.
Synthetic division as the shortcut
For a linear divisor, synthetic division does the same arithmetic with only the coefficients. It is faster and less error-prone, but it applies to linear divisors only — a limitation worth remembering.
Step 2: Try It Yourself
Tap and try it out.
Step 3: Watch an Example
One step at a time.
Watch Rosa Use the Remainder Theorem
Rosa needs the remainder when x³ − 2x + 5 is divided by (x − 2).
- Step 1
The theorem says the remainder is f(2), so no division is needed.
Step 4: Your Turn
Practice makes it stick.
The Remainder
Problem 1 of 2
f(x) = x³ − 2x + 5 divided by (x − 2). What is the remainder?
The Factor
Problem 2 of 2
f(3) = 0. Is (x − 3) a factor? 1 yes, 0 no.
Divide and Evaluate
1 of 8
f(x) = x² + 3x. Remainder on division by (x − 1)?
2 of 8
f(x) = x³ − 8. Remainder on division by (x − 2)?
3 of 8
That remainder means (x − 2) is a factor? 1 yes, 0 no.
4 of 8
A degree-4 polynomial divided by a linear factor. Degree of the quotient?
5 of 8
f(x) = x² − 5x + 6. Remainder on division by (x − 3)?
6 of 8
f(x) = 2x + 7. Remainder on division by (x + 1)?
7 of 8
Match each statement with what it tells you.
Tap a card on the left to start.
8 of 8
f(x) = x³ + 1. Remainder on division by (x + 1)?
Step 5: Quick Check
Show what you know.
Question 1 of 2
f(x) = x² + 2x. Remainder on division by (x − 3)?
Question 2 of 2
Why is the Factor Theorem a special case of the Remainder Theorem?
What You Learned
- Dividing by (x − a) leaves a remainder of f(a).
- A zero remainder means (x − a) is a factor.
- Division rewrites a top-heavy rational function as a polynomial plus a proper fraction.