Theorem

Division algorithm for polynomials

Algebra • Polynomial • Factor and Remainder Theorem

Formula

f (x) = d (x) q (x) + r (x), de g r < de g d

For polynomials

f (x)

and

d (x)

with

d (x)

not the zero polynomial, there are unique polynomials

q (x)

and

r (x)

with

f (x) = d (x) q (x) + r (x)

and either

r (x) = 0

de g r < de g d

How to use it

1Read the shape of the remainder off the divisor: a quadratic divisor leaves $a x + b$ , a cubic leaves $a x^{2} + b x + c$ .
2Write the remainder with unknown coefficients, then substitute the divisor's roots, where $d (x) q (x)$ vanishes and $q$ drops out.
3A linear divisor leaves a constant, which is why the remainder theorem is a single evaluation.

Worked example

On dividing

f (x)

x^{2} - 1

the remainder is

2 x + 5

. What is

f (1)

Step-by-step

1
$f (x) = (x^{2} - 1) q (x) + 2 x + 5$
Why: The divisor is quadratic, so the remainder has degree at most $1$ , and $2 x + 5$ has that shape.
2
$f (1) = (1 - 1) q (1) + 2 (1) + 5 = 7$
Why: At $x = 1$ the divisor vanishes, so the quotient never has to be found.

Answer

7

Watch out

The degree condition is what makes

q

and

r

unique: without it

f (x) = d (x) \cdot 0 + f (x)

would also qualify. The bound is on the DEGREE, not the number of terms, and a remainder of degree one less than the divisor is the general case, not an exception.

Also known as

polynomial division algorithm, degree of the remainder, remainder degree bound, divide a polynomial by a quadratic, quotient and remainder of polynomials, linear remainder

Appears in past papers

2020 (I)

Q22