MHF4U The Remainder and Factor Theorems: Notes and Practice

Key things to remember

1Division statement:

P (x) = D (x) Q (x) + R (x)

, with

de g R < de g D

2Remainder theorem: dividing

P (x)

x - a

leaves remainder

P (a)

; dividing by

a x - b

leaves

P (\frac{b}{a})

3Factor theorem:

x - a

is a factor of

P (x)

if and only if

P (a) = 0

4Integer zero theorem: an integer zero of a polynomial with integer coefficients must divide the constant term.

5Rational zero theorem: a rational zero

\frac{p}{q}

has

p ∣ a_{0}

and

q ∣ a_{n}

6Synthetic division: bring down, multiply by

a

, add. Last entry is the remainder; the rest are the quotient's coefficients.

Everything in the previous topic assumed somebody handed you the factored form. Usually nobody does — you get $x^{3} - 4 x^{2} + x + 6$ and you are expected to find the factors yourself. For a quadratic you had the quadratic formula, but there is no such formula you are expected to use for a cubic. Instead you use a search: guess a zero, confirm it instantly, and divide it out to reduce the problem to a quadratic you can already handle. The remainder and factor theorems are what make the "confirm it instantly" step possible.

Dividing polynomials

Long division of polynomials works exactly like long division of numbers. Divide the leading term of what is left by the leading term of the divisor, write that on top, multiply back, subtract, and bring down the next term. You stop when the degree of what remains is smaller than the degree of the divisor. The result is always expressible as a division statement, and you should be able to write it both ways.

P (x) = D (x) Q (x) + R (x) ⟺ \frac{P ( x )}{D ( x )} = Q (x) + \frac{R ( x )}{D ( x )}

Dividend equals divisor times quotient plus remainder. The remainder has lower degree than the divisor.

The remainder theorem

Here is the shortcut. When the divisor is linear, say $D (x) = x - a$ , the remainder is just a number $R$ , so $P (x) = (x - a) Q (x) + R$ . Now substitute $x = a$ : the first term vanishes because $(a - a) = 0$ , and you are left with $P (a) = R$ . So the remainder can be found by a single substitution — no division at all.

Remainder theorem

When

P (x)

is divided by

x - a

, the remainder is

P (a)

. When it is divided by

a x - b

, the remainder is

P (\frac{b}{a})

— substitute the value that makes the divisor zero, whatever that value is.

VideoKhan Academy explains why dividing P(x) by x minus a leaves the remainder P(a), so you can find a remainder by substitution instead of dividing.Watch on YouTube

The factor theorem

A divisor divides evenly exactly when the remainder is $0$ . Combine that with the remainder theorem and you get the factor theorem: $x - a$ is a factor of $P (x)$ if and only if $P (a) = 0$ . This is an "if and only if", so it works in both directions — finding a zero hands you a factor, and knowing a factor hands you a zero. It also turns factoring into a guessing game with a very short list of guesses.

Fig 1From a candidate value to a confirmed factor to a reduced quotient.

Where to look for zeros

P (x)

has integer coefficients, any integer zero must divide the constant term. More generally, any rational zero

\frac{p}{q}

in lowest terms has

p

dividing the constant term and

q

dividing the leading coefficient. For

2 x^{3} + x^{2} - 8 x - 4

the candidates are

\pm 1, \pm 2, \pm 4

over

\pm 1, \pm 2

— a list of eight, not infinity. Try the small integers first.

Synthetic division

Synthetic division is a compressed version of long division that works when the divisor is $x - a$ . Write the coefficients of $P (x)$ in a row — including a $0$ for every missing degree — and put $a$ off to the left. Bring the first coefficient straight down. Multiply it by $a$ , write the product under the next coefficient, add, and repeat. The last number you produce is the remainder; the numbers before it are the coefficients of the quotient, whose degree is one less than $P$ .

Fig 2The bring-down, multiply, add cycle of synthetic division, with the remainder isolated at the end.

VideoA step-by-step run through the bring-down, multiply and add pattern of synthetic division, and how to read the quotient and remainder from the bottom row.Watch on YouTube

Fully factoring and solving

List the candidate zeros: factors of the constant term over factors of the leading coefficient.
Test candidates in $P (x)$ until one gives $0$ . Start with $\pm 1$ and $\pm 2$ — they are by far the most common.
Divide $P (x)$ by the corresponding factor using synthetic division to get a quotient of one degree lower.
If the quotient is quadratic, factor it (or use the quadratic formula). If it is still cubic, repeat the search on the quotient. If the quadratic has a negative discriminant it does not factor over the real numbers — leave it as it is and say so, rather than forcing a factoring that is not there.
Write $P (x)$ as the product of all the factors; to solve $P (x) = 0$ , set each factor to zero.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

Find the remainder when

P (x) = 2 x^{3} - 3 x^{2} + 4 x - 5

is divided by

x - 2

Example 2

Fully factor

P (x) = x^{3} - 4 x^{2} + x + 6

Example 3

Solve

2 x^{3} + x^{2} - 8 x - 4 = 0

Example 4

Determine the value of

k

so that

x + 1

is a factor of

P (x) = 2 x^{3} + k x^{2} - 5 x + 3

Where marks get lost

Substituting the wrong sign. For the divisor

x + 3

you test

x = - 3

, because

- 3

is what makes the bracket zero. The rule is "the value that zeroes the divisor", never "the number you see".

Forgetting placeholder zeros in synthetic division. Dividing

x^{3} - 7 x + 6

means using the coefficients

1, 0, - 7, 6

— the missing

x^{2}

term must appear as a

0

Testing only positive candidates. Negative zeros are just as common; work through

\pm 1

\pm 2

and so on in pairs.

For a divisor like

2 x + 1

, testing

x = - 2

instead of

x = - \frac{1}{2}

. Solve

2 x + 1 = 0

properly before substituting.

Stopping after one factor.

P (x) = (x - 2) (2 x^{2} + 5 x + 2)

is not fully factored — the quadratic still breaks into

(2 x + 1) (x + 2)

Key things to remember

Worked examples

Where marks get lost

Flashcards 8

Practice