The Factor Theorem

You know that $P(a) = 0$ means the remainder on dividing by $x - a$ is zero. Write out the division identity for that case and simplify it. What kind of statement have you ended up with, and what does it say about $x - a$?

$x - a$ is a factor of $P(x)$ if and only if $P(a) = 0$. The two halves are both worth having: a zero gives you a factor, and a factor guarantees a zero. Together they turn factorising into a search for numbers rather than a search for brackets.

Start from the remainder theorem: the remainder on dividing $P(x)$ by $x - a$ is $P(a)$. Now ask what happens when that remainder is zero.

The division identity reads

$$P(x) = (x-a)Q(x) + P(a)$$

If $P(a) = 0$, the last term disappears and

$$P(x) = (x-a)Q(x)$$

which says precisely that $x - a$ is a factor of $P(x)$: the polynomial has been written as $(x-a)$ times something.

The converse is just as quick. If $x - a$ is a factor, then $P(x) = (x-a)Q(x)$ for some polynomial $Q$, and substituting $x = a$ gives $P(a) = 0 \times Q(a) = 0$.

So the two statements imply each other:

$$(x-a) \text{ is a factor of } P(x) \quad \iff \quad P(a) = 0$$

The theorem tells you what to look for. It does not, by itself, tell you where to look, and testing every real number is not a plan. This does:

If a monic polynomial with integer coefficients has an integer zero, that zero divides the constant term.

Here is why. Suppose $P(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_1x + c_0$ has integer coefficients and $P(a) = 0$ with $a$ an integer. Then

$$a^n + c_{n-1}a^{n-1} + \cdots + c_1a = -c_0$$

Every term on the left contains a factor of $a$, so the left side is $a$ times an integer. Therefore $a$ divides $-c_0$, and hence divides $c_0$.

For $P(x) = x^3 - 6x^2 + 11x - 6$ the constant term is $-6$, so the only integer zeroes possible are

$$\pm 1, \ \pm 2, \ \pm 3, \ \pm 6$$

Eight numbers, and $P(1) = 1 - 6 + 11 - 6 = 0$ succeeds on the first attempt. These are candidates, not answers: the result says every integer zero is on the list, not that every number on the list is a zero.

If the leading coefficient is not $1$, the zeroes need not be integers. Take $P(x) = 2x^3 - 3x^2 - 3x + 2$. Testing $\pm 1$ and $\pm 2$ gives $P(-1) = -2 - 3 + 3 + 2 = 0$, so $x + 1$ is a factor and one zero is an integer. Dividing by $x+1$ leaves the quotient $2x^2 - 5x + 2$, whose zeroes are $2$ and $\tfrac{1}{2}$, and the half would never appear on a list of integers.

The general version covers it. If $\dfrac{p}{q}$ is a zero in lowest terms, then

$$p \text{ divides the constant term} \qquad \text{and} \qquad q \text{ divides the leading coefficient}$$

For $2x^3 - 3x^2 - 3x + 2$ that gives $p \in \{\pm 1, \pm 2\}$ and $q \in \{1, 2\}$, so the candidates are $\pm 1, \pm 2, \pm \tfrac{1}{2}$.

In practice, Year 10 questions are built so that at least one zero is a small integer, and testing $1$, $-1$, $2$ and $-2$ finds it almost every time. Knowing the general rule matters mainly so that a failure to find an integer zero is recognised as a possibility rather than as a mistake.

Factorise $P(x) = x^3 - 6x^2 + 11x - 6$ completely.

Step 1. Candidates. The constant term is $-6$, so test $\pm 1, \pm 2, \pm 3, \pm 6$.

Step 2. Find one zero. $P(1) = 1 - 6 + 11 - 6 = 0$. So $x - 1$ is a factor, by the factor theorem.

Step 3. Divide. Dividing $P(x)$ by $x - 1$ gives quotient $x^2 - 5x + 6$, so

$$P(x) = (x-1)(x^2 - 5x + 6)$$

Step 4. Factorise the quotient. $x^2 - 5x + 6 = (x-2)(x-3)$, by ordinary quadratic factorising.

$$P(x) = (x-1)(x-2)(x-3)$$

Two things are worth noticing. Once one factor is found, the problem drops a degree and becomes a quadratic, which needs no new theory. And the answer can be checked instantly: the product of the constant terms is $(-1)(-2)(-3) = -6$, matching $P$'s constant term.

Questions often hand you a candidate rather than asking you to find one. "Show that $x + 4$ is a factor of $P(x)$" is a one-line task.

Solve $x + 4 = 0$ to get $x = -4$, evaluate $P(-4)$, and if it is zero, cite the factor theorem. If it is not zero, you have shown the opposite, and the value you found is the remainder.

The sign flip is where marks are lost. The factor $x + 4$ corresponds to the zero $-4$, and the factor $x - 4$ corresponds to $+4$. Writing the factor down from a zero requires the same flip in reverse: a zero of $-5$ gives the factor $(x + 5)$.

For a non-monic factor such as $2x - 1$, the zero is $\tfrac{1}{2}$, and finding $P\!\left(\tfrac{1}{2}\right) = 0$ shows $2x - 1$ is a factor. Note the factor is $2x - 1$, not $x - \tfrac{1}{2}$, although both are valid factors differing by a constant multiple.

Five items on the factor theorem. Work each one, then reveal the answer.

1. 1 Is $x - 1$ a factor of $x^3 - 2x^2 + 3x - 2$?

   Show answer
   $P(1) = 1 - 2 + 3 - 2 = 0$.Yes

2. 2 $P(5) = 0$. Write down a factor of $P(x)$.

   Show answer
   A zero of $5$ gives the factor with the opposite sign inside.$x - 5$

3. 3 List the integer candidates for a zero of $x^3 + 2x^2 - x - 10$.

   Show answer
   Divisors of the constant term $-10$.$\pm 1, \pm 2, \pm 5, \pm 10$

4. 4 Is $x + 3$ a factor of $x^3 + 27$?

   Show answer
   $P(-3) = -27 + 27$.Yes

5. 5 $x^3 - 7x + 6 = (x-1)(x^2 + x - 6)$. Finish the factorisation.

   Show answer
   Factorise the quadratic: two numbers multiplying to $-6$ and adding to $1$.$(x-1)(x+3)(x-2)$