Skill: finding zeroes by testing a few numbers

Do not guess. Test.

You are asked to factorise $x^3 - 6x^2 + 11x - 6$. A guess may take a long time. There is a short list of numbers that can be zeroes, and you can test them all in a minute. In an olympiad, where time is short, this one skill saves many minutes.

The list of candidates

Suppose that a polynomial with whole-number coefficients and $1$ as the coefficient of its highest power has a whole-number zero. Then that zero must divide the constant term. (Reason: when you put $x = a$ in $x^3 + bx^2 + cx + d = 0$, you get $d = -a\,(a^2 + ba + c)$, so $a$ is a factor of $d$.)

For $x^3 - 6x^2 + 11x - 6$, the constant term is $-6$. Its divisors are $\pm 1, \pm 2, \pm 3, \pm 6$. Only eight numbers can be whole-number zeroes.

Test with the factor theorem

The factor theorem: $x - a$ is a factor of $p(x)$ if and only if $p(a) = 0$.

• $p(1) = 1 - 6 + 11 - 6 = 0$. So $x - 1$ is a factor.
• $p(2) = 8 - 24 + 22 - 6 = 0$. So $x - 2$ is a factor.
• $p(3) = 27 - 54 + 33 - 6 = 0$. So $x - 3$ is a factor.

A cubic has at most three zeroes, so we have them all:

Check. The constant term: $(-1)(-2)(-3) = -6$. ✓ The coefficient of $x^2$: $-(1 + 2 + 3) = -6$. ✓

Start with $x^3 - 7x + 6$. Of the divisors of $6$, only $1$, $2$ and $-3$ give $p(t) = 0$. Then set your own coefficients and hunt again.

When only one zero is found

Suppose that you find only $p(1) = 0$. Then divide $p(x)$ by $x - 1$. The quotient is a quadratic, and you can factorise that in the usual way. You do not need to test any more numbers.

When the leading coefficient is not 1

Now take $2x^3 - 3x^2 - 3x + 2$. The zeroes can be fractions. A rational zero $\dfrac{p}{q}$, in lowest terms, has $p$ dividing the constant term ($2$) and $q$ dividing the leading coefficient ($2$). The candidates are $\pm 1, \pm 2, \pm \dfrac{1}{2}$.

• $p(2) = 16 - 12 - 6 + 2 = 0$.
• $p(-1) = -2 - 3 + 3 + 2 = 0$.
• $p\left(\dfrac{1}{2}\right) = \dfrac{1}{4} - \dfrac{3}{4} - \dfrac{3}{2} + 2 = 0$.

The three zeroes are $2$, $-1$ and $\dfrac{1}{2}$, and $2x^3 - 3x^2 - 3x + 2 = (x - 2)(x + 1)(2x - 1)$.

Finding an unknown from a factor

A problem may say: "$x - 2$ is a factor of $x^3 - kx^2 + 6x - 8$. Find $k$." Put $x = 2$ and make it zero:

If the problem gives a remainder, make the value equal to the remainder. "When $x^3 + kx + 5$ is divided by $x - 1$ the remainder is $8$." Then $1 + k + 5 = 8$, so $k = 2$.

The remainder is the same trick

The remainder theorem: the remainder when $p(x)$ is divided by $x - a$ is $p(a)$. When the divisor is $x + 2$, use $a = -2$. This gives the remainder at once, with no long division.

A plan for a problem

1. Write down the constant term and its divisors.
2. Test the small ones first: $1$, $-1$, $2$, $-2$.
3. When you find a zero, divide it out.
4. Factorise what is left.
5. Check by multiplying out, or by the sum and the product of the zeroes.

Practise with new numbers

Try it in Python

This program tests the divisors of the constant term for you.

Common mistakes

• Testing only the positive divisors. Test both signs.
• Using the wrong sign for the factor. If $p(-2) = 0$, the factor is $x + 2$.
• Stopping when one zero is found. Divide and continue.
• Forgetting the fractions when the leading coefficient is not $1$.

Remember these

• A whole-number zero divides the constant term. List the divisors, and test them with the factor theorem.
• $x - a$ is a factor if and only if $p(a) = 0$. The remainder on dividing by $x - a$ is $p(a)$.
• To find an unknown, put in the zero and make the value $0$ (or the given remainder).
• With a leading coefficient other than $1$, the zeroes can be fractions $\dfrac{p}{q}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Which numbers can be whole-number zeroes of $x^2 - 5x + 6$?

2. Find all the zeroes of $x^3 - 7x + 6$.

3. Which fractions can be zeroes of $3x^2 + x - 2$?

4. Factorise $x^3 - 3x^2 - 4x + 12$.