The Rational Root Theorem: Practice

Factor by grouping.

$3x^3 - x^2 + 3x - 1 = x^2(3x - 1) + (3x - 1) = (3x - 1)\left(x^2 + 1\right)$

The factor $3x - 1$ gives the rational root $\tfrac{1}{3}$, and $x^2 + 1 = 0$ gives $x = \pm i$. Only $\tfrac{1}{3}$ could ever appear on the theorem's candidate list; the pair $\pm i$ is nonreal and invisible to it.

Test $x = 1$: $P(1) = 2 + 3 - 8 + 3 = 0$. Dividing by $x - 1$ leaves $2x^2 + 5x - 3 = (2x - 1)(x + 3)$.

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

The roots are $1$, $\tfrac{1}{2}$, and $-3$: exactly $3$ of the $8$ candidates. The other five pass the divisibility filter but fail the substitution test, which is the normal state of affairs.

The polynomial $x^2 - 6$ is monic with integer coefficients, so any rational root is an integer dividing $6$. Testing the candidates (squaring kills the sign):

$1 - 6 = -5, \quad 4 - 6 = -2, \quad 9 - 6 = 3, \quad 36 - 6 = 30$

None is zero, so $x^2 - 6$ has no rational roots; since $\sqrt{6}$ is a root, it is irrational. The other arguments fail: exhibiting finitely many decimal digits proves nothing, non-primeness does not force irrationality ($\sqrt{4} = 2$), and plenty of rational numbers lie between $2$ and $3$.

Test the easiest candidates first: $P(1) = 1 - 1 - 7 + 1 + 6 = 0$ and $P(-1) = 1 + 1 - 7 - 1 + 6 = 0$, so both $1$ and $-1$ are roots and $(x - 1)(x + 1) = x^2 - 1$ divides $P$. Dividing gives

$P(x) = \left(x^2 - 1\right)\left(x^2 - x - 6\right) = (x - 1)(x + 1)(x - 3)(x + 2)$

since $(x^2 - 1)(x^2 - x - 6) = x^4 - x^3 - 7x^2 + x + 6$. The roots are $1, -1, 3, -2$.

Numerators divide $10$, so $p \in \{1, 2, 5, 10\}$; denominators divide $4$, so $q \in \{1, 2, 4\}$.

$\tfrac{5}{4}\text{: } 5 \mid 10 \ \checkmark \quad 4 \mid 4 \ \checkmark$

So $\tfrac{5}{4}$ is a legitimate candidate, and so is $\tfrac{5}{2}$. But the theorem is a one-way implication: it never promises that a candidate is a root, and it never promises any rational root exists at all.

Test $x = 1$: $P(1) = 4 - 8 + 5 - 1 = 0$. Synthetic division by $1$ on $4, -8, 5, -1$ gives the quotient row $4, -4, 1$.

$P(x) = (x - 1)\left(4x^2 - 4x + 1\right) = (x - 1)(2x - 1)^2$

The quadratic is a perfect square, so $\tfrac{1}{2}$ is a root of multiplicity two. This is why you re-test a successful candidate on the quotient: roots can repeat.

Substitute $x = \tfrac{1}{2}$ and solve for $k$, using $\left(\tfrac{1}{2}\right)^3 = \tfrac{1}{8}$ and $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$.

$P\left(\tfrac{1}{2}\right) = \tfrac{1}{4} + \tfrac{k}{4} - \tfrac{5}{2} + 6 = \frac{1 + k + 14}{4} = \frac{k + 15}{4}$

Setting this to zero gives $k = -15$. Note that $\tfrac{1}{2}$ is consistent with the theorem for any integer $k$, since $1 \mid 6$ and $2 \mid 2$; the divisibility filter alone could never determine $k$.

The candidates are $\pm 1, \pm 2, \pm 3, \pm 6$. Test $x = 1$: $P(1) = 1 - 6 + 11 - 6 = 0$. Synthetic division by $1$ on $1, -6, 11, -6$ gives the quotient row $1, -5, 6$.

$P(x) = (x - 1)\left(x^2 - 5x + 6\right) = (x - 1)(x - 2)(x - 3)$

The roots are $1$, $2$, and $3$. The negative candidates could have been skipped from the start: for $x < 0$ each term $x^3$, $-6x^2$, $11x$, $-6$ is negative, so $P(x) < 0$ there.

For $x^2 - 10$, any rational root must be an integer dividing $10$, so test $\pm 1, \pm 2, \pm 5, \pm 10$ (squaring kills the sign):

$1 - 10 = -9, \quad 4 - 10 = -6, \quad 25 - 10 = 15, \quad 100 - 10 = 90$

All fail, so $x^2 - 10$ has no rational roots (its roots are $\pm\sqrt{10}$). The others factor readily: $x^2 - 9 = (x-3)(x+3)$, $2x^2 - x - 1 = (2x+1)(x-1)$, and $x^2 - 5x + 6 = (x-2)(x-3)$.

Test small integers first: $P(-1) = 2 + 1 - 11 - 4 + 12 = 0$ and $P(2) = 32 - 8 - 44 + 8 + 12 = 0$, so $-1$ and $2$ are roots. $P(-2) = 32 + 8 - 44 - 8 + 12 = 0$ as well. Dividing out $(x + 1)(x - 2)(x + 2)$, whose product is $x^3 + x^2 - 4x - 4$, leaves the factor $2x - 3$:

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

Expanding confirms the match, and the roots are $\tfrac{3}{2}, -1, 2, -2$.

Test $x = \tfrac{1}{2}$: $12 \cdot \tfrac{1}{8} + 20 \cdot \tfrac{1}{4} - \tfrac{1}{2} - 6 = \tfrac{3}{2} + 5 - \tfrac{1}{2} - 6 = 0$, a root. Synthetic division by $\tfrac{1}{2}$ on $12, 20, -1, -6$ gives the quotient row $12, 26, 12$.

$12x^3 + 20x^2 - x - 6 = \left(x - \tfrac{1}{2}\right)\left(12x^2 + 26x + 12\right) = (2x - 1)(3x + 2)(2x + 3)$

since $12x^2 + 26x + 12 = 2(3x + 2)(2x + 3)$. The roots are $\tfrac{1}{2}$, $-\tfrac{2}{3}$, and $-\tfrac{3}{2}$.

The theorem is exactly one implication.

$\text{rational root } \tfrac{p}{q} \text{ (lowest terms)} \;\Rightarrow\; p \mid a_0 \text{ and } q \mid a_n$

Its converse is false (most candidates fail when tested), it guarantees no rational root exists ($x^2 - 2$ has none), and irrational roots can never appear on a list made of fractions. The only guarantee is that no rational root escapes the list.