The Rational Root Theorem: Additional Practice (Set 2)

The candidates are $\pm 1, \pm 3, \pm\tfrac{1}{2}, \pm\tfrac{3}{2}$. Test $x = 1$: $P(1) = 2 - 9 + 10 - 3 = 0$, a root. Synthetic division by $1$ on $2, -9, 10, -3$ gives the quotient row $2, -7, 3$.

$P(x) = (x - 1)\left(2x^2 - 7x + 3\right) = (x - 1)(2x - 1)(x - 3)$

The factorization of the quadratic checks out: $(2x - 1)(x - 3) = 2x^2 - 7x + 3$. The roots are $1, \tfrac{1}{2}, 3$.

Test $x = 2$: $P(2) = 8 - 4 - 16 + 12 = 0$, a root. Synthetic division by $2$ on $1, -1, -8, 12$ gives the quotient row $1, 1, -6$.

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

The root $2$ appears again in the quotient, so it is a double root; the remaining root is $-3$. Re-testing a successful candidate on the quotient is what catches repeated roots.

Factor by grouping.

$x^4 - 2x^3 - x + 2 = x^3(x - 2) - (x - 2) = (x - 2)\left(x^3 - 1\right) = (x - 2)(x - 1)\left(x^2 + x + 1\right)$

The rational roots are $2$ and $1$. The factor $x^2 + x + 1$ has discriminant $1 - 4 = -3 < 0$, so its two roots are nonreal (not irrational real numbers). A direct check also rules out $-1$: $P(-1) = 1 + 2 + 1 + 2 = 6$.

A rational root $\tfrac{p}{q}$ in lowest terms needs $p \mid 4$, so $p \in \{1, 2, 4\}$, and $q \mid 6$, so $q \in \{1, 2, 3, 6\}$.

$\tfrac{3}{2}\text{: the numerator } 3 \text{ does not divide } 4$

So $\tfrac{3}{2}$ is excluded. The other three pass both checks: $\tfrac{4}{3}$ ($4 \mid 4$, $3 \mid 6$), $-\tfrac{1}{6}$ ($1 \mid 4$, $6 \mid 6$), and $\tfrac{2}{3}$ ($2 \mid 4$, $3 \mid 6$).

Testing the candidates finds $P\left(\tfrac{2}{3}\right) = \tfrac{8}{9} - \tfrac{20}{9} + \tfrac{30}{9} - \tfrac{18}{9} = 0$. Synthetic division by $\tfrac{2}{3}$ on $3, -5, 5, -2$ gives the quotient row $3, -3, 3$, so

$P(x) = \left(x - \tfrac{2}{3}\right)\left(3x^2 - 3x + 3\right) = (3x - 2)\left(x^2 - x + 1\right)$

after pulling the factor $3$ out of the quadratic. The discriminant of $x^2 - x + 1$ is $-3 < 0$, so it does not factor further over the reals. Expanding confirms $(3x - 2)(x^2 - x + 1) = 3x^3 - 5x^2 + 5x - 2$.

The polynomial $x^2 - 10$ is monic with integer coefficients, so any rational root is an integer dividing $10$. Testing (signs vanish on squaring):

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

No candidate works, so $x^2 - 10$ has no rational roots; since $\sqrt{10}$ is one of its roots, $\sqrt{10}$ is irrational. The other options are wrong: $\sqrt{10}$ is not a root of $x^2 + 10$, the last-digit claim proves nothing, and $P(10) = 90 \ne 0$.

Numerators divide $6$: $p \in \{1, 2, 3, 6\}$. Denominators divide $4$: $q \in \{1, 2, 4\}$. The twelve raw quotients contain duplicates: $\tfrac{2}{2} = 1$, $\tfrac{6}{2} = 3$, $\tfrac{2}{4} = \tfrac{1}{2}$, and $\tfrac{6}{4} = \tfrac{3}{2}$ all repeat other entries.

$\text{distinct positives: } 1, 2, 3, 6, \tfrac{1}{2}, \tfrac{3}{2}, \tfrac{1}{4}, \tfrac{3}{4}$

Eight distinct positive values, so with signs the list has $16$ members, not the naive $4 \times 3 \times 2 = 24$.

The polynomial is monic with constant term $1$, so the entire candidate list is $\pm 1$. Test both:

$P(1) = 1 + 1 + 1 + 1 + 1 = 5, \qquad P(-1) = 1 - 1 + 1 - 1 + 1 = 1$

Neither is zero, so the quartic has no rational roots at all. Two evaluations settled the question for every rational number, which is the theorem at its most efficient. (The theorem applies to any degree; nothing is special about quartics.)

The candidates are $\pm 1, \pm 2, \pm 5, \pm 10$. Testing $x = 2$:

$P(2) = 8 + 2 - 10 = 0$

so $2$ is a root, and dividing gives $P(x) = (x - 2)(x^2 + 2x + 5)$, whose quadratic factor has discriminant $4 - 20 = -16 < 0$: the other two roots are nonreal. In particular $P(5) = 125 + 5 - 10 = 120 \ne 0$, so $5$ is a candidate that fails, exactly the converse error the theorem never licenses.

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

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

The quadratic's discriminant is $4 - 8 = -4$, so the quadratic formula gives $x = -1 \pm i$. The only rational root is $1$; the conjugate pair could never appear on the candidate list.

The candidates are $\pm 1, \pm 2$. Testing: $P(1) = 9$, $P(-1) = 1$, $P(2) = 28$, and

$P(-2) = -8 + 12 - 6 + 2 = 0$

so $-2$ is a root. Synthetic division by $-2$ on $1, 3, 3, 2$ gives the quotient row $1, 1, 1$, so $P(x) = (x + 2)(x^2 + x + 1)$. The discriminant of $x^2 + x + 1$ is $-3 < 0$, so the remaining two roots are nonreal and $-2$ is the only real root.

The list is built as $\pm\tfrac{p}{q}$ over all valid $p$ and $q$: whenever a number is on the list, so is its negative.

$\tfrac{p}{q} + \left(-\tfrac{p}{q}\right) = 0$

Pairing each candidate with its negative cancels the entire sum to $0$, for every polynomial. The sum carries no information; only testing individual candidates does.