The Rational Root Theorem: Additional Practice (Set 1)

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

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

The factorization checks: $(2x - 1)(x - 2) = 2x^2 - 5x + 2$. The roots are $-1$, $\tfrac{1}{2}$, and $2$.

Factor by grouping.

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

The roots are $-1$, $\tfrac{1}{2}$, and $-\tfrac{1}{2}$, all on the candidate list $\pm 1, \pm\tfrac{1}{2}, \pm\tfrac{1}{4}$.

Factor by grouping.

$3x^3 + x^2 - 12x - 4 = x^2(3x + 1) - 4(3x + 1) = (3x + 1)\left(x^2 - 4\right) = (3x + 1)(x - 2)(x + 2)$

The roots are $-\tfrac{1}{3}$, $2$, and $-2$. Note how the denominator $3$ of the fractional root divides the leading coefficient $3$, and every numerator divides the constant term $4$, exactly as the theorem requires.

Test $x = 2$: $P(2) = 96 + 40 - 152 + 10 + 6 = 0$, a root. Synthetic division by $2$ on $6, 5, -38, 5, 6$ gives the row $6, 17, -4, -3$. Test $x = -3$ on the cubic $6x^3 + 17x^2 - 4x - 3$: $-162 + 153 + 12 - 3 = 0$, another root, and dividing gives $6x^2 - x - 1$.

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

The four roots are $2$, $-3$, $-\tfrac{1}{3}$, and $\tfrac{1}{2}$, all rational. (The coefficients read the same forwards and backwards, which is why the roots pair off into reciprocals: $2$ with $\tfrac{1}{2}$, and $-3$ with $-\tfrac{1}{3}$.)

The candidates are $\pm 1, \pm 2, \pm 4$. Test $x = 2$: $P(2) = 8 - 4 - 4 = 0$, a root. Synthetic division by $2$ on $1, 0, -2, -4$ gives the quotient row $1, 2, 2$, so

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

The quadratic has discriminant $4 - 8 = -4$, so the quadratic formula gives $x = \tfrac{-2 \pm 2i}{2} = -1 \pm i$. The roots are $2$ and the conjugate pair $-1 \pm i$; only the rational root was visible to the theorem.

The theorem is one-directional: roots must appear on the list, but list membership guarantees nothing.

$\text{on the list} \;\not\Rightarrow\; \text{root}$

For instance, $2$ is on the list for $x^2 - 2$ yet $2^2 - 2 = 2 \ne 0$. The other statements are true: the list captures every rational root, a nonzero constant term has finitely many divisors, and every list entry is a fraction, hence rational.

The constant term is $1$ and the leading coefficient is $8$, so the candidates are $\pm 1, \pm\tfrac{1}{2}, \pm\tfrac{1}{4}, \pm\tfrac{1}{8}$. Testing the halves:

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

Similarly $P(1) = 3$, $P(-1) = -1$, $P\left(\pm\tfrac{1}{4}\right) = \pm\tfrac{1}{8} \mp \tfrac{3}{2} + 1 \ne 0$, and $P\left(\pm\tfrac{1}{8}\right) = \pm\tfrac{1}{64} \mp \tfrac{3}{4} + 1 \ne 0$. Every candidate fails, so the cubic has no rational roots. It does have real roots (they are all irrational), so concluding "no real roots" would be overreach.

The constant term is $0$, so first factor out the largest power of $x$.

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

The roots are $0$ with multiplicity two, plus $1$ and $2$. Applying the divisibility filter to the original polynomial would have been useless: with $a_0 = 0$, every integer satisfies $p \mid 0$.

The candidates are $\pm 1, \pm\tfrac{1}{2}$. Testing: $P(1) = 3$, $P(-1) = -3$, and

$P\left(\tfrac{1}{2}\right) = \tfrac{1}{4} + \tfrac{1}{4} + \tfrac{1}{2} - 1 = 0$

Synthetic division by $\tfrac{1}{2}$ on $2, 1, 1, -1$ gives the quotient row $2, 2, 2$, so $P(x) = \left(x - \tfrac{1}{2}\right)\left(2x^2 + 2x + 2\right) = (2x - 1)\left(x^2 + x + 1\right)$. The quadratic's discriminant is $1 - 4 = -3 < 0$, so it is irreducible over the reals and the factorization stops there.

Monic with constant term $-1$, so the only candidates are $\pm 1$. Testing: $P(1) = 0$ and $P(-1) = -2$. Dividing by $x - 1$:

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

The quadratic has discriminant $-3 < 0$, so its two roots are the nonreal conjugate pair $\tfrac{-1 \pm i\sqrt{3}}{2}$. Calling them irrational real numbers would be wrong: they are not real at all.

Candidates use $p \mid 2$ and $q \mid 6$. Testing $x = \tfrac{2}{3}$: $6 \cdot \tfrac{4}{9} - \tfrac{2}{3} - 2 = \tfrac{8}{3} - \tfrac{2}{3} - 2 = 0$, a root. Factoring confirms it and finds the partner:

$6x^2 - x - 2 = (3x - 2)(2x + 1)$

The roots are $\tfrac{2}{3}$ and $-\tfrac{1}{2}$. Both denominators ($3$ and $2$) divide the leading coefficient $6$, and both numerators divide the constant term $2$.

Numerators are positive divisors of the constant term $3$, so $p \in \{1, 3\}$; denominators are positive divisors of the leading coefficient $5$, so $q \in \{1, 5\}$.

$\tfrac{1}{1}, \ \tfrac{3}{1}, \ \tfrac{1}{5}, \ \tfrac{3}{5}$

The first option is the flipped-fraction trap, with the roles of $3$ and $5$ exchanged; the others mix divisors of the two coefficients into the wrong places.