The Rational Root Theorem: Practice

The polynomial is monic with constant term $3$, so the candidates are $\pm 1, \pm 3$. Note that $5$ is not even a candidate, since $5 \nmid 3$. Test $x = 1$:

$P(1) = 1 + 1 - 5 + 3 = 0$

So $1$ is a root. Checking the others: $P(3) = 27 + 9 - 15 + 3 = 24$ and $P(-1) = -1 + 1 + 5 + 3 = 8$, so neither is a root. (In fact $P(x) = (x - 1)^2(x + 3)$.)

Treat it as a quadratic in $x^2$.

$x^4 - 5x^2 + 4 = \left(x^2 - 1\right)\left(x^2 - 4\right) = (x-1)(x+1)(x-2)(x+2)$

The four roots are $\pm 1$ and $\pm 2$, all rational and all on the candidate list built from the constant term $4$. The candidates $\pm 4$ are on the list too, but they fail the test.

Factor by grouping.

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

The factor $x + 2$ gives the root $-2$. The factor $x^2 + 2$ is never zero for real $x$ (its roots are $\pm i\sqrt{2}$), so $-2$ is the only rational root. Direct checks agree: $P(-1) = 3$, $P(2) = 24$, $P(-4) = -36$.

When $a_0 = 0$, the condition $p \mid 0$ is satisfied by every integer, so the theorem filters nothing. Factor out the largest power of $x$ first.

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

The roots are $x = 0$ and $x = 2$ (a double root). Testing only $\pm 1, \pm 2, \pm 4$ on the original polynomial would find $2$ but miss the root $0$.

Factor by grouping.

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

Since $4x^2 - 1 = (2x - 1)(2x + 1)$, the roots are $x = 1$, $x = \tfrac{1}{2}$, and $x = -\tfrac{1}{2}$. All three sit on the candidate list $\pm 1, \pm\tfrac{1}{2}, \pm\tfrac{1}{4}$.

For $\tfrac{3}{4}$ to survive the filter, the numerator $3$ must divide the constant term and the denominator $4$ must divide the leading coefficient.

$8x^3 - 2x + 3\text{: } \ 3 \mid 3 \ \checkmark \quad 4 \mid 8 \ \checkmark$

The others fail: $4x^3 + x - 4$ has constant $-4$ with $3 \nmid 4$; $3x^3 + x + 8$ has $3 \nmid 8$ and $4 \nmid 3$; $6x^2 + x - 3$ has $4 \nmid 6$. Surviving the filter only makes $\tfrac{3}{4}$ a candidate; testing would still be required.

Factor the quadratic.

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

Setting each factor to zero gives $x = -\tfrac{1}{2}$ and $x = 1$. The other two candidates fail: $P(-1) = 2 + 1 - 1 = 2$ and $P\left(\tfrac{1}{2}\right) = \tfrac{1}{2} - \tfrac{1}{2} - 1 = -1$.

Monic means $q = 1$, so every candidate is an integer dividing $7$. Since $7$ is prime, its positive divisors are just $1$ and $7$, and each carries both signs.

$\pm 1, \ \pm 7$

Forgetting the negative candidates is a common slip; roots are often negative.

The theorem requires integer coefficients, and multiplying an equation by a nonzero constant changes no roots.

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

Now the theorem applies, and the roots $1$ and $-\tfrac{1}{2}$ both appear on its candidate list. Rounding coefficients would change the equation entirely.

Substitute, using $\left(\tfrac{3}{2}\right)^3 = \tfrac{27}{8}$ and $\left(\tfrac{3}{2}\right)^2 = \tfrac{9}{4}$.

$P\left(\tfrac{3}{2}\right) = \tfrac{27}{4} - \tfrac{9}{4} - 9 + 3 = \tfrac{18}{4} - 6 = \tfrac{9}{2} - 6 = -\tfrac{3}{2}$

The value is not zero, so $\tfrac{3}{2}$ is a candidate that fails, a reminder that the list only narrows the search.

From the factorization $(2x - 1)(x^2 - 3)$, the remaining roots satisfy $x^2 = 3$.

$x = \pm\sqrt{3}$

These are irrational. The candidate list is built entirely from fractions $\pm\tfrac{p}{q}$, so an irrational root can never appear on it; the theorem is structurally blind to such roots.

A root makes the polynomial evaluate to zero, so substitute $x = 2$ and solve for $k$.

$P(2) = 8 + 2k + 6 = 0 \;\Rightarrow\; 2k = -14 \;\Rightarrow\; k = -7$

With $k = -7$ the polynomial is $x^3 - 7x + 6 = (x - 1)(x - 2)(x + 3)$, so its roots are $1$, $2$, and $-3$, all on the candidate list for constant term $6$.