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The Rational Root Theorem: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A list to prepare

List all distinct rational-root candidates for $P(x)=3x^4-2x^2+6$ , reduced to lowest terms and with both signs.
1. Hint 1
  Possible numerators divide the constant term and denominators divide the leading coefficient.
2. Hint 2
  Use positive numerators $1,2,3,6$ and denominators $1,3$ , then remove duplicates.
Answer

$\pm1,\pm2,\pm3,\pm6,\pm\frac13,\pm\frac23$ .

Full solution

The numerator divisors are $1,2,3,6$ and the denominator divisors are $1,3$ .
Dividing by $1$ gives $1,2,3,6$ ; dividing by $3$ adds $\frac13,\frac23$ while $\frac33$ and $\frac63$ duplicate $1$ and $2$ .
Attach both signs to these six positive values.
This is a candidate list, not a list of verified roots.

Answer

$\pm1,\pm2,\pm3,\pm6,\pm\frac13,\pm\frac23$ .

Key idea

Reduction and removal of duplicates make the rational-root candidate list complete without repeated entries.
Problem 1: Mark complete
Problem 2 Two unknown coefficients

An integer-coefficient polynomial has $\frac76$ as a root. Its leading coefficient $A$ and nonzero constant term $B$ are both positive. Find the smallest positive values of $A$ and $B$ permitted by the Rational Root Theorem.
1. Hint 1
  The given fraction is already in lowest terms.
2. Hint 2
  The denominator must divide $A$ and the numerator must divide $B$ .
Answer

$A=6$ , $B=7$ .

Full solution

Since $\frac76$ is reduced, the theorem requires
$6\mid A$
and
$7\mid B$
The least positive permitted values are $6$ and $7$ .
These are necessary coefficient restrictions; they do not by themselves determine the intervening coefficients or guarantee a root.

Answer

$A=6$ , $B=7$ .

Key idea

A known reduced rational root restricts both endpoint coefficients by divisibility.
Problem 2: Mark complete
Problem 3 A fractional coefficient rule

List the complete rational-root candidate set for $\frac12x^3-\frac34x+\frac14=0$ .
1. Hint 1
  The theorem first needs integer coefficients.
2. Hint 2
  Multiply the equation by $4$ , a nonzero constant that changes no roots.
Answer

$\pm1,\pm\frac12$ .

Full solution

Clearing the numerical denominators gives
$2x^3-3x+1=0$
The constant term permits numerator magnitude $1$ , and the leading coefficient permits denominators $1$ and $2$ .
Thus the full list is $\pm1,\pm\frac12$ .

Answer

$\pm1,\pm\frac12$ .

Key idea

Clearing constant denominators gives an equivalent integer-coefficient equation suitable for the theorem.
Problem 3: Mark complete
Problem 4 A zero constant term

Find every rational root of $P(x)=2x^4+3x^3-4x^2$ , and show why the remaining factor contributes no rational root.
1. Hint 1
  First remove the largest common power of $x$ .
2. Hint 2
  List the candidates from the cofactor's own leading and constant coefficients, then test the small ones first.
Answer

Only $x=0$ , occurring twice.

Full solution

The polynomial is
$P(x)=x^2(2x^2+3x-4)$
This gives the root $0$ twice.
For the cofactor, candidates are $\pm1,\pm2,\pm4,\pm\frac12$ .
Its values at $1,-1,2,-2$ are respectively $1,-5,10,-2$ ; at $4,-4,\frac12,-\frac12$ they are $40,16,-2,-5$ .
None is zero.
Thus the cofactor has no rational root, so $0$ is the complete rational-root list.

Answer

Only $x=0$ , occurring twice.

Key idea

A zero constant term calls for removing powers of $x$ before making a finite candidate list.
Problem 4: Mark complete
Problem 5 An interval and a list

For $F(x)=x^4-3x^2-1$ , prove that it has an irrational real zero between $1$ and $2$ . You do not need its exact value.
1. Hint 1
  Use the endpoints to establish a real zero, then address whether it could be rational.
2. Hint 2
  The polynomial is monic with constant term $-1$ , so the full rational candidate list is short.
Answer

At least one irrational real zero lies in $(1,2)$ .

Full solution

The endpoint values are
$F(1)=-3$
and
$F(2)=3$
A polynomial graph is unbroken, so the sign change gives a real zero between them.
The only possible rational roots are $\pm1$ , and both give
$F(\pm1)=-3.$
The full list fails, proving that any real zero in the interval is irrational.

Answer

At least one irrational real zero lies in $(1,2)$ .

Key idea

A sign change establishes a real zero, and a failed full rational list establishes its irrationality.
Problem 5: Mark complete
Problem 6 An unsupplied root

Find all complex roots of $P(x)=3x^3+x^2+4x-4$ , showing a complete rational candidate list and a factorization that accounts for all roots.
1. Hint 1
  List candidates from the leading and constant coefficients, then test manageable entries.
2. Hint 2
  A successful linear division leaves a quadratic you can solve over the complex numbers.
Answer

$x=\frac23,\ \frac{-1+i\sqrt7}{2},\ \frac{-1-i\sqrt7}{2}$ .

Full solution

The reduced candidate list is $\pm1,\pm2,\pm4,\pm\frac13,\pm\frac23,\pm\frac43$ .
At $\frac23$ the polynomial is zero.
Dividing by $3x-2$ gives $x^2+x+2$ , so
$P(x)=(3x-2)(x^2+x+2)$
The quadratic formula gives
$x=\frac{-1\pm i\sqrt7}{2}.$
These three roots account for degree $3$ , and expanding the factorization restores all coefficients.

Answer

$x=\frac23,\ \frac{-1+i\sqrt7}{2},\ \frac{-1-i\sqrt7}{2}$ .

Key idea

A rational-root search can reveal one linear factor while a quadratic supplies the remaining nonreal roots.
Problem 6: Mark complete
Problem 7 A cubic to fully factor

Find all rational roots of $P(x)=3x^3+5x^2+x-1$ , including how many times each occurs as a linear factor.
1. Hint 1
  The complete candidate list is determined by constant term $-1$ and leading coefficient $3$ .
2. Hint 2
  After a root is found, test it again in the quotient.
Answer

$x=-1$ twice and $x=\frac13$ once.

Full solution

The candidates are $\pm1,\pm\frac13$ .
The value $P(-1)=0$ leads to quotient $3x^2+2x-1$ after division by $x+1$ .
That quotient also vanishes at $-1$ , leaving $3x-1$ .
Thus
$P(x)=(x+1)^2(3x-1)$
It gives $-1$ twice and $\frac13$ once.
Expansion gives $3x^3+5x^2+x-1$ , checking both the roots and their repetitions.

Answer

$x=-1$ twice and $x=\frac13$ once.

Key idea

A rational root can repeat, so the quotient should be tested after each successful division.
Problem 7: Mark complete
Problem 8 A list and its opposite

Consider this argument: $5x^2+7x-6$ has candidate list $\pm1,\pm2,\pm3,\pm6,\pm\frac15,\pm\frac25,\pm\frac35,\pm\frac65$ , and $\frac35$ turns out to be a root, so its negative $-\frac35$ must be a root too, because the list is symmetric. Evaluate the polynomial at both signed values to judge the argument.
1. Hint 1
  The list contains possibilities, which still need evaluation.
2. Hint 2
  Compute the polynomial at both signed candidates.
Answer

The argument fails: $P(\frac35)=0$ but $P(-\frac35)=-\frac{42}{5}$ .

Full solution

At the positive candidate,
$P\left(\frac35\right)=\frac95+\frac{21}{5}-6$
This is zero.
At the negative candidate,
$P\left(-\frac35\right)=\frac95-\frac{21}{5}-6$
This equals $-\frac{42}{5}$ .
So both signs occur as candidates, but only one is a root in this example.

Answer

The argument fails: $P(\frac35)=0$ but $P(-\frac35)=-\frac{42}{5}$ .

Key idea

Symmetry of a candidate list does not imply symmetry of the actual roots.
Problem 8: Mark complete
Problem 9 A positive-root claim

Let $P(x)=x^4+2x^2+3x+1$ . A student says that since every term of $P$ is positive when $x>0$ , the candidate $+1$ never needs to be evaluated directly, though it still belongs on the theorem's candidate list. Confirm the full candidate list, explain the student's shortcut, and decide whether $P$ has a rational root by evaluating what remains.
1. Hint 1
  The full list comes from the leading coefficient and constant term.
2. Hint 2
  For a positive input, every term of this polynomial is positive.
Answer

Candidate list $\pm1$ ; $P$ has no rational root.

Full solution

The theorem gives only $1$ and $-1$ as candidates, from the divisors of the constant term $1$ and leading coefficient $1$ .
For $x>0$ , every term $x^4$ , $2x^2$ , $3x$ , and $1$ is positive, so $P(x)>0$ for every positive $x$ ; this rules out $+1$ as a root without evaluating it, though it still counts as one of the two candidates the theorem lists.
Testing the remaining candidate gives
$P(-1)=1+2-3+1$
This is $1$ , so the full rational search has failed.
The conclusion concerns rational roots only.

Answer

Candidate list $\pm1$ ; $P$ has no rational root.

Key idea

A sign argument can rule out a candidate before evaluation, but the theorem's candidate list itself is unchanged; only which candidates need testing shrinks.
Problem 9: Mark complete
Problem 10 A reduced fraction condition

A worked solution rejects the proposed root $\frac46$ of $3x^2+x-2$ , reasoning that $6$ does not divide the leading coefficient $3$ . Identify whether that reasoning is valid, and determine if the proposed value is actually a root.
1. Hint 1
  The theorem restricts a fraction in lowest terms.
2. Hint 2
  Reduce the proposed fraction before testing its denominator or evaluating it.
Answer

Invalid; $\frac46=\frac23$ is a root.

Full solution

The fraction reduces to $\frac23$ , whose denominator $3$ divides the leading coefficient.
Evaluate the reduced value:
$3\left(\frac23\right)^2+\frac23-2=0$
Thus it really is a root.
A nonreduced denominator is not the denominator constrained by the theorem, so testing $6$ against $3$ was never the right comparison.

Answer

Invalid; $\frac46=\frac23$ is a root.

Key idea

Lowest terms is essential when applying the denominator restriction in the Rational Root Theorem.
Problem 10: Mark complete

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