The rational zero theorem and the finite candidate list it produces, testing candidates with synthetic division, narrowing a long list with a graph, repeating on the quotient until it is quadratic, and solving a pyramid-volume model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Find Rational Zeros
Objectives
Five outcomes. The first makes an infinite search finite; the rest carry it out.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 370-378 — the lesson these objectives are drawn from
Warm-up
Lesson 5.5 could factor a cubic completely — but only after being told one of its zeros.
Discussion prompt
Suppose nobody tells you a zero of x cubed minus 8x squared plus 11x plus 20. How would you look for one, and what makes that search hard?
Hint: There are infinitely many numbers to try.
Answer:
Testing values one at a time is hopeless if any real number is a candidate: there is no place to start and no way to finish.
But there is a pattern hiding in the coefficients. If the zeros are rational, their numerators must divide the constant term and their denominators must divide the leading coefficient — which turns an infinite search into a list of six numbers for this cubic.
Concept
If a polynomial has integer coefficients, every rational zero it has must be a factor of the constant term divided by a factor of the leading coefficient. That produces a finite list of candidates, and one synthetic tableau tests each.
rational zero theorem — If a polynomial with integer coefficients has a rational zero p over q in lowest terms, then p is a factor of the constant term and q is a factor of the leading coefficient.
\[ \frac{p}{q} = \frac{\text{factor of } a_0}{\text{factor of } a_n} \]
The theorem says only that every rational zero is on the list, not that everything on the list is a zero. Most candidates fail, and a failure costs one tableau.
Figure (svg): The rational zero theorem shown as a fraction built from two lists of factors
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 370-370
Section
Section 1
Concept
Write every factor of the constant term, positive and negative, and every factor of the leading coefficient. Every possible rational zero is one of the first divided by one of the second, and the list is simplified by removing duplicates.
\[ f(x) = 4x^4 - x^3 - 3x^2 + 9x - 10 \]
Both lists must include negative factors. Leaving them out halves the list and usually removes the zero you were looking for, which is why the textbook flags it.
Figure (svg): The rational zero theorem shown as a fraction built from two lists of factors
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 370-370 — The Rational Zero Theorem
Picture it
Example 1b: a quartic with leading coefficient 4.
Figure (svg): The rational zero theorem shown as a fraction built from two lists of factors
Four factors above and three below give twelve fractions, of which eight are distinct after simplifying. A leading coefficient of 1 would have given only four.
Worked example
Example 1, both parts.
\[ \text{List the possible rational zeros of } x^3 + 2x^2 - 11x + 12 \text{ and } 4x^4 - x^3 - 3x^2 + 9x - 10. \]
First: factor the constant term
Why: Twelve has factors 1, 2, 3, 4, 6 and 12, each with both signs.
\[ +- 1, +- 2, +- 3, +- 4, +- 6, +- 12 \]
First: factor the leading coefficient
Why: The leading coefficient is 1, whose only factors are 1 and negative 1.
\[ +- 1 \]
First: form the list
Why: Dividing by 1 leaves the numerators unchanged, so the candidates are the factors of 12.
Second: both lists
Why: Ten has factors 1, 2, 5 and 10; 4 has factors 1, 2 and 4.
Second: form and simplify
Why: Twelve fractions reduce to eight distinct values once duplicates such as 2 over 2 are removed.
\[ +- 1, +- 2, +- 5, +- 10, +- \frac{1}{2}, +- \frac{5}{2}, +- \frac{1}{4}, +- \frac{5}{4} \]
Figure (svg): The solution to Worked example list the candidates shown as a ladder of expressions, one row per algebraic move
\[ \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12; \quad \pm 1, \pm 2, \pm 5, \pm 10, \pm\tfrac{1}{2}, \pm\tfrac{5}{2}, \pm\tfrac{1}{4}, \pm\tfrac{5}{4} \]
Verify: check the theorem against a known example
Why: The function 64x cubed plus 152x squared minus 62x minus 105 has zeros negative five halves, negative three quarters and seven eighths. Every numerator — 5, 3 and 7 — divides 105, and every denominator — 2, 4 and 8 — divides 64. The theorem is describing a pattern that really is there in the coefficients.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 370-370
Matching
Factor the constant, then the leading coefficient.
Match the pairs
Why: The first and last share a list because both have constant term 12 and leading coefficient 1 — the middle coefficients play no part at all. Only the third has fractional candidates, because only its leading coefficient is not 1.
Worked example
Guided Practice 1 and 2. One list is far shorter than the other.
\[ \text{List the candidates for } x^3 + 9x^2 + 23x + 15 \text{ and } 2x^3 + 3x^2 - 11x - 6. \]
First: the constant is 15
Why: Its factors are 1, 3, 5 and 15, with both signs.
\[ +- 1, +- 3, +- 5, +- 15 \]
First: the leading coefficient is 1
Why: Dividing by 1 changes nothing.
Second: the constant is negative 6
Why: Its factors are 1, 2, 3 and 6.
\[ +- 1, +- 2, +- 3, +- 6 \]
Second: the leading coefficient is 2
Why: Dividing each by 2 adds the halves, and only the odd numerators give new values.
\[ +- 1, +- 2, +- 3, +- 6, +- \frac{1}{2}, +- \frac{3}{2} \]
Figure (svg): The solution to Worked example two more lists shown as a ladder of expressions, one row per algebraic move
\[ \pm 1, \pm 3, \pm 5, \pm 15; \quad \pm 1, \pm 2, \pm 3, \pm 6, \pm\tfrac{1}{2}, \pm\tfrac{3}{2} \]
Verify: check which fractions are genuinely new
Why: In the second, 2 over 2 and 6 over 2 give 1 and 3, both already on the list, so only 1 over 2 and 3 over 2 are new. Simplifying before listing keeps the search shorter, and it is worth doing because every duplicate would otherwise cost a tableau.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 371-371
Trap
\[ f(x) = x^3 - 8x^2 + 11x + 20 \]
List the factors of 20 and of 1
Why: Only the positive factors are written down.
\[ 1, 2, 4, 5, 10, 20 \quad \text{(half the list)} \]
The zeros of this function are negative 1, 4 and 5, and negative 1 is not on the shortened list at all.
\[ \pm 1, \pm 2, \pm 4, \pm 5, \pm 10, \pm 20 \]
Include both signs for every factor of both numbers
Why: A factor of 20 is any integer dividing it, and negative 1 divides 20 exactly as 1 does.
\[ f(-1) = -1 - 8 - 11 + 20 = 0 \]
The book flags this omission in an Avoid Errors note. Writing the plus-or-minus symbol in front of each factor as you list it makes the doubling automatic.
Fill the middle
Example 1b.
Fill in the blanks
p \text4 -10, \; q \text___ 4 \;\Longrightarrow\; \tfrac______ \text___ \pm 5, \pm\tfrac______, \pm\tfrac______}
Why: The factors of 4 are 1, 2 and 4, so the numerator 5 produces 5, five halves and five quarters, each with both signs. Working numerator by numerator rather than at random is the way to be sure the list is complete.
Prediction
Commit before reasoning.
Predict first
If the leading coefficient is 1, what do the candidates look like?
Correct: Only integers — the factors of the constant term.
\[ a_n = 1 \;\Longrightarrow\; \tfrac{p}{q} = \tfrac{p}{\pm 1} = \pm p \]
Why: Dividing by 1 or negative 1 leaves the numerator unchanged, so every candidate is an integer factor of the constant term. That is why Example 2 has six candidates while Example 3 has sixteen, and it is a good reason to divide out a common factor from the coefficients before starting, when one exists.
Two truths and a lie
All three are about the theorem.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one, and the textbook flags it directly. In Example 2 the candidate 1 fails and only three of the twelve values are zeros. The theorem narrows the search to a finite list; it does not identify which members of that list work, and testing is what does.
Section
Section 2
Concept
Test each candidate with synthetic division. A non-zero remainder rules it out; a zero remainder identifies a zero and hands you the quotient at the same time, so nothing is wasted.
\[ f(x) = x^3 - 8x^2 + 11x + 20 \;\Longrightarrow\; (x+1)(x^2 - 9x + 20) \]
Once the quotient is quadratic, stop testing and use Chapter 4's methods. There is no need to hunt for the remaining zeros one at a time.
Figure (svg): Two synthetic tableaux side by side, one with a non-zero remainder and one with zero
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 371-371 — Find zeros when the leading coefficient is 1
Picture it
Example 2: testing 1, then negative 1.
Figure (svg): Two synthetic tableaux side by side, one with a non-zero remainder and one with zero
The failed test cost one row and removed a candidate permanently. The successful one produced both a zero and the quadratic that supplies the other two.
Worked example
Example 2, in the book's three steps.
\[ \text{Find all real zeros of } f(x) = x^3 - 8x^2 + 11x + 20. \]
List the candidates
Why: The leading coefficient is 1 and the constant term is 20.
\[ +- 1, +- 2, +- 4, +- 5, +- 10, +- 20 \]
Test x equal to 1
Why: The tableau's bottom row ends at 24, so 1 is not a zero.
\[ \text{remainder } 24 \]
Test x equal to negative 1
Why: The bottom row is 1, negative 9, 20, 0, so negative 1 is a zero.
\[ \text{remainder } 0 \]
Write the factorisation
Why: The quotient is x squared minus 9x plus 20.
\[ (x + 1) (x ^{2} - 9 x + 20) \]
Factor the quadratic
Why: A product of 20 and a sum of negative 9 give the pair negative 4 and negative 5.
\[ (x + 1) (x - 4) (x - 5) \]
Figure (svg): The solution to Worked example find all real zeros shown as a ladder of expressions, one row per algebraic move
\[ x = -1, \; 4, \; 5 \]
Verify: check that all three are on the candidate list
Why: Negative 1, 4 and 5 are all factors of 20, as the theorem requires. And their product is negative 20, which for a cubic with leading coefficient 1 must equal the negative of the constant term — a useful check that no zero has been missed or invented.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 371-371
Ranking
Finding all real zeros when the leading coefficient is 1.
Put in order
Why: The last step is where answers get lost: the tested zero is easy to forget once attention moves to the quotient. Step four does not require factoring — if the quotient will not factor, the quadratic formula finishes it and may produce irrational zeros the candidate list could never have contained.
Worked example
Guided Practice 3 and 4.
\[ \text{Find all real zeros of } x^3 - 4x^2 - 15x + 18 \text{ and } x^3 - 8x^2 + 5x + 14. \]
First: test 1
Why: One minus 4 minus 15 plus 18 is zero, so 1 is a zero on the first try.
\[ 1\text{ is } a\text{ zero} \]
First: factor the quotient
Why: The tableau gives 1, negative 3, negative 18, so the quotient is x squared minus 3x minus 18.
\[ (x - 6) (x + 3) \]
Second: test negative 1
Why: Negative 1 minus 8 minus 5 plus 14 is zero.
\[ -1\text{ is } a\text{ zero} \]
Second: factor the quotient
Why: The tableau gives 1, negative 9, 14, so the quotient is x squared minus 9x plus 14.
\[ (x - 7) (x - 2) \]
Figure (svg): The solution to Worked example two more cubics shown as a ladder of expressions, one row per algebraic move
\[ \{1, 6, -3\}; \qquad \{-1, 7, 2\} \]
Verify: check the products
Why: For the first, 1 times 6 times negative 3 is negative 18, which is the negative of the constant term 18. For the second, negative 1 times 7 times 2 is negative 14, the negative of 14. Both check. Testing small candidates such as 1 and negative 1 first is worthwhile precisely because they are so quick to evaluate.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 371-371
Error analysis
A student finds one zero of a cubic and reports it.
Annotate
On: \( f(x) = x^3 - 8x^2 + 11x + 20, \; f(-1) = 0, \text{ so the zero is } -1 \)
The quotient is handed to you for free by the same tableau. Finding one zero is the start of the problem, not the end of it.
Fill the middle
Example 2, at the successful tableau.
Fill in the blanks
\text20 1, -9, 20, 0 \;\Longrightarrow\; f(x) = (x + 1)(x^2 - 9x + ___)
Why: The first three entries are the quotient's coefficients and the final zero confirms the division was exact. Note that the divisor is x plus 1 because the value tested was negative 1 — the factor is x minus the zero, and the zero was negative.
Sorting
A candidate is a zero exactly when the remainder is 0.
Sort into buckets
For f(x) = x^3 - 8x^2 + 11x + 20, sort each candidate.
Three of the twelve candidates are zeros, which is the maximum a cubic allows. Once three are found, testing can stop.
Prediction
Commit before reasoning.
Predict first
You are finding the zeros of a cubic and one candidate works. Must you keep testing candidates?
Correct: No — the quotient is quadratic and Chapter 4 finishes it.
\[ \text{degree } 3 \;\to\; \text{one test} \;\to\; \text{degree } 2 \;\to\; \text{the formula} \]
Why: After one successful division the remaining polynomial has degree 2, and the quadratic formula finds its zeros whether they are rational, irrational or imaginary. Continuing to test candidates would find only the rational ones, and would miss the irrational zeros of Example 3 entirely. Testing is for getting down to degree 2, not for finding every zero.
Section
Section 3
Concept
When the leading coefficient is not 1, the candidate list can run to sixteen values or more. A quick sketch shows roughly where the graph crosses the axis, and only the candidates near those crossings are worth testing.
\[ f(x) = 10x^4 - 11x^3 - 42x^2 + 7x + 12 \]
The graph does not prove anything: a crossing near 0.6 is a reason to test three fifths, not evidence that three fifths is a zero. The tableau still does the proving.
Figure (svg): A quartic graph with candidate zeros marked, showing which are worth testing
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 372-372 — Find zeros when the leading coefficient is not 1
Picture it
Example 3: a quartic with leading coefficient 10.
Figure (svg): A quartic graph with candidate zeros marked, showing which are worth testing
The graph crosses near negative 1.5, negative 0.5, 0.6 and 2.4. Two of those are rational candidates and two are not, so only two of the sixteen values need testing.
Worked example
Example 3, steps 1 to 4.
\[ \text{Find a rational zero of } 10x^4 - 11x^3 - 42x^2 + 7x + 12. \]
List the candidates
Why: The constant 12 has six positive factors and the leading coefficient 10 has four, giving sixteen distinct values.
Sketch the graph and read the crossings
Why: The curve meets the axis near negative 1.5, negative 0.5, 0.6 and 2.4.
Pick the candidates nearest those crossings
Why: Negative three halves, negative one half, three fifths and twelve fifths are all on the list.
Test negative three halves
Why: The tableau's last entry is not zero, so it fails.
Test negative one half
Why: The bottom row is 10, negative 16, negative 34, 24, 0.
\[ -\frac{1}{2}\text{ is } a\text{ zero} \]
Figure (svg): The solution to Worked example narrow, then test shown as a ladder of expressions, one row per algebraic move
\[ x = -\tfrac{1}{2}, \quad q(x) = 10x^3 - 16x^2 - 34x + 24 \]
Verify: check the value directly
Why: At x equal to negative one half: 10 times one sixteenth is 0.625; negative 11 times negative one eighth is 1.375; negative 42 times one quarter is negative 10.5; 7 times negative one half is negative 3.5; plus 12. The total is zero. Note that the crossing near 2.4 was tested as twelve fifths and failed — the graph suggests, it does not decide.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 372-372
Prediction
Commit before reasoning.
Predict first
The candidate list has sixteen values. What does the graph save?
Correct: Most of the tableaux, by pointing at three or four worth trying.
\[ \text{16 candidates} \;\to\; \text{4 plausible} \;\to\; \text{2 zeros} \]
Why: Sixteen tableaux is a great deal of arithmetic, and each is a chance to make an error. Four is manageable. The graph cannot prove anything, but it can rule out candidates that are nowhere near a crossing — a candidate at 12 is plainly not a zero of a function whose graph is far from the axis there.
Worked example
Example 3, step 4 in detail.
\[ \text{Rewrite } \left(x + \tfrac{1}{2}\right)(10x^3 - 16x^2 - 34x + 24) \text{ without fractions.} \]
Notice the common factor in the quotient
Why: Ten, 16, 34 and 24 are all even, so 2 comes out.
\[ 2(5 x ^{3} - 8 x ^{2} - 17 x + 12) \]
Move the 2 into the first factor
Why: Multiplying x plus one half by 2 gives 2x plus 1.
\[ (2 x + 1) (5 x ^{3} -...) \]
Check the product is unchanged
Why: Halving one factor and doubling the other leaves the product the same.
Read the zero from the new factor
Why: Two x plus 1 vanishes at negative one half, as it must.
Figure (svg): The solution to Worked example clear the fraction from the factor shown as a ladder of expressions, one row per algebraic move
\[ (2x+1)(5x^3 - 8x^2 - 17x + 12) \]
Verify: confirm the leading coefficient
Why: Two times 5 is 10, matching the original leading coefficient. Clearing the fraction is not required, but it keeps every later coefficient an integer, which matters because the rational zero theorem needs integer coefficients to apply to the quotient.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 372-372
Error analysis
A student reads a quartic's graph and reports its zeros.
Annotate
On: \( \text{the graph crosses near } -1.5, -0.5, 0.6, 2.4, \text{ so those are the zeros} \)
A graph locates crossings to within about a tenth; a tableau proves a zero exactly. Use the first to choose what to test and the second to decide.
Fill the middle
Example 3, step 4.
Fill in the blanks
\left(x + \tfrac2___\right) \cdot 2 = ___x + 1
Why: Doubling x plus one half gives 2x plus 1, which has the same zero, negative one half. The 2 came out of the quotient's coefficients, so the product is unchanged: one factor was halved and the other doubled.
Sorting
Compare each candidate with where the graph crosses.
Sort into buckets
For a graph crossing near -1.5, -0.5, 0.6 and 2.4, sort each candidate.
Skipping is a judgement, not a proof. If none of the promising candidates works, the skipped ones are still on the list and can be tested after all.
Comparison
Fill the blanks. One number changes the whole search.
Comparison matrix
| Question | Leading coefficient 1 | Leading coefficient 10 |
|---|---|---|
| Candidates | integer factors of the constant | those over factors of 10 as well |
| Typical list length | six to twelve | sixteen or more |
| Is a graph needed? | usually not | yes, to narrow the list |
| Are the zeros integers? | if rational, yes | often fractions |
The last row is the practical difference: a fractional zero such as three fifths is invisible to guessing and only a systematic list will find it.
Section
Section 4
Concept
After each successful division the quotient is a new polynomial of lower degree, and the rational zero theorem applies to it too. Repeat until the quotient is quadratic, then finish with factoring or the quadratic formula.
\[ 10x^4 - \cdots = (2x+1)(5x-3)(x^2 - x - 4) \]
The quotient's candidate list is usually shorter than the original's, because its constant term and leading coefficient are smaller. Each round of the search is easier than the last.
Figure (svg): A quartic reduced step by step to a quadratic by two successful divisions
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 372-372 — Example 3, step 5
Picture it
Example 3: two divisions reduce a quartic to a quadratic.
Figure (svg): A quartic reduced step by step to a quadratic by two successful divisions
The final quadratic does not factor over the integers, and its zeros are irrational — values no candidate list could ever have contained.
Worked example
Example 3, step 5 onward.
\[ \text{Finish finding the real zeros of } 10x^4 - 11x^3 - 42x^2 + 7x + 12. \]
Start again on the cubic quotient
Why: For 5x cubed minus 8x squared minus 17x plus 12, the constant is 12 and the leading coefficient 5.
Use the graph again
Why: The graph of the cubic suggests three fifths as a zero.
\[ \text{test } \frac{3}{5} \]
Divide and clear the fraction
Why: The quotient is 5x squared minus 5x minus 20, and taking out 5 turns the factor into 5x minus 3.
\[ (5 x - 3) (x ^{2} - x - 4) \]
Solve the quadratic
Why: It does not factor, so the formula gives 1 plus or minus root 17, over 2.
\[ \text{about } 2.56\text{ and } -1.56 \]
Collect every zero
Why: Two rational and two irrational.
Figure (svg): The solution to Worked example finish the quartic shown as a ladder of expressions, one row per algebraic move
\[ -\tfrac{1}{2}, \; \tfrac{3}{5}, \; \tfrac{1 \pm \sqrt{17}}{2} \]
Verify: count against the degree
Why: A quartic has at most four zeros and four were found, so the search is complete. The two irrational ones explain the crossings near negative 1.56 and 2.56 that the graph showed but no candidate matched — which is why testing had to stop at degree 2 rather than continue.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 372-372
Ranking
Finding all real zeros with no zero given.
Put in order
Why: Step two is optional when the leading coefficient is 1 and nearly essential otherwise. Step four is a loop, not a single action, and the exit condition is degree 2 rather than a fixed number of rounds.
Worked example
Guided Practice 5 and 6.
\[ \text{Find all real zeros of } 48x^3 + 4x^2 - 20x + 3 \text{ and } 2x^4 + 5x^3 - 18x^2 - 19x + 42. \]
First: try a small fraction
Why: One half is a candidate, and the tableau gives a remainder of zero.
\[ \frac{1}{2}\text{ is } a\text{ zero} \]
First: factor the quotient
Why: The quotient 48x squared plus 28x minus 6 is 2 times 24x squared plus 14x minus 3, which factors as 4x plus 3 times 6x minus 1.
\[ (2 x - 1) (4 x + 3) (6 x - 1) \]
Second: test negative 2
Why: The tableau gives a remainder of zero, leaving 2x cubed plus x squared minus 20x plus 21.
\[ -2\text{ is } a\text{ zero} \]
Second: test three halves on the cubic
Why: Another zero remainder, leaving 2x squared plus 4x minus 14.
\[ \frac{3}{2}\text{ is } a\text{ zero} \]
Second: solve the quadratic
Why: Dividing by 2 gives x squared plus 2x minus 7, whose zeros are negative 1 plus or minus 2 root 2.
\[ \text{about } 1.83\text{ and } -3.83 \]
Figure (svg): The solution to Worked example two more with awkward coefficients shown as a ladder of expressions, one row per algebraic move
\[ \{\tfrac{1}{2}, -\tfrac{3}{4}, \tfrac{1}{6}\}; \quad \{-2, \tfrac{3}{2}, -1 \pm 2\sqrt{2}\} \]
Verify: check that every rational zero is on its list
Why: For the first, the numerators 1, 3 and 1 all divide 3, and the denominators 2, 4 and 6 all divide 48 — as the theorem requires. All three zeros are fractions, which is exactly the case guessing would never find and a systematic list always does.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 373-373
Trap
\[ q(x) = x^2 - x - 4 \]
Keep testing rational candidates on the quadratic
Why: The candidates are the factors of 4 over the factors of 1.
\[ \pm 1, \pm 2, \pm 4 \text{ all fail} \;\Longrightarrow\; \text{no more zeros} \quad \text{(wrong)} \]
The quadratic does have two real zeros, at about 2.56 and negative 1.56 — they are simply irrational, so no rational candidate could ever find them.
\[ x^2 - x - 4 = 0 \;\Longrightarrow\; x = \frac{1 \pm \sqrt{17}}{2} \]
Stop testing at degree 2 and use the quadratic formula
Why: The formula finds every zero, rational or not, which the candidate list cannot.
\[ \approx 2.56 \text{ and } -1.56 \]
The rational zero theorem finds only rational zeros, and that is not a defect — it is what the theorem is for. Its job is to get the degree down to where a complete method exists.
Fill the middle
Example 3, at the end.
Fill in the blanks
x^2 - x - 4 = 0 \;\Longrightarrow\; x = \frac17}___ = \frac___}}}___
Why: The discriminant is 1 minus 4 times 1 times negative 4, which is 1 plus 16, or 17. Seventeen is not a perfect square, so the zeros are irrational — and that is why no candidate on the rational list could have found them.
Two truths and a lie
All three are about the search.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. A failed candidate list means no RATIONAL zeros, which is a much weaker statement. The polynomial x squared minus 2 has candidates 1 and 2 with both signs, all of which fail, yet it has two perfectly good real zeros at plus and minus root 2.
Prediction
Commit before reasoning.
Predict first
After dividing, the quotient has different coefficients. What happens to its candidate list?
Correct: It is usually shorter, since the constant and leading coefficient are smaller.
\[ 10x^4 - \cdots \;\to\; 5x^3 - 8x^2 - 17x + 12: \; \text{fewer candidates} \]
Why: In Example 3 the quartic's constant was 12 and its leading coefficient 10, giving sixteen candidates; the cubic quotient had constant 12 and leading coefficient 5, giving twelve. Each round of the search is therefore a little easier than the last — provided the quotient still has integer coefficients, which is why clearing the fraction from a factor is worth doing.
Section
Section 5
Concept
A volume or area model of degree 3 rarely factors by inspection. Build the equation, put it in standard form, list the candidates, and test only the ones the situation allows — which usually means only the positive ones.
\[ 4 = \tfrac{1}{3}x^2(x+1) \;\Longrightarrow\; x^3 + x^2 - 12 = 0 \]
Restricting to positive candidates halves the list before any testing, which is one of the few cases where the situation makes the algebra easier rather than harder.
Figure (svg): A square pyramid whose height exceeds its base side by one, with the volume equation beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 373-373 — Solve a multi-step problem
Picture it
Example 4: a square pyramid whose height is one foot more than its base side.
Figure (svg): A square pyramid whose height exceeds its base side by one, with the volume equation beside it
The candidate 2 works on the second test, and the remaining quadratic has imaginary roots — so the situation and the algebra agree that there is exactly one answer.
Worked example
Example 4, in the book's four steps.
\[ \text{A square pyramid has height } x+1 \text{ and base side } x, \text{ with volume } 4 \text{ ft}^3. \text{ Find } x. \]
Write the volume equation
Why: A pyramid's volume is a third of the base area times the height.
\[ 4 = (\frac{1}{3}) x ^{2}(x + 1) \]
Clear the fraction and rearrange
Why: Multiplying by 3 gives 12 equals x cubed plus x squared.
\[ 0 = x ^{3} + x ^{2} - 12 \]
List the candidates and restrict them
Why: The factors of 12 give twelve candidates, but only the positive six make sense as a length.
\[ 1, 2, 3, 4, 6, 12 \]
Test 1, then 2
Why: One gives a remainder of negative 10; 2 gives 0.
\[ x = 2 \]
Check for other solutions
Why: The quotient is x squared plus 3x plus 6, whose discriminant is negative 15, so its roots are imaginary.
Figure (svg): The solution to Worked example the ice sculpture shown as a ladder of expressions, one row per algebraic move
\[ x = 2 \;\Longrightarrow\; 2 \times 2 \times 3 \text{ ft} \]
Verify: compute the volume directly
Why: A third of 2 squared times 3 is a third of 12, which is 4 cubic feet — the required volume. The imaginary roots of the quotient are worth noticing: they mean the answer is unique not just physically but mathematically, which is not always the case in these models.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 373-373
Fill the middle
Example 4, step 1.
Fill in the blanks
4 = \tfrac12___x^2(x+1) \;\Longrightarrow\; ___ = x^3 + x^2
Why: Multiplying both sides by 3 gives 12 on the left and clears the third on the right. Doing this before anything else keeps the coefficients integers, which the rational zero theorem requires and which also makes the candidate list readable.
Worked example
Guided Practice 7. The roles of the two dimensions swap.
\[ \text{The base sides are } 1 \text{ ft longer than the height, and the volume is } 6 \text{ ft}^3. \text{ Find the dimensions.} \]
Name the variable carefully
Why: Let x be the height, so each base side is x plus 1.
\[ \text{height } x,\text{ base } x + 1 \]
Write and clear the equation
Why: Six equals a third of the quantity x plus 1, squared, times x.
\[ 18 = x(x + 1) ^{2} \]
Expand and rearrange
Why: The product is x cubed plus 2x squared plus x.
\[ 0 = x ^{3} + 2 x ^{2} + x - 18 \]
Test positive candidates
Why: One gives negative 14; 2 gives 8 plus 8 plus 2 minus 18, which is zero.
\[ x = 2 \]
State the dimensions
Why: The height is 2 feet and each base side is 3 feet.
\[ 3\text{ by } 3\text{ by } 2 \text{ft} \]
Figure (svg): The solution to Worked example the base one foot longer shown as a ladder of expressions, one row per algebraic move
\[ 3 \times 3 \times 2 \text{ ft} \]
Verify: compute the volume
Why: A third of 3 squared times 2 is a third of 18, which is 6 cubic feet. Note that naming the variable was the whole difficulty here: calling x the base side instead would have given a different and much messier equation for the same situation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 373-373
Error analysis
A student sets up the ice sculpture equation.
Annotate
On: \( 4 = \tfrac{1}{3}x^2(x+1) \;\Longrightarrow\; 0 = \tfrac{1}{3}x^3 + \tfrac{1}{3}x^2 - 4 \)
Clear every fraction before listing candidates. The theorem has a hypothesis, and integer coefficients is it.
Sorting
The situation restricts the search.
Sort into buckets
For a length x with x cubed plus x squared minus 12 equal to zero, sort each candidate.
Restricting to positive values halves the list before any arithmetic. Very few problems let the situation shorten the algebra like this.
Prediction
Commit before reasoning.
Predict first
Example 4 finds x equal to 2 and then examines the quotient anyway. Why?
Correct: Because a cubic can have up to three solutions, and another might also be sensible.
\[ x^2 + 3x + 6 = 0 \;\Longrightarrow\; x = \frac{-3 \pm i\sqrt{15}}{2} \]
Why: Had the quotient factored into two positive roots, there would have been three possible moulds and the problem would need more information to choose between them. Here the quotient's discriminant is negative 15, so its roots are imaginary and 2 is the only real solution at all — but that had to be checked rather than assumed.
Comparison
Fill the blanks. The same shape, the roles swapped.
Comparison matrix
| Question | Height one more than base | Base one more than height |
|---|---|---|
| Let x be | the base side | the height |
| The equation | 0 = x^3 + x^2 - 12 | 0 = x^3 + 2x^2 + x - 18 |
| The solution | x = 2 | x = 2 |
| The dimensions | 2 by 2 by 3 feet | 3 by 3 by 2 feet |
Both give x equal to 2, but x means something different in each, so the moulds are different. Writing down what the variable stands for is what keeps the two apart.
Comparison
Fill the blanks. No single method finds everything.
Comparison matrix
| Tool | Finds | Misses |
|---|---|---|
| Rational zero theorem | a finite list of candidates | irrational and imaginary zeros |
| Synthetic division | whether a candidate is a zero | nothing: it is decisive |
| A graph | roughly where the crossings are | exact values, and imaginary zeros |
| The quadratic formula | every zero of a quadratic | nothing, once the degree is 2 |
The strategy is to use the first three to get down to degree 2, and the fourth to finish.
Pattern
One routine for finding every real zero.
Stop testing at degree 2. Continuing would find only rational zeros and would miss any irrational ones the quadratic has.
OpenStax Algebra and Trigonometry 2e, §5.5 Zeros of Polynomial Functions §5.5
Check
Listing candidates. Include both signs.
Check your understanding
List the possible rational zeros of f(x) = 2x^3 + 3x^2 - 11x - 6.
Answer: A
Why: The constant -6 has factors 1, 2, 3, 6 and the leading coefficient 2 has factors 1 and 2.
Check
Finding all zeros. Do not stop at one.
Check your understanding
Find all real zeros of f(x) = x^3 - 8x^2 + 11x + 20.
Answer: A
Why: Testing -1 gives a remainder of 0 and a quotient of x^2 - 9x + 20, which factors as (x - 4)(x - 5).
Check
A model. Clear fractions first.
Check your understanding
A square pyramid has base side x, height x + 1 and volume 4 cubic feet. What is x?
Answer: A
Why: The equation x^3 + x^2 - 12 = 0 has 2 as its only real solution.
Real world
A box is made from a sheet of card 10 inches by 12 inches by cutting a square of side x from each corner and folding up the sides. The finished box is to hold 96 cubic inches.
Discussion prompt
Write the volume equation, list the candidates, and find every value of x that works. Which are physically possible?
Hint: The base measures 10 minus 2x by 12 minus 2x, and the height is x.
Answer:
\[ x(10-2x)(12-2x) = 96 \;\Longrightarrow\; 4x^3 - 44x^2 + 120x - 96 = 0 \;\Longrightarrow\; x^3 - 11x^2 + 30x - 24 = 0 \]
\[ \text{candidates } \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, \pm24; \; \text{test } 4: \; 64 - 176 + 120 - 24 = -16 \]
\[ \text{test } 2: \; 8 - 44 + 60 - 24 = 0 \;\Longrightarrow\; (x-2)(x^2 - 9x + 12) = 0 \]
\[ x = \frac{9 \pm \sqrt{33}}{2} \approx 1.63 \text{ or } 7.37 \]
Three real solutions: x equal to 2, about 1.63, and about 7.37. Only the first two are possible, because a cut of 7.37 inches would remove more than the 10-inch side allows.
Two things are worth noticing. Dividing through by 4 first made the leading coefficient 1 and cut the candidate list from dozens to sixteen. And two different cuts really do give the same volume — a squat wide box and a taller narrow one — so the situation does not pick a unique answer here, unlike the pyramid.
Commit first
Answer, then rate your confidence honestly.
Predict first
Every candidate on a polynomial's rational zero list fails. What follows?
Correct: It has no rational zeros, but may have irrational or imaginary ones.
\[ x^2 - 2: \; \text{candidates } \pm1, \pm2 \text{ all fail; zeros } \pm\sqrt{2} \]
Why: The theorem constrains rational zeros only. The polynomial x squared minus 2 has candidates 1 and 2 with both signs, every one of which fails, and yet it has two real zeros at plus and minus root 2. An odd-degree polynomial always has at least one real zero by the end-behaviour argument of Lesson 5.2, so a failed list on a cubic guarantees an irrational zero exists.
Explain it
They can factor with a given zero and do not know where the first zero comes from.
Discussion prompt
In four sentences or fewer, explain how to find a zero when none is given, and why the search is finite.
Hint: Talk about the two coefficients at the ends.
Answer:
If the polynomial has integer coefficients, any rational zero has to be a factor of the constant term divided by a factor of the leading coefficient. That turns an infinite search into a list of a dozen or so numbers, which you test one at a time with synthetic division.
A zero remainder means you have found a zero and also hands you a polynomial one degree lower to work on. Keep going until what is left is a quadratic, and then use the formula, which finds every remaining zero whether it is rational or not.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the list, write the plus-or-minus in front of every factor as you go, and work numerator by numerator. For choosing, sketch the graph or restrict to what the situation allows. For tableaux, write the coefficient row once at the top and reuse it. For stopping, the exit condition is always degree 2. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take f of x equals 10x to the fourth minus 11x cubed minus 42x squared plus 7x plus 12 and work the whole search on one page. Top left: build the candidate list in full, showing the two factor lists and the fractions they produce, and count how many distinct values there are. Top right: sketch the graph well enough to mark the four crossings, and circle the candidates nearest each one. Middle: run the two successful tableaux, showing how each quotient is read and how the fraction is cleared from each factor. Bottom: solve the final quadratic with the formula and list all four zeros, marking which two the candidate list could never have found. In a margin, write one sentence saying why the search stops at degree 2.
If your candidate list has more than sixteen distinct values, check for duplicates: 2 over 2 and 10 over 5 are both just 2 and should appear only once.
Recap
Five things, and together they remove the last crutch from Lesson 5.5.
| If you see | Then |
|---|---|
| Integer coefficients | The rational zero theorem applies |
| Fractional coefficients | Clear them first |
| A leading coefficient of 1 | The candidates are all integers |
| A long candidate list | Sketch the graph and narrow it |
| A zero remainder | You have a zero and a quotient |
| A quadratic quotient | Stop testing; use the formula |
| Every candidate failing | No rational zeros; others may exist |
This lesson finds real zeros. Lesson 5.7 counts them properly: the fundamental theorem of algebra guarantees exactly n zeros for degree n, once imaginary zeros are included.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.6 Find Rational Zeros §5.6, pp. 370-378 — everything on these slides traces back here
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