The fundamental theorem of algebra and its corollary counting zeros with multiplicity, finding every zero including imaginary ones, the complex and irrational conjugate theorems, building a polynomial from given zeros, Descartes' rule of signs, and approximating real zeros in a model.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 5 — Polynomials and Polynomial Functions
Apply the Fundamental Theorem of Algebra
Objectives
Five outcomes. The first tells you how many to look for; the rest find them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 379-385 — the lesson these objectives are drawn from
Warm-up
Lesson 5.6 found real zeros by testing candidates. It never said how many there were supposed to be.
Discussion prompt
Factor x cubed minus 5x squared minus 8x plus 48. How many distinct solutions does the equation have, and does that sit comfortably with the degree?
Hint: Test 4 as a candidate, twice.
Answer:
\[ x^3 - 5x^2 - 8x + 48 = (x+3)(x-4)^2 \]
There are only two distinct solutions, negative 3 and 4 — which looks like a cubic falling short. But the factor x minus 4 appears twice, so counting 4 twice gives three solutions in all. Counting that way is what makes the number of solutions exactly the degree, every time.
Concept
Every polynomial of degree n has exactly n zeros, provided repeated zeros are counted once for each time their factor appears and imaginary zeros are allowed. That turns a vague at most into an exact count, so you always know when the search is finished.
repeated solution — A solution that arises from a factor appearing more than once. A factor appearing twice contributes two solutions to the count, even though they are the same number.
\[ \deg f = n \;\Longrightarrow\; f \text{ has exactly } n \text{ zeros in } \mathbb{C} \]
Two things had to be added for this to work: Lesson 4.6's imaginary numbers, and the convention of counting multiplicity. Neither is a trick; both were introduced precisely so that the count could come out right.
Figure (svg): A cubic factored into three linear factors, with a repeated factor counted twice
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 379-379
Section
Section 1
Concept
The fundamental theorem of algebra says every polynomial of positive degree has at least one complex zero. Its corollary says a polynomial of degree n has exactly n zeros, with a zero repeated k times counted k times.
fundamental theorem of algebra — If f is a polynomial of degree n greater than zero, then the equation f of x equals zero has at least one solution among the complex numbers. Its corollary gives exactly n solutions when repeated solutions are counted with multiplicity.
\[ x^3 - 5x^2 - 8x + 48 = 0 \;\Longrightarrow\; -3, \; 4, \; 4 \]
The theorem was first proved by Gauss. Notice how much it needs: without imaginary numbers a quadratic could have no zeros, and without multiplicity a cubic could have two.
Figure (svg): A cubic factored into three linear factors, with a repeated factor counted twice
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 379-379 — The Fundamental Theorem of Algebra
Picture it
The warm-up cubic, split into its three factors.
Figure (svg): A cubic factored into three linear factors, with a repeated factor counted twice
Three linear factors, three solutions. Two of them happen to be the same number, and the count treats them as separate because the factorisation does.
Worked example
Example 1. The degree alone answers it.
\[ \text{How many solutions has } x^3 + 5x^2 + 4x + 20 = 0, \text{ and how many zeros has } x^4 - 8x^3 + 18x^2 - 27? \]
Read the first degree
Why: The highest power is 3, so the equation has exactly three solutions.
Note what they turn out to be
Why: They are negative 5, 2i and negative 2i — one real and a conjugate pair.
\[ -5, 2 i, -2 i \]
Read the second degree
Why: The highest power is 4, so the function has exactly four zeros.
Note what they turn out to be
Why: They are negative 1, 3, 3 and 3 — one simple zero and one repeated three times.
\[ -1, 3, 3, 3 \]
Figure (svg): The solution to Worked example count without solving shown as a ladder of expressions, one row per algebraic move
\[ 3 \text{ solutions}; \quad 4 \text{ zeros} \]
Verify: check that both counts need the conventions
Why: The first has only one real solution, so without imaginary numbers the count would be 1 rather than 3. The second has only two distinct zeros, so without multiplicity the count would be 2 rather than 4. Both conventions are doing real work in these two examples.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 379-379
Matching
Reorder if necessary, then read the degree.
Match the pairs
Why: The last one arrives out of standard form, so the degree has to be found rather than read from the first term. Two of these share a count of 4 despite looking nothing alike, which is the point: only the degree matters.
Worked example
Guided Practice 1 and 2, plus two exercises.
\[ \text{Count for } x^4 + 5x^2 - 36 = 0, \; x^3 + 7x^2 + 8x - 16, \; 5y^3 - 3y^2 + 8y = 0, \; 6x^6 + 19x^5 - 22x^3 + 16x - 3. \]
Read the degrees in order
Why: Four, then 3, then 3.
\[ 4, 3, 3 \]
Watch the fourth
Why: The terms are out of order, so reorder before reading: the highest power is 6.
\[ ^\circ 6 \]
State the counts
Why: Four solutions, three zeros, three solutions and six zeros.
\[ 4, 3, 3, 6 \]
Note what is irrelevant
Why: The number of terms, the coefficients and the constant play no part at all.
Figure (svg): The solution to Worked example four more counts shown as a ladder of expressions, one row per algebraic move
\[ 4, \; 3, \; 3, \; 6 \]
Verify: check the third for a hidden zero
Why: The third factors as y times the quantity 5y squared minus 3y plus 8, so one solution is y equal to zero and the quadratic supplies two more with a negative discriminant. Three solutions, one real and two imaginary — matching the count from the degree alone.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 379-383
Trap
\[ f(x) = x^4 - 8x^3 + 18x^2 - 27 = (x+1)(x-3)^3 \]
Count the different numbers that appear
Why: Negative 1 and 3 are the only values that make f zero.
\[ \text{two zeros} \quad \text{(wrong count)} \]
The corollary promises four zeros for a quartic, and this factorisation has four linear factors — three of them identical.
\[ (x+1)(x-3)(x-3)(x-3) \;\Longrightarrow\; -1, \; 3, \; 3, \; 3 \]
Count one zero for each linear factor
Why: A factor appearing three times contributes three to the count.
\[ \text{four zeros, two distinct} \]
Both statements are true and they answer different questions. How many zeros counts factors; how many distinct zeros counts values, and only the first is fixed by the degree.
Fill the middle
The warm-up cubic.
Fill in the blanks
(x+3)(x-4)^2 = 0 \;\Longrightarrow\; \text4 -3, \; 4, \; ___
Why: The factor x minus 4 appears twice, so 4 is counted twice and the total is three — matching the degree. Writing the repeated solution out rather than mentioning it once is what makes the count come out right.
Two truths and a lie
All three are about the count.
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. Distinct is the word that breaks it: a quintic could have a single zero repeated five times, as x to the fifth does. The count of n applies to zeros with multiplicity, and the number of distinct zeros can be anything from 1 to n.
Prediction
Commit before reasoning.
Predict first
A quadratic with a negative discriminant has no real zeros. How does the corollary still promise two?
Correct: The two zeros are imaginary, and the theorem counts complex zeros.
\[ x^2 + 1 = 0 \;\Longrightarrow\; x = i, \; -i: \; \text{two zeros} \]
Why: The theorem is stated over the complex numbers, so a quadratic missing the axis still has two zeros, forming a conjugate pair. Without Lesson 4.6's imaginary unit the count would fail immediately at degree 2. That is a good part of why complex numbers were introduced at all: they make the count exact rather than merely an upper bound.
Section
Section 2
Concept
Use the rational zero theorem to find rational zeros and divide them out. Once the quotient is a quadratic, the quadratic formula supplies the remaining zeros, which may be real or imaginary, completing the count promised by the degree.
\[ f(x) = (x+1)^2(x-2)(x^2 - 4x + 7) \]
Only real zeros show up on a graph. A factor raised to an even power makes the graph tangent to the axis, and one raised to an odd power makes it cross.
Figure (svg): A quintic graph showing only its real zeros, with the repeated zero touching the axis
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 380-380 — Find the zeros of a polynomial function
Picture it
Example 2: a quintic with a repeated real zero and a conjugate pair.
Figure (svg): A quintic graph showing only its real zeros, with the repeated zero touching the axis
The curve touches the axis at negative 1 without crossing, because that factor is squared, and crosses at 2, where the factor is to the first power. The imaginary pair is nowhere on the picture.
Worked example
Example 2, in the book's three steps.
\[ \text{Find all zeros of } f(x) = x^5 - 4x^4 + 4x^3 + 10x^2 - 13x - 14. \]
Count first, then list candidates
Why: The degree is 5, so there are five zeros; the candidates are the factors of 14.
\[ +- 1, +- 2, +- 7, +- 14 \]
Test candidates and divide
Why: Negative 1 is a zero twice over, and 2 is a zero.
\[ -1, -1, 2 \]
Read the remaining quotient
Why: After three divisions the quotient is x squared minus 4x plus 7.
\[ x ^{2} - 4 x + 7 \]
Use the quadratic formula
Why: The discriminant is 16 minus 28, or negative 12, so the roots are 2 plus or minus i root 3.
\[ 2 + - i \sqrt{3} \]
Collect all five
Why: Two copies of negative 1, then 2, then the conjugate pair.
Figure (svg): The solution to Worked example find all five zeros shown as a ladder of expressions, one row per algebraic move
\[ -1, \; -1, \; 2, \; 2 \pm i\sqrt{3} \]
Verify: check the count and the graph
Why: Five zeros for a degree-5 polynomial, as the corollary promises. The graph meets the axis at only two places, negative 1 and 2, because the imaginary pair cannot appear there — and it is tangent at negative 1 because that factor is squared.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 380-380
Sorting
Look at the power on each factor.
Sort into buckets
For f(x) = (x + 1)^2 (x - 2)(x - 5)^3 (x^2 + 1), sort each real zero.
Even power touches, odd power crosses, imaginary shows nothing. Three rules that let a factorisation be read straight into a sketch.
Worked example
Guided Practice 3 and 4.
\[ \text{Find all zeros of } x^3 + 7x^2 + 15x + 9 \text{ and } x^5 - 2x^4 + 8x^2 - 13x + 6. \]
First: test negative 1
Why: Negative 1 plus 7 minus 15 plus 9 is zero.
\[ -1\text{ is } a\text{ zero} \]
First: factor the quotient
Why: The quotient is x squared plus 6x plus 9, a perfect square.
\[ (x + 3) ^{2} \]
Second: note the missing power
Why: The x cubed term is absent, so its coefficient is 0 in every tableau.
\[ 1, -2, 0, 8, -13, 6 \]
Second: test 1 twice
Why: One is a zero, and it is a zero of the quotient too; then negative 2 is a zero of the next quotient.
\[ 1, 1, -2 \]
Second: solve the final quadratic
Why: The quotient is x squared minus 2x plus 3, whose discriminant is negative 8.
\[ 1 + - i \sqrt{2} \]
Figure (svg): The solution to Worked example two more full searches shown as a ladder of expressions, one row per algebraic move
\[ \{-1, -3, -3\}; \quad \{1, 1, -2, 1 \pm i\sqrt{2}\} \]
Verify: check both counts
Why: The first has degree 3 and three zeros; the second has degree 5 and five. Both have a repeated real zero, which the graph would show as a tangency rather than a crossing — the first touches the axis at negative 3 and crosses at negative 1.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 380-380
Error analysis
A student reads a quintic's graph and reports its zeros.
Annotate
On: \( \text{the graph meets the axis at } -1 \text{ and } 2, \text{ so } f \text{ has two zeros} \)
A graph undercounts twice over: it hides imaginary zeros and shows a repeated zero as one point. Count from the degree, not from the picture.
Fill the middle
Example 2, at the quadratic.
Fill in the blanks
x^2 - 4x + 7 = 0 \;\Longrightarrow\; x = \frac3}___ = 2 \pm i\sqrt___}
Why: The radicand is negative 12, whose root is 2i root 3, and dividing by 2 leaves i root 3. The two zeros differ only in the sign of that imaginary part, which is the complex conjugate theorem showing itself.
Ranking
Finding every zero of a polynomial.
Put in order
Why: Step one is new in this lesson and it is what makes step five possible: knowing the total in advance turns the final check into a proof that nothing was missed. Without it there is no way to know whether the search is finished.
Prediction
Commit before reasoning.
Predict first
Could a quartic have the zeros i, i, negative i and negative i?
Correct: Yes — it is the quantity x squared plus 1, squared.
\[ (x^2+1)^2 = x^4 + 2x^2 + 1: \; \text{zeros } i, i, -i, -i \]
Why: Expanding gives x to the fourth plus 2x squared plus 1, a quartic with real coefficients whose four zeros are i, i, negative i and negative i. Its graph never meets the axis at all, and it has no real zeros of any kind. Multiplicity and imaginariness are independent properties, and either can happen with or without the other.
Section
Section 3
Concept
If a polynomial has real coefficients and an imaginary zero, its conjugate is a zero too. If it has rational coefficients and an irrational zero of the form a plus root b, then a minus root b is also a zero. So a polynomial of least degree must include both members of every such pair.
irrational conjugates — The pair a plus root b and a minus root b, where a and b are rational and root b is irrational. If one is a zero of a polynomial with rational coefficients, so is the other.
\[ a + bi \text{ a zero} \;\Longrightarrow\; a - bi \text{ a zero} \]
The two theorems have different hypotheses: real coefficients for the complex version and rational coefficients for the irrational one. Reading which applies is part of the problem.
Figure (svg): The complex and irrational conjugate theorems, each with a pair of zeros
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 380-381 — Complex and Irrational Conjugates Theorems
Picture it
Complex conjugates and irrational conjugates side by side.
Figure (svg): The complex and irrational conjugate theorems, each with a pair of zeros
Both say the same kind of thing: a coefficient restriction forces zeros to come in partners. Neither says anything about rational zeros, which can appear alone.
Worked example
Example 3. The missing conjugate is supplied first.
\[ \text{Write a polynomial of least degree with rational coefficients, leading coefficient } 1, \text{ and zeros } 3 \text{ and } 2+\sqrt{5}. \]
Supply the missing conjugate
Why: The coefficients are rational and 2 plus root 5 is a zero, so 2 minus root 5 must be one too.
Write the factors
Why: Each zero k gives a factor x minus k, by the factor theorem.
\[ (x - 3) [x - (2 + \sqrt{5})] [x - (2 - \sqrt{5})] \]
Regroup the conjugate pair
Why: Writing each as x minus 2, minus or plus root 5, makes the pair a difference of squares.
\[ [(x - 2) ^{2} - 5] \]
Expand and simplify
Why: The bracket is x squared minus 4x plus 4 minus 5, or x squared minus 4x minus 1.
\[ (x - 3) (x ^{2} - 4 x - 1) \]
Multiply out
Why: The product is x cubed minus 7x squared plus 11x plus 3.
\[ x ^{3} - 7 x ^{2} + 11 x + 3 \]
Figure (svg): The solution to Worked example build from given zeros shown as a ladder of expressions, one row per algebraic move
\[ f(x) = x^3 - 7x^2 + 11x + 3 \]
Verify: evaluate at the given zeros
Why: At x equal to 3: 27 minus 63 plus 33 plus 3 is zero. At 2 plus root 5 the terms combine to 38 plus 17 root 5, minus 63 minus 28 root 5, plus 22 plus 11 root 5, plus 3 — the rational parts give zero and the root 5 parts give zero. Both check, and the third zero follows from the irrational conjugates theorem without further work.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 381-381
Matching
Only irrational and imaginary zeros need partners.
Match the pairs
Why: A rational zero stands alone, which is why the rational zero theorem of Lesson 5.6 could hunt for them one at a time. Irrational and imaginary zeros cannot, and that difference is what makes the least degree of a polynomial larger than the number of zeros you were told about.
Worked example
Guided Practice 5 to 8. Count the conjugates before starting.
\[ \text{Build from zeros } \{-1,2,4\}, \; \{4, 1+\sqrt{5}\}, \; \{2, 2i, 4-\sqrt{6}\}, \; \{3, 3-i\}. \]
First: no conjugates needed
Why: All three zeros are rational, so the degree is 3.
\[ (x + 1) (x - 2) (x - 4) \]
Second: add the irrational partner
Why: One minus root 5 joins, giving degree 3.
\[ (x - 4) [(x - 1) ^{2} - 5] \]
Third: add two partners
Why: Negative 2i joins 2i, and 4 plus root 6 joins 4 minus root 6, giving degree 5.
\[ (x - 2) (x ^{2} + 4) (x ^{2} - 8 x + 10) \]
Fourth: add the complex partner
Why: Three plus i joins 3 minus i, giving degree 3.
\[ (x - 3) [(x - 3) ^{2} + 1] \]
Multiply each out
Why: The four results are cubics, cubics, a quintic and a cubic.
Figure (svg): The solution to Worked example four more constructions shown as a ladder of expressions, one row per algebraic move
\[ x^3-5x^2+2x+8; \; x^3-6x^2+4x+16; \; x^3-9x^2+28x-30 \]
Verify: check one constant term
Why: For the last, the three zeros 3, 3 minus i and 3 plus i multiply to 3 times the quantity 9 plus 1, which is 30; for a monic cubic the constant term is the negative of that, namely negative 30 — matching. Conjugate pairs always multiply to a real number, which is why the coefficients come out real.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 381-381
Trap
\[ \text{zeros } 3 \text{ and } 2 + \sqrt{5}, \text{ rational coefficients} \]
Use only the two zeros given
Why: The polynomial is built as a product of two linear factors.
\[ (x-3)[x - (2+\sqrt{5})] = x^2 - (5 + \sqrt{5})x + 6 + 3\sqrt{5} \quad \text{(wrong)} \]
The coefficients contain root 5, so they are not rational — but rational coefficients were required.
\[ (x-3)[(x-2)^2 - 5] = x^3 - 7x^2 + 11x + 3 \]
Include the conjugate of every irrational or imaginary zero
Why: The pair multiplies to a rational quadratic, which is what keeps the coefficients rational.
\[ [x-(2+\sqrt5)][x-(2-\sqrt5)] = (x-2)^2 - 5 = x^2 - 4x - 1 \]
The conjugate is not an extra requirement invented for the exercise; it is forced by the coefficients being rational. Leaving it out makes the requested polynomial impossible.
Fill the middle
Example 3, at the regrouping.
Fill in the blanks
[(x-2) - \sqrt5][(x-2) + \sqrt___] = (x-2)^2 - ___
Why: The pair is a difference of squares with a equal to x minus 2 and b equal to root 5, so the product is x minus 2, squared, minus 5. That is why the radical vanishes: squaring root 5 makes it rational, exactly as it did for the conjugate denominators of Lesson 4.5.
Prediction
Commit before reasoning.
Predict first
A polynomial with real coefficients has zeros 2, 2i and 4 minus root 6. What is its least possible degree?
Correct: Five, once both missing conjugates are added.
\[ (x-2)(x^2+4)(x^2-8x+10): \; \text{degree } 1+2+2 = 5 \]
Why: The imaginary zero 2i forces negative 2i, and if the coefficients are also rational then 4 minus root 6 forces 4 plus root 6. Three given zeros become five, so the least degree is 5. Counting the forced partners before writing anything is the way to know how large the answer will be.
Comparison
Fill the blanks. Similar statements, different hypotheses.
Comparison matrix
| Question | Complex conjugates | Irrational conjugates |
|---|---|---|
| Coefficients must be | real | rational |
| The pair | a + bi and a - bi | a + sqrt(b) and a - sqrt(b) |
| Their product is | a^2 + b^2, real | a^2 - b, rational |
| Contributes degree | 2 | 2 |
The hypothesis difference matters: a polynomial with real but irrational coefficients may have root 5 as a zero without having negative root 5, but one with rational coefficients cannot.
Section
Section 4
Concept
The number of positive real zeros equals the number of sign changes in the coefficients of f, or is less than that by an even number. The number of negative real zeros is found the same way from the coefficients of f of negative x.
Descartes' rule of signs — For a polynomial with real coefficients, the number of positive real zeros equals the number of sign changes in its coefficients or is less by an even number, and the number of negative real zeros is found the same way from f of negative x.
\[ + \; - \; + \; - \; - \; - \; -: \; \text{three changes} \]
The less-by-an-even-number clause is exactly the conjugate theorem in disguise: real zeros can only be lost two at a time, because they are replaced by a conjugate pair.
Figure (svg): Sign changes counted in a polynomial and in the same polynomial with x negated, with a table of possible zero counts
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 381-382 — Descartes' Rule of Signs
Picture it
Example 4: a sextic, with three sign changes each way.
Figure (svg): Sign changes counted in a polynomial and in the same polynomial with x negated, with a table of possible zero counts
Four possible splits, all totalling six. The rule does not decide between them, but it rules out a great deal — six positive zeros, for instance, is impossible.
Worked example
Example 4. Two counts and a table.
\[ \text{Classify the zeros of } f(x) = x^6 - 2x^5 + 3x^4 - 10x^3 - 6x^2 - 8x - 8. \]
Count sign changes in f
Why: The signs are plus, minus, plus, minus, minus, minus, minus: changes at positions 1 to 2, 2 to 3 and 3 to 4.
Conclude about positive zeros
Why: There are 3 or 1 positive real zeros, since the count drops by even amounts.
\[ 3\text{ or } 1 \]
Compute f of negative x
Why: Even powers keep their signs and odd powers flip: plus, plus, plus, plus, minus, plus, minus.
Conclude about negative zeros
Why: There are 3 or 1 negative real zeros.
\[ 3\text{ or } 1 \]
Tabulate the possibilities
Why: Each row's imaginary count makes the total 6.
Figure (svg): The solution to Worked example apply the rule shown as a ladder of expressions, one row per algebraic move
\[ (3,3,0), \; (3,1,2), \; (1,3,2), \; (1,1,4) \]
Verify: check the totals
Why: Every row sums to 6, which the corollary requires. Note that the imaginary count is always even, because imaginary zeros of a real polynomial come in conjugate pairs — so an odd total for the imaginary column would signal an error immediately.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 382-382
Fill the middle
Example 4.
Fill in the blanks
f(x) = x^6 - 2x^5 + 3x^4 - \cdots \;\Longrightarrow\; f(-x) = x^6 + 2x^5 + 3x^4 - \cdots
Why: The fifth power of negative x is negative x to the fifth, and the coefficient of negative 2 turns that into positive 2x to the fifth. Even powers are unchanged, so the sixth and fourth power terms keep their signs exactly.
Worked example
The rule applied to two functions already met in this chapter.
\[ \text{Classify the zeros of } x^3 + 5x^2 + 4x + 20 \text{ and } x^3 - 8x^2 + 11x + 20. \]
First: count changes in f
Why: All four coefficients are positive, so there are no sign changes.
\[ 0\text{ positive zeros} \]
First: count changes in f of negative x
Why: The signs become minus, plus, minus, plus: three changes.
\[ 3\text{ or } 1\text{ negative} \]
First: reconcile with the actual zeros
Why: The zeros are negative 5, 2i and negative 2i, so 1 negative and 2 imaginary.
\[ (0, 1, 2) \]
Second: count both ways
Why: For f the signs are plus, minus, plus, plus: two changes; for f of negative x they are minus, minus, minus, plus: one change.
Second: reconcile
Why: The zeros are negative 1, 4 and 5, which is 2 positive and 1 negative.
\[ (2, 1, 0) \]
Figure (svg): The solution to Worked example two shorter cases shown as a ladder of expressions, one row per algebraic move
\[ (0,1,2); \qquad (2,1,0) \]
Verify: notice what the rule proved
Why: For the first, no sign changes means no positive zeros at all — a definite conclusion reached without any testing, and one that halves the candidate list of Lesson 5.6 before it is even written down. That is the rule's practical value.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 382-382
Error analysis
A student computes f of negative x to apply the rule.
Annotate
On: \( f(x) = x^3 - 8x^2 + 11x + 20 \;\Longrightarrow\; f(-x) = -x^3 + 8x^2 - 11x - 20 \)
Substituting negative x flips only the odd-power terms. Writing the exponents down and marking which are odd takes a second and settles it.
Sorting
Substitute negative x and see.
Sort into buckets
In f(-x), sort each term of f(x) = x^6 - 2x^5 + 3x^4 - 10x^3 - 6x^2 - 8x - 8.
Exactly half the terms flip in a polynomial with all powers present, and the pattern alternates — which is why the sign count for f of negative x can differ so much from the one for f.
Prediction
Commit before reasoning.
Predict first
Descartes' rule says the count of positive zeros may be less than the sign-change count by an even number. Why even?
Correct: Real zeros are lost two at a time, replaced by a conjugate pair.
\[ \text{imaginary zeros come in pairs} \;\Longrightarrow\; \text{real count changes by } 2 \]
Why: A polynomial with real coefficients has imaginary zeros only in conjugate pairs, so the number of imaginary zeros is always even. Since the total is fixed by the degree, the real count can only change in steps of two. The even in Descartes' rule is the conjugate theorem restated, which is why the two ideas belong in the same lesson.
Comparison
Fill the blanks. Three tools, three kinds of information.
Comparison matrix
| Tool | Tells you | Does not tell you |
|---|---|---|
| The corollary | the exact total number of zeros | how they split into real and imaginary |
| Descartes' rule | the possible splits | which split is correct |
| Rational zero theorem | a finite list of rational candidates | which candidates work |
| Synthetic division | whether a candidate is a zero | nothing: it settles the question |
Only the last is decisive. The other three narrow the search, which is what makes an otherwise unbounded problem finite.
Section
Section 5
Concept
A model's coefficients are often decimals, so the rational zero theorem does not apply. The real zeros are then found graphically, using a calculator's zero feature, which locates a crossing to several decimal places.
\[ s(x) = 0.00547x^3 - 0.225x^2 + 3.62x - 11.0 \]
Only real zeros can be approximated this way, since imaginary zeros do not appear on a graph. The corollary still says how many zeros there are in total.
Figure (svg): A cubic boat-speed model with the real zero located where the speed reaches fifteen miles per hour
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 382-383 — Approximate real zeros
Picture it
Example 6: finding the shaft speed that gives 15 miles per hour.
Figure (svg): A cubic boat-speed model with the real zero located where the speed reaches fifteen miles per hour
One real zero at about 19.9, and x is measured in hundreds of revolutions per minute, so the tachometer reads about 1990.
Worked example
Example 5. Two crossings, both irrational.
\[ \text{Approximate the real zeros of } x^6 - 2x^5 + 3x^4 - 10x^3 - 6x^2 - 8x - 8. \]
Recall what the rule predicted
Why: Descartes' rule allowed 3 or 1 positive and 3 or 1 negative real zeros.
Graph and count the crossings
Why: The curve meets the axis exactly twice.
Use the calculator's zero feature
Why: The crossings are at about negative 0.73 and about 2.73.
\[ -0.73\text{ and } 2.73 \]
Reconcile with the rule
Why: One positive and one negative real zero means four imaginary ones, which is the last row of the table.
\[ (1, 1, 4) \]
Figure (svg): The solution to Worked example approximate the zeros shown as a ladder of expressions, one row per algebraic move
\[ x \approx -0.73, \; 2.73 \]
Verify: check the total
Why: Two real and four imaginary is six, matching the degree. Descartes' rule had allowed four splits and the graph decided between them — the two tools together give a complete classification that neither could give alone.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 382-382
Fill the middle
Example 6, at the substitution.
Fill in the blanks
15 = 0.00547x^3 - 0.225x^2 + 3.62x - 11.0 \;\Longrightarrow\; 0 = \cdots - 26
Why: Subtracting 15 from both sides turns the constant negative 11.0 into negative 26.0. Only the constant changes, since the other terms contain x, and the zeros of the rearranged function are the x values giving a speed of exactly 15.
Worked example
Example 6 and Guided Practice 12.
\[ \text{With } s(x) = 0.00547x^3 - 0.225x^2 + 3.62x - 11.0, \text{ find } x \text{ when } s = 15 \text{ and when } s = 20. \]
Set the speed and rearrange
Why: Substituting 15 and moving it across gives a constant of negative 26.0.
\[ 0 =... - 26.0 \]
Approximate the real zero
Why: The graph crosses once, at about 19.9.
\[ x\text{ about } 19.9 \]
Convert to the model's units
Why: X is measured in hundreds of revolutions per minute.
\[ \text{about } 1990 RPM \]
Repeat for 20 miles per hour
Why: The constant becomes negative 31.0 and the crossing moves to about 23.1.
\[ \text{about } 2310 RPM \]
Figure (svg): The solution to Worked example the tachometer shown as a ladder of expressions, one row per algebraic move
\[ 1990 \text{ RPM}; \qquad 2310 \text{ RPM} \]
Verify: substitute back
Why: At x equal to 19.9: 0.00547 times 7881 is about 43.1; minus 0.225 times 396 is about negative 89.1; plus 3.62 times 19.9 is 72.0; minus 11.0 gives about 15.0 miles per hour. The reading checks. Note that raising the speed by a third raised the shaft rate by only about a sixth, because the model is not proportional.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 383-383
Error analysis
A student starts on a model with decimal coefficients.
Annotate
On: \( 0 = 0.00547x^3 - 0.225x^2 + 3.62x - 26.0: \text{ candidates are factors of } 26 \)
Check the hypothesis before using a theorem. When it fails, a graph is the honest tool and an approximation is the honest answer.
Sorting
Look at the coefficients and the kind of zero.
Sort into buckets
Sort each situation by the method that will find the zero.
The last bucket is honest rather than defeatist: the corollary still says how many zeros there are, even when this course cannot produce them.
Prediction
Commit before reasoning.
Predict first
The calculator reports a zero at 19.863247. How should the tachometer reading be given?
Correct: About 1990 RPM, matching the model's precision.
\[ 19.863247 \;\to\; 19.9 \;\to\; 1990 \text{ RPM} \]
Why: The model's coefficients carry three significant digits, so its output cannot be trusted to seven. Reporting 1986.3247 claims a precision the model does not have, exactly as multiplying models did in Lesson 5.3. The third option confuses the units — x is in hundreds of revolutions per minute — and the fourth forgets to convert at all.
Ranking
Answering a question with a decimal-coefficient model.
Put in order
Why: Counting the crossings before approximating matters: if there were two, the situation would have to decide which one is meant. Step five does two jobs — converting hundreds of RPM into RPM, and cutting a seven-digit calculator reading back to three significant digits.
Comparison
Fill the blanks. Each answers a different question about the same zeros.
Comparison matrix
| Theorem | Answers | Lesson |
|---|---|---|
| Factor theorem | is x - k a factor? | 5.5 |
| Rational zero theorem | which rational numbers could be zeros? | 5.6 |
| Fundamental theorem | how many zeros are there in all? | 5.7 |
| Conjugate theorems | which zeros force partners? | 5.7 |
Together they turn finding zeros from a search with no end into a procedure with a known finish line.
Pattern
One routine for classifying and finding every zero.
The search is finished when the number of zeros found, counted with multiplicity, equals the degree — not when the graph runs out of crossings.
OpenStax Algebra and Trigonometry 2e, §5.5 Zeros of Polynomial Functions §5.5
Check
Counting. Read the degree.
Check your understanding
How many zeros does f(x) = x^4 - 8x^3 + 18x^2 - 27 have?
Answer: A
Why: The degree is 4, so there are exactly four zeros counted with multiplicity: -1, 3, 3 and 3.
Check
Conjugates. Count the forced partners.
Check your understanding
A polynomial with rational coefficients has zeros 3 and 2 + sqrt(5). What is its least degree?
Answer: A
Why: The irrational conjugates theorem forces 2 - sqrt(5) to be a zero too, giving three zeros in all.
Check
Descartes' rule. Count sign changes.
Check your understanding
How many positive real zeros can f(x) = x^3 + 5x^2 + 4x + 20 have?
Answer: A
Why: All four coefficients are positive, so there are no sign changes and therefore no positive real zeros.
Real world
An engineer needs a polynomial with real coefficients, leading coefficient 1, whose zeros describe a system's natural frequencies: 0 with multiplicity 2, and 3 plus 4i.
Discussion prompt
Write the polynomial of least degree, state that degree, and say how many of its zeros would show up on a graph.
Hint: Real coefficients force a partner for the imaginary zero, and a repeated zero contributes its factor twice.
Answer:
\[ \text{zeros: } 0, 0, 3+4i, 3-4i \;\Longrightarrow\; \text{degree } 4 \]
\[ f(x) = x^2[(x-3)^2 + 16] = x^2(x^2 - 6x + 25) = x^4 - 6x^3 + 25x^2 \]
The degree is 4 and only one point appears on the graph: the origin, where the curve is tangent to the axis because the factor is squared.
Two things are worth noticing. The imaginary pair multiplied to x squared minus 6x plus 25, a real quadratic, which is why the coefficients came out real — that is the complex conjugate theorem doing its job. And the graph shows one point for four zeros, which is exactly why the corollary counts factors rather than crossings.
Commit first
Answer, then rate your confidence honestly.
Predict first
A polynomial of degree 7 with real coefficients has how many real zeros, at minimum?
Correct: At least one, because imaginary zeros come in pairs and 7 is odd.
\[ 7 = \text{even} + \text{real} \;\Longrightarrow\; \text{real count is odd, so at least } 1 \]
Why: Seven zeros in total, with imaginary ones arriving strictly in conjugate pairs, so the number of imaginary zeros is even — at most 6. That leaves at least one real zero. It is the same conclusion the end-behaviour argument of Lesson 5.2 reached for odd-degree polynomials, arrived at from a completely different direction, and the agreement is a good sign that both arguments are sound.
Explain it
They can find real zeros and are puzzled that a cubic sometimes seems to have only one.
Discussion prompt
In four sentences or fewer, explain why every cubic has exactly three zeros even when its graph crosses the axis once.
Hint: Two things get counted that the graph does not show.
Answer:
A graph shows only real zeros, and a cubic can have one real zero and two imaginary ones. Imaginary zeros are perfectly good numbers; they just have no place on a picture of real x against real y.
A repeated zero also hides: if a factor appears twice, the graph touches the axis at one point but the zero counts twice. Once you count imaginary zeros and count repeats properly, the total is always exactly the degree.
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 counting, write every linear factor out separately rather than using an exponent. For conjugates, ask of each given zero whether it is irrational or imaginary, and pair it if so. For f of negative x, mark the odd exponents and flip only those. For approximating, check whether the coefficients are integers before reaching for a candidate list. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take f of x equals x to the fifth minus 4x to the fourth plus 4x cubed plus 10x squared minus 13x minus 14 and classify it completely. Top left: state the degree and how many zeros to expect, then apply Descartes' rule to f and to f of negative x and tabulate every possible split. Top right: find all five zeros, showing the divisions and the quadratic formula step. Bottom left: sketch the graph, marking where it crosses and where it is tangent, and write beside each why. Bottom right: list the five zeros in a column and label each as rational, irrational or imaginary, circling the conjugate pair. In a margin, write the one sentence explaining why the imaginary count in Descartes' table is always even.
If your graph crosses at negative 1, check the factorisation: that zero comes from a squared factor, so the curve must touch and turn rather than pass through.
Recap
Five things, and the first tells you when the other four are finished.
| If you see | Then |
|---|---|
| A polynomial of degree n | It has exactly n zeros |
| A repeated factor | Count that zero once per factor |
| An even power on a factor | The graph is tangent there |
| An odd power on a factor | The graph crosses there |
| An imaginary zero, real coefficients | Its conjugate is a zero too |
| No sign changes in f | No positive real zeros |
| Decimal coefficients | Approximate graphically |
Lesson 5.8 puts all of this on a graph: turning points, local maxima and minima, and how the zeros and end behaviour together determine a polynomial's shape.
McDougal Littell Algebra 2 (Texas Edition), Ch. 5 Polynomials and Polynomial Functions — Lesson 5.7 Apply the Fundamental Theorem of Algebra §5.7, pp. 379-385 — everything on these slides traces back here
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