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Polynomial Functions

How to read a polynomial's graph from its formula: where the ends go, where it meets the x-axis, and how to find its zeros.

You already know how to add, multiply and factor polynomials. This unit treats them as functions, like f(x) = x³ − 4x, and asks what their graphs look like. A few facts about the formula tell you a lot: the degree and the leading coefficient fix the ends of the graph, and the factors fix where it crosses or touches the x-axis.

Start: polynomial functions Jump to the theorems Jump to the self-guided quiz

The graph of f(x) = (x + 2)(x − 1)² crosses the x-axis at x = −2 and touches it and turns around at x = 1. The y-intercept is f(0) = 2.

Polynomial functions

A polynomial function is a function like f(x) = 2x³ − 3x² + 5: a sum of terms, each a number times a whole-number power of x. Written from the highest power down, it is in standard form. The highest power is the degree, and the number in front of that term is the leading coefficient. Degree 1 gives a line, degree 2 a parabola, degree 3 a cubic and degree 4 a quartic.

Example 1: Degree and leading coefficient in standard form

f(x) = 2x³ − 3x² + 5The highest power is x³2x³Degree 3, leading coefficient 2

Example 2: Without multiplying out

f(x) = (2x − 1)(x + 3)²Multiply only the first terms of the factors2x · x · x = 2x³There are three factors with an x2x³Degree 3, leading coefficient 2

Example 3: Evaluate a polynomial

f(2) = 2(2)³ − 3(2)² + 5Replace each x with 2f(2) = 16 − 12 + 5f(2) = 9

End behavior

The end behavior of a graph is what happens as x gets very large (x → ∞) or very negative (x → −∞). For large x the leading term is much bigger than all the other terms, so it alone decides where the ends go. Two questions settle it: is the degree odd or even, and is the leading coefficient positive or negative?

End behavior from the degree and the sign of the leading coefficient
Leading term	Left end, as x → −∞	Right end, as x → ∞
Even degree, positive	Up	Up
Even degree, negative	Down	Down
Odd degree, positive	Down	Up
Odd degree, negative	Up	Down

The cubic y = x³ − 4x (odd degree, positive lead) falls on the left and rises on the right. The quartic y = −x⁴ + 4x² (even degree, negative lead) falls at both ends.

Example 4: Read the ends

f(x) = −3x⁵ + 7x² − 1The leading term is −3x⁵f(x) → ∞asx → −∞Odd degree, negative lead: up on the leftf(x) → −∞asx → ∞and down on the right

Zeros and multiplicity

A zero of f is an x-value with f(x) = 0. Real zeros are the x-intercepts of the graph. When f is factored, each factor (x − r) gives the zero x = r. If a factor appears k times, the zero has multiplicity k. At a zero with odd multiplicity the graph crosses the x-axis. At a zero with even multiplicity it touches the axis and turns around.

Example 5: Find the zeros and their behavior

f(x) = (x + 2)(x − 1)²Set each factor equal to 0x = −2Multiplicity 1 (odd): the graph crossesx = 1Multiplicity 2 (even): the graph touches and turns around

Example 6: Factor first

x³ − 6x² + 9x = x(x² − 6x + 9)Take out the common factor xx³ − 6x² + 9x = x(x − 3)²A perfect squarex = 0 or x = 30 has multiplicity 1 and 3 has multiplicity 2

Turning points

A turning point is where the graph changes from rising to falling or from falling to rising: a peak or a valley. A polynomial of degree n has at most n − 1 turning points. It can have fewer. A cubic can have 2 turning points or none, and a quartic can have 3 or 1.

Turning points of some polynomials
Function	Degree	Turning points
y = x^3 − 4x	3	2
y = x^3 + x	3	0
y = −x^4 + 4x^2	4	3
y = x^4	4	1

Example 7: Use the limit to rule out a degree

5turningpointsAt most n − 1 turning pointsn − 1 ≥ 5n ≥ 6The degree is at least 6

The remainder and factor theorems

The remainder theorem says: when a polynomial f(x) is divided by x − a, the remainder is f(a). So you can find a remainder without dividing. The factor theorem follows from it: x − a is a factor of f(x) exactly when f(a) = 0, that is, when the remainder is 0.

Example 8: Find a remainder

f(x) = 2x³ − 3x² + 5Divide by x − 2, so a = 2f(2) = 16 − 12 + 5f(2) = 9The remainder is 9

Example 9: Test a factor

f(x) = x³ − 4x² + x + 6Is x − 2 a factor?f(2) = 8 − 16 + 2 + 6f(2) = 0Yes: x − 2 is a factor

Example 10: Watch the sign of a

f(−1) = −1 − 4 − 1 + 6Dividing by x + 1 means a = −1f(−1) = 0So x + 1 is a factor too

Synthetic division

Synthetic division is a short way to divide a polynomial by x − a. Write a, then the coefficients (with 0 for any missing power). Bring down the first coefficient. Then repeat: multiply by a, write the product under the next coefficient, and add. The last number is the remainder, and the others are the coefficients of the quotient, one degree lower.

Here is x³ − 4x² + x + 6 divided by x − 2:

Synthetic division of x cubed minus 4 x squared plus x plus 6 by x minus 2
Row	x³	x²	x	Constant
Coefficients	1	−4	1	6
Multiply by 2		2	−4	−6
Add	1	−2	−3	0

Example 11: Read the result

x³ − 4x² + x + 6 = (x − 2)(x² − 2x − 3)Remainder 0, quotient x² − 2x − 3x³ − 4x² + x + 6 = (x − 2)(x − 3)(x + 1)Factor the quadraticx = 2 or x = 3 or x = −1The three zeros

Rational roots

Where do you find a first zero to divide by? The rational root theorem says: if a polynomial has integer coefficients, every rational zero p/q in lowest terms has p dividing the constant term and q dividing the leading coefficient. That gives a short list of candidates to test.

Example 12: List the candidates and test them

f(x) = 2x³ − 3x² − 11x + 6p divides 6 and q divides 2, so the candidates are ±1, ±2, ±3, ±6, ±1/2 and ±3/2f(3) = 54 − 27 − 33 + 6 = 0So x − 3 is a factorf(x) = (x − 3)(2x² + 3x − 2)Synthetic division by 3 gives 2, 3, −2f(x) = (x − 3)(2x − 1)(x + 2)x = 3 or x = 12 or x = −2

The fundamental theorem of algebra

Some polynomials have no real zeros at all: x² + 4 is never 0 for real x. Using complex numbers, every polynomial has zeros. The fundamental theorem of algebra says that a polynomial of degree n ≥ 1 has exactly n complex zeros, if you count each one as many times as its multiplicity. For polynomials with real coefficients, non-real zeros come in conjugate pairs a + bi and a − bi.

Example 13: Find all the complex zeros

x³ − x² + 4x − 4 = x²(x − 1) + 4(x − 1)Factor by groupingx³ − x² + 4x − 4 = (x − 1)(x² + 4)x = 1 or x = 2i or x = −2iDegree 3, three zeros: one real, a conjugate pair

Sketching a graph from factored form

To sketch a polynomial in factored form: (1) mark the zeros on the x-axis; (2) decide at each zero whether the graph crosses or touches; (3) find the y-intercept f(0); (4) use the end behavior to start and finish the curve; (5) plot a point or two between zeros to see how high or low the graph goes.

For f(x) = x(x − 2)(x + 3): zeros at −3, 0 and 2, all crossing. Degree 3 with a positive lead, so the graph starts low and ends high. Between the zeros, f(−2) = 8 and f(1) = −4.

Example 14: Points between the zeros

f(−2) = (−2)(−4)(1) = 8Above the axis between −3 and 0f(1) = (1)(−1)(4) = −4Below the axis between 0 and 2

Common mistakes

✗Reading the degree from the first term written: f(x) = 5x + x⁴ is not degree 1.✓Look for the highest power: f(x) = 5x + x⁴ has degree 4 and leading coefficient 1.

✗Using the wrong sign in the remainder theorem: dividing by x + 3 and working out f(3).✓x + 3 = x − (−3), so the remainder is f(−3).

✗Leaving out a missing power in synthetic division: writing 1, 0, −7 for x³ − 7x + 6.✓Write a 0 for every missing power: x³ − 7x + 6 has coefficients 1, 0, −7, 6.

✗Saying (x − 1)² makes the graph cross at x = 1.✓An even multiplicity means the graph touches the x-axis and turns around at x = 1.

✗Expecting exactly n − 1 turning points.✓n − 1 is only the most a degree n polynomial can have: y = x³ has none.

Practice problems

Try each one, then open the answer to check your work.

Give the degree and leading coefficient of f(x) = −4x²(x − 1)(x + 5).
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−4x² · x · x = −4x⁴Degree 4, leading coefficient −4
Describe the end behavior of f(x) = −2x⁴ + x³ + 9.
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f(x) → −∞asx → ∞Even degree, negative leadf(x) → −∞asx → −∞Both ends go down
Find the zeros of f(x) = (x − 4)³(x + 1)² and say whether the graph crosses or touches at each.
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x = 4Multiplicity 3 (odd): crossesx = −1Multiplicity 2 (even): touches
Find the remainder when x³ + 2x² − 5x + 1 is divided by x − 3.
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f(3) = 27 + 18 − 15 + 1f(3) = 31
Is x + 2 a factor of x³ + 8?
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f(−2) = −8 + 8f(−2) = 0Yes, x + 2 is a factor
Find k so that x − 1 is a factor of x³ + 3x² − x + k.
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f(1) = 1 + 3 − 1 + k = 3 + kk = −3Set 3 + k = 0
Divide x³ − 7x + 6 by x − 1 with synthetic division, then find all the zeros.
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x³ − 7x + 6 = (x − 1)(x² + x − 6)Coefficients 1, 0, −7, 6 give 1, 1, −6 and remainder 0x³ − 7x + 6 = (x − 1)(x − 2)(x + 3)x = 1 or x = 2 or x = −3
List the possible rational zeros of 3x³ − x² + 6x − 2, then find the real zero.
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f(x) = 3x³ − x² + 6x − 2p divides 2 and q divides 3, so the candidates are ±1, ±2, ±1/3 and ±2/33x³ − x² + 6x − 2 = (3x − 1)(x² + 2)Group the termsx = 1/3x² + 2 has no real zeros
How many complex zeros does f(x) = x⁵ − x have, and what are they?
Show answer
x⁵ − x = x(x² − 1)(x² + 1)x = 0 or x = 1 or x = −1 or x = i or x = −iFive zeros, as the degree is 5

History: how many roots?

The question of how many solutions a polynomial equation has is old. According to the MacTutor History of Mathematics archive, the first claim that an equation of degree n always has n solutions was made by the Flemish mathematician Albert Girard in 1629, in L'invention en algèbre (Invention in algebra).

The idea behind the factor theorem, that a polynomial which is 0 at x = t has a factor x − t, was known to Thomas Harriot, MacTutor says, but did not become well known until René Descartes stated it in La géométrie in 1637.

Proving that every polynomial really has a root took much longer. Carl Friedrich Gauss presented a proof in his doctoral thesis of 1799, along with objections to earlier attempts. MacTutor notes that this first proof of Gauss's itself had some serious gaps.

Synthetic division is closely related to what is called Horner's method, after the English mathematician William George Horner (1786–1837). MacTutor points out that the method was ascribed to him by Augustus De Morgan and others, and that its history is more tangled than the name suggests.

Connections

This unit builds on Polynomials and Factoring, where you learned to multiply and factor, and on Quadratic Equations, the degree 2 case. The idea of a zero as an x-intercept comes from Functions.

The fundamental theorem of algebra explains why Complex Numbers were worth inventing: with them, a degree n polynomial always has n zeros.

Self-guided quiz

Pick a level, or Mixed for a bit of everything. Most answers are a whole number, like −7 or 12, and you can type x = or k = in front. For end behavior, type ∞ or −∞ (or inf and -inf). Your score is just for you: nothing is sent anywhere.

This self-guided quiz needs JavaScript turned on. Here are three to try on paper: (1) find the remainder when x cubed minus 2 x plus 4 is divided by x minus 2; (2) find the zero of the quantity x plus 5, squared, times the quantity x minus 1, where the graph touches the x-axis; (3) find the y-intercept of 2 times the quantity x minus 1, times the quantity x plus 3. Answers: (1) 8; (2) negative 5; (3) negative 6.

Sources

J J O'Connor and E F Robertson, “The fundamental theorem of algebra”, MacTutor History of Mathematics, University of St Andrews: history topic (Girard's claim of 1629, Harriot and Descartes on the factor x − t, Gauss's thesis of 1799 and its gaps)
J J O'Connor and E F Robertson, “Albert Girard”, MacTutor: Girard biography
J J O'Connor and E F Robertson, “René Descartes”, MacTutor: Descartes biography
J J O'Connor and E F Robertson, “Carl Friedrich Gauss”, MacTutor: Gauss biography
J J O'Connor and E F Robertson, “William George Horner”, MacTutor: Horner biography (Horner's method, ascribed to him by De Morgan and others)
Common Core State Standards for Mathematics: Arithmetic with Polynomials and Rational Expressions (A-APR.B.2, A-APR.B.3), Interpreting Functions (F-IF.C.7c) and The Complex Number System (N-CN.C.9)

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