Polynomials and Their Graphs

Learning goals

  • Identify a polynomial, and read its degree, leading term and constant from standard form
  • Find a product's degree and leading term without expanding, and see degree drop in a sum
  • Evaluate f(0)f(0) as the constant term and f(1)f(1) as the sum of the coefficients
  • Predict a graph's far-end behavior from the leading term alone
  • Bound the real zeros by nn, the turning points by n−1n - 1 (none if constant)

What counts as a polynomial

Look at three expressions: 3x2−2x+73x^2 - 2x + 7, 3x+1\dfrac{3}{x} + 1, and 12x3+2\tfrac{1}{2}x^3 + \sqrt{2}. The first and the third are polynomials. The middle one is not, because xx sits in a denominator. What separates them is not whether a fraction or a square root appears anywhere in the expression: it is whether the variable xx is ever divided into, rooted, or placed in an exponent.

A term in the variable xx is an expression of the form c xkc\,x^k, where the coefficient cc is a real number and the exponent kk is a whole number: 0,1,2,3,…0, 1, 2, 3, \ldots. A polynomial in xx is a finite sum of such terms. Writing aka_k for the coefficient that sits on xkx^k, the general shape is

f(x)=anxn+an−1xn−1+⋯+a1x+a0,f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0,

and the two right-hand terms fit the pattern too, since x1=xx^1 = x and x0=1x^0 = 1: the constant a0a_0 is the term a0x0a_0 x^0. You have been working with polynomials all along. A constant function is a polynomial, a linear function mx+bmx + b is one, and a quadratic ax2+bx+cax^2 + bx + c is one; what is new is the license to keep climbing past the square.

This matches the description above exactly. Take any expression built from numbers and xx using only addition, subtraction, and multiplication, and multiply out every parenthesis. Multiplying powers adds their exponents (x2⋅x3=x5x^2 \cdot x^3 = x^5), so the exponents stay whole numbers, and a subtraction is just an addition with a minus sign built into the coefficient. Everything collapses into a sum of terms c xkc\,x^k. So a polynomial is exactly the algebra you can do with xx without ever dividing by it or taking a root of it.

Two more expressions fail the same test:

But the restriction falls only on xx, never on the coefficients. The expression x23−x5+2\dfrac{x^2}{3} - \dfrac{x}{5} + \sqrt{2} is a perfectly good polynomial: the fractions and the square root are applied to numbers. Each term is still a real coefficient times a whole-number power of xx.

Check your understanding

Which of these is a polynomial?

Answer choices

Degree, leading coefficient, and standard form

Before naming a polynomial’s parts, tidy it: combine like terms, then write the terms in standard form, with the exponents descending from left to right. Once that is done, the degree is the highest power of xx that survives with a nonzero coefficient, and the leading term is the term carrying that power. The leading coefficient is that term’s coefficient, and the constant term is the term of degree zero, sign included.

Anatomy of a polynomial in standard formThe expression f(x) = 3x^4 - 5x^3 + 2x - 7 with the leading coefficient, degree, leading term, and constant term labeled.f(x) =3x⁴- 5x³+ 2x- 7leading coefficientdegreeleading termconstant term(sign included)
The anatomy of a standard-form polynomial. Here f(x) = 3x⁴ - 5x³ + 2x - 7 has degree 4, leading term 3x⁴, leading coefficient 3, and constant term -7. Each sign belongs to the term it precedes.

The low degrees have names you know, and the next few are worth learning:

DegreeNameExample
00constantf(x)=12f(x) = 12
11linearf(x)=3x−1f(x) = 3x - 1
22quadraticf(x)=x2−5x+6f(x) = x^2 - 5x + 6
33cubicf(x)=x3−4xf(x) = x^3 - 4x
44quarticf(x)=x4−5x2+4f(x) = x^4 - 5x^2 + 4
55quinticf(x)=2x5+xf(x) = 2x^5 + x

A second naming scheme counts terms instead of degree: a monomial has one term, a binomial two, a trinomial three. So x4−5x2+4x^4 - 5x^2 + 4 is a quartic trinomial, and 2x5+x2x^5 + x is a quintic binomial. One special case sits outside the scheme: the zero polynomial f(x)=0f(x) = 0 has no nonzero term at all, so it gets no leading term and no degree.

Both tidying steps are load-bearing, and skipping either one produces wrong answers. Read the degree only after combining like terms: (x3+2x)−(x3−x)(x^3 + 2x) - (x^3 - x) wears two cubes on its sleeve, but the cubes cancel and the difference is 3x3x, a polynomial of degree 11. And read the leading coefficient only after ordering the terms: the leading coefficient of 5x−2x4+35x - 2x^4 + 3 is not 55 but −2-2, because standard form is −2x4+5x+3-2x^4 + 5x + 3 and the x4x^4 term leads.

Check your understanding

What is the degree of (2x4−3x)−(2x4+x)(2x^4 - 3x) - (2x^4 + x)?

Answer choices

How does degree behave when polynomials combine? Addition is tame, multiplication is tamer, and both facts get used constantly for the rest of the course.

Degrees add when polynomials multiply#

Concretely, (2x3)(−5x4)=−10x7(2x^3)(-5x^4) = -10x^7: the exponents added and the coefficients multiplied. The same pairing argument works for any two polynomials.

Let ff have degree mm with leading term axma x^m, and let gg have degree nn with leading term bxnb x^n. When you multiply ff by gg and expand, every term of the product comes from pairing one term of ff with one term of gg. Each such pairing multiplies out to (c xi)(d xj)=cd xi+j(c\,x^i)(d\,x^j) = cd\,x^{i+j}.

The largest exponent any pairing can reach is m+nm + n, and exactly one pairing reaches it: the two leading terms. Every other pairing uses an exponent below mm from ff or below nn from gg, so it falls short of m+nm + n. The coefficient on xm+nx^{m+n} is therefore exactly abab, and a product of two nonzero real numbers is nonzero. So fgfg has degree exactly m+nm + n, with leading coefficient abab: degrees add, and leading coefficients multiply.

Addition is looser. Every exponent appearing in f+gf + g already appears in ff or in gg, so the degree of a sum is at most the larger of the two degrees. It can be smaller, but only through a cancellation of leading terms: (x3+2x)+(−x3+x)=3x(x^3 + 2x) + (-x^3 + x) = 3x drops from degree 33 to degree 11 because the cubes annihilate each other.

The multiplication rule means factored expressions surrender their degree and leading term without being expanded. The product (1−2x)(x2+3)2(1 - 2x)(x^2 + 3)^2 has degree 1+2+2=51 + 2 + 2 = 5, and its leading term is the product of the leading pieces, (−2x)(x2)(x2)=−2x5(-2x)(x^2)(x^2) = -2x^5. Keep that trick close; it powers the end-behavior readings later in this lesson.

Check your understanding

Without expanding, what is the leading term of (3x2−1)(x5+4x)(3x^2 - 1)(x^5 + 4x)?

Answer choices

Check your understanding

What is the leading coefficient of p(x)=5x−2x4+3p(x) = 5x - 2x^4 + 3?

Answer choices

A polynomial is a function

A polynomial is not just a string of symbols; it is a rule. Feed it an input and it returns an output, exactly the function machinery from Functions and Their Graphs. Powers, multiples, and sums accept every real number, so the domain of every polynomial function is all real numbers. There is no input to exclude, no division to go wrong, no root to turn negative.

Evaluating is substitution plus order of operations, and the one discipline worth drilling is to wrap negative inputs in parentheses before touching anything. Two evaluations are so cheap they are worth knowing by name. First, f(0)f(0) kills every term carrying an xx, so f(0)f(0) is the constant term, and the graph’s yy-intercept sits at that height. Second, every power of 11 equals 11, so f(1)f(1) is the sum of the coefficients, a one-line check you will use more often than you expect.

Worked example 1 Three quick reads of f(x)=2x4−x3+3x−9f(x) = 2x^4 - x^3 + 3x - 9

Evaluate f(−2)f(-2), f(0)f(0), and f(1)f(1).

For f(−2)f(-2), parenthesize the input and take the powers first. Since (−2)4=16(-2)^4 = 16 and (−2)3=−8(-2)^3 = -8,

f(−2)=2(16)−(−8)+3(−2)−9=32+8−6−9=25.f(-2) = 2(16) - (-8) + 3(-2) - 9 = 32 + 8 - 6 - 9 = 25.

The middle step is where sign errors live: the term −x3-x^3 becomes −(−8)=+8-(-8) = +8, a subtraction of a negative.

For f(0)f(0), every term with an xx vanishes and only the constant term survives:

f(0)=−9,f(0) = -9,

so the graph crosses the yy-axis at (0,−9)(0, -9). For f(1)f(1), every power of 11 is 11, so the value is just the coefficients added up, minding that the missing x2x^2 term contributes 00:

f(1)=2−1+0+3−9=−5.f(1) = 2 - 1 + 0 + 3 - 9 = -5.

Check your understanding

For h(x)=3x5−x2+7h(x) = 3x^5 - x^2 + 7, what are h(0)h(0) and h(1)h(1)?

Answer choices

Smooth and unbroken

Graph any polynomial and two qualities appear that no list of coefficients would lead you to expect. The curve is unbroken: you can draw it without lifting your pencil, with no gaps, jumps, or holes. That is because every real number is a legal input, and nudging the input a little nudges each term, and hence the output, only a little. And the curve is smooth: it bends, sometimes sharply, but it never creases into a corner the way y=∣x∣y = |x| does at the origin.

Both claims are proved properly in calculus; for now, trust them the way Graphs of Functions trusted smooth interpolation between plotted points. The contrast is what matters: no choice of coefficients gives a polynomial graph a sharp corner, and none gives it a break like the one y=1xy = \dfrac{1}{x} has at x=0x = 0.

Smooth and unbroken, versus a corner and a breakA smooth polynomial curve, a curve with a sharp corner, and a curve with a break at a dashed vertical line.a polynomial graphsmooth and unbrokena sharp cornernever on a polynomial grapha break in the curvenever on a polynomial graph
Polynomial graphs are smooth, unbroken curves. A sharp corner, like the one on the graph of y = |x|, or a break, like the one the graph of y = 1/x has at x = 0, rules a curve out as a polynomial's graph.

Unbrokenness has one consequence this lesson will lean on twice. An unbroken curve cannot get from below the xx-axis to above it without touching it on the way. If f(a)f(a) is negative and f(b)f(b) is positive, then somewhere between aa and bb the polynomial takes the value 00. A sign change traps a zero.

End behavior: the leading term takes over

Now for the promise in the opening: one term controls both far ends of the graph. (This section is about polynomials with degree n≥1n \ge 1; a constant polynomial like f(x)=12f(x) = 12 has a horizontal graph with no far ends to speak of.) Watch it happen numerically for f(x)=x3−100xf(x) = x^3 - 100x, whose two terms pull in opposite directions.

xxx3x^3100x100xf(x)=x3−100xf(x) = x^3 - 100x
−100-100−1,000,000-1{,}000{,}000−10,000-10{,}000−990,000-990{,}000
−20-20−8,000-8{,}000−2,000-2{,}000−6,000-6{,}000
−5-5−125-125−500-500375375
−1-1−1-1−100-1009999
1111100100−99-99
55125125500500−375-375
20208,0008{,}0002,0002{,}0006,0006{,}000
1001001,000,0001{,}000{,}00010,00010{,}000990,000990{,}000

Near the origin the −100x-100x term bullies the cube on both sides: at x=±1x = \pm 1 and x=±5x = \pm 5 the output has the opposite sign you would expect from a positive leading coefficient. But the cube grows like a cube while 100x100x grows only like a line, so by x=±20x = \pm 20 the cube has taken over. And at x=±100x = \pm 100 the correction is a one percent nick out of a million. The middle of a graph belongs to all the terms; the far ends belong to the leading term alone, one sign at each end here because x3x^3 itself is negative on the left and positive on the right.

Say the graph rises on an end when the outputs grow past any bound as you move along that end, and falls when they drop below any bound. Here is why the leading term decides both ends. Factor the highest power out of every term:

f(x)=xn(an+an−1x+an−2x2+⋯+a0xn).f(x) = x^n \left( a_n + \frac{a_{n-1}}{x} + \frac{a_{n-2}}{x^2} + \cdots + \frac{a_0}{x^n} \right).

As ∣x∣|x| grows, each fraction akxn−k\dfrac{a_k}{x^{n-k}} shrinks toward 00, because dividing by an enormous number crushes anything finite. So far from the origin the bracket is close to ana_n alone, and f(x)f(x) behaves like anxna_n x^n: the leading term, and nothing else.

Read off the sign of anxna_n x^n and two conclusions drop out. When nn is even, xn>0x^n > 0 on both far ends, so both ends carry the sign of ana_n. When nn is odd, xnx^n keeps the sign of xx itself, so the far right carries the sign of ana_n and the far left carries the opposite sign.

Two switches, four silhouettes. The parity of the degree decides whether the two ends agree, and the sign of the leading coefficient decides which way the right end points:

Degree nnLeading coefficientFar leftFar right
evenan>0a_n > 0risesrises
evenan<0a_n < 0fallsfalls
oddan>0a_n > 0fallsrises
oddan<0a_n < 0risesfalls

The first two floors of the tower are the two positive-coefficient rows in miniature. A line with positive slope is the odd case (n=1n = 1), falling on the left and rising on the right. Every upward parabola is the even case (n=2n = 2), rising on both ends, exactly as chapter 4 read off the sign of aa.

The four end-behavior casesFour schematic polynomial graphs showing the end-behavior cases by degree parity and leading-coefficient sign.even degree, positive leadboth ends riseeven degree, negative leadboth ends fallodd degree, positive leadfalls on the left, rises on the rightodd degree, negative leadrises on the left, falls on the right
The four end-behavior silhouettes. An even degree makes the far ends agree, an odd degree makes them disagree, and the sign of the leading coefficient decides which way the right end points. Only the ends are dictated; the wiggles in the middle belong to the lower terms.

The odd rows hide a gift. An odd-degree polynomial falls without bound on one end and rises without bound on the other. So its unbroken graph starts below the xx-axis and finishes above it, or the reverse. By the sign-change principle from the last section, it must touch the axis somewhere: every polynomial of odd degree has at least one real zero. No even-degree polynomial can make that promise; x2+1x^2 + 1 stays at height 11 or above and never comes down to the axis.

Worked example 2 Reading g(x)=(1−2x)(x2+3)2g(x) = (1 - 2x)(x^2 + 3)^2 at sight

Describe the far ends of the graph of gg, its yy-intercept, and what the shape forces about its real zeros.

No expansion is needed. Degrees add across the product, so the degree is 1+2+2=51 + 2 + 2 = 5, and the leading term is the product of the leading pieces:

(−2x)(x2)(x2)=−2x5.(-2x)(x^2)(x^2) = -2x^5.

The degree 55 is odd and the leading coefficient −2-2 is negative, so this is the fourth silhouette. The graph rises on the far left and falls on the far right. The yy-intercept is

g(0)=(1)(0+3)2=9.g(0) = (1)(0 + 3)^2 = 9.

Being of odd degree, gg must have at least one real zero, and here the factored form shows it exactly once: x2+3≥3x^2 + 3 \ge 3 never vanishes, while 1−2x=01 - 2x = 0 at x=12x = \tfrac{1}{2}. So the graph crosses the axis at 12\tfrac{1}{2} and nowhere else, comfortably within the ceiling of 55 that the next section explains.

Check your understanding

Which silhouette matches f(x)=−3x4+5x3−1f(x) = -3x^4 + 5x^3 - 1 far from the origin?

Answer choices

Counting zeros and turning points

A real zero of ff is an input with f(x)=0f(x) = 0; on the graph it is an xx-intercept, exactly as in Graphs of Functions. The degree does not tell you how many real zeros a polynomial has, but it does set a hard ceiling:

A polynomial of degree nn has at most nn real zeros.

You have already proved the first two cases in earlier chapters. A linear polynomial mx+bmx + b with m≠0m \ne 0 has exactly one zero, x=−bmx = -\tfrac{b}{m}, and no second one, because a nonzero slope never returns to a height it has left. A quadratic has at most two, as the quadratic formula and the discriminant made precise. The pattern behind the general bound is already visible in a factored cubic like x(x−2)(x+2)=x3−4xx(x - 2)(x + 2) = x^3 - 4x: a product is zero exactly when a factor is. So the three linear factors hand over the three zeros 00, 22, and −2-2, and since degrees add, a degree-nn polynomial has room for at most nn linear factors. What is missing is the reverse direction, that every zero of a polynomial really does yield a linear factor. That missing link is the Factor Theorem, proved later in this chapter once long division is in hand. For now, use the ceiling as a counting tool, not a promise: x2+1x^2 + 1 has no real zeros at all, and x4x^4 has just the one, x=0x = 0.

A turning point is a point where the graph switches between rising and falling. Such a point is the top of a local hill or the bottom of a local valley, like the vertex of a parabola. Turning points obey a twin ceiling:

The graph of a nonconstant polynomial of degree nn has at most n−1n - 1 turning points.

(A constant polynomial, degree 00, is the excluded case: its graph is a flat horizontal line with no turning points at all.) Check it against everything you know: a line (n=1n = 1) never turns, and a parabola (n=2n = 2) turns exactly once at its vertex. Meanwhile y=x3y = x^3 has no turning point at all, and y=x3−4xy = x^3 - 4x has two, one hill and one valley. The full proof of the ceiling belongs to calculus, where you will meet a companion polynomial of degree n−1n - 1, the derivative, whose zeros include every turning point. The zeros ceiling applied to that companion gives the turning-point ceiling at once. What is within reach right now is the floor. Between two neighboring xx-intercepts the graph leaves the axis and must come back to it, so it stops rising or stops falling somewhere in between. Each consecutive pair of zeros forces at least one turning point, so a polynomial with kk distinct real zeros carries at least k−1k - 1 turning points.

A quartic with four real zeros and three turning pointsThe W-shaped graph of y = x^4 - 5x^2 + 4 with its four x-intercepts and three turning points marked.xyy = x⁴ - 5x² + 4x-intercepts (4)turning points (3)
The quartic y = x⁴ - 5x² + 4 hits both ceilings for degree 4. The four accent dots on the axis are its real zeros, the most a quartic can have, and the three solid dots mark its turning points, also the most a quartic can have. Both far ends rise, matching an even degree with a positive leading coefficient.

The quartic in the figure is the same x4−5x2+4x^4 - 5x^2 + 4 whose end behavior you worked out in Graphs of Functions, now with its whole census on display. The census is four real zeros at ±1\pm 1 and ±2\pm 2 (it factors as (x2−1)(x2−4)(x^2 - 1)(x^2 - 4)), and three turning points, the maximum 4−14 - 1 allows.

One more fact ties the ends to the turns, for any nonconstant polynomial (a constant graph has no ends that rise or fall, so it sits outside this count). Trace a polynomial’s graph from far left to far right: every turning point switches you from rising to falling or back, and nothing else does. When both ends point the same way, as in every even degree n≥2n \ge 2, you finish moving the opposite way you started, and only an odd number of switches can do that. So a graph of even degree n≥2n \ge 2 has an odd number of turning points, at least one. When the ends disagree, as in every odd degree, you finish moving the same way you started, which takes an even number of switches, including zero. That is exactly what y=x3y = x^3 does: it rises the whole way, with no turning point at all.

Worked example 3 The least degree a described graph allows

A polynomial has 55 different real zeros, its graph rises on both far ends, and that graph has exactly 55 turning points. What is the least possible degree?

Collect what each clue forces. Five real zeros need n≥5n \ge 5 by the zeros ceiling. Five turning points need n−1≥5n - 1 \ge 5, so n≥6n \ge 6, a stronger demand. And both ends rising means the ends agree, so the degree is even, which 66 satisfies:

n≥6,n even  ⇒  n=6 is the least candidate.n \ge 6, \qquad n \text{ even} \;\Rightarrow\; n = 6 \text{ is the least candidate.}

So far every clue has only ruled degrees OUT, which leaves the question of whether 66 is actually reachable. Ceilings never answer that on their own, so finish by exhibiting one:

f(x)=x2(x2−1)(x2−4)=x6−5x4+4x2f(x) = x^2\left(x^2 - 1\right)\left(x^2 - 4\right) = x^6 - 5x^4 + 4x^2

has degree 66 with a positive leading coefficient, so both ends rise, and its zeros are 00, ±1\pm 1, and ±2\pm 2, five different values. Those five zeros force at least four turning points, and an even degree forces an odd count of them, so with a ceiling of 6−1=56 - 1 = 5 the only option between four and five is five itself. Every clue holds, so the least possible degree is 66.

Check your understanding

A polynomial's graph crosses the xx-axis 33 times, has exactly 22 turning points, and its two far ends point in opposite directions. What is the least possible degree?

Answer choices

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Why the leading term decides the far ends

The leading term wins#

Let f(x)=anxn+an−1xn−1+⋯+a1x+a0f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 with an≠0a_n \ne 0 and n≥1n \ge 1. For x≠0x \ne 0, factor the highest power out of every term:

f(x)=xn(an+an−1x+an−2x2+⋯+a0xn).f(x) = x^n \left( a_n + \frac{a_{n-1}}{x} + \frac{a_{n-2}}{x^2} + \cdots + \frac{a_0}{x^n} \right).

Write D=∣an−1∣+∣an−2∣+⋯+∣a0∣D = |a_{n-1}| + |a_{n-2}| + \cdots + |a_0| for the combined size of the lower coefficients. If D=0D = 0 the bracket is exactly ana_n and there is nothing to prove, so suppose D>0D > 0 and take any xx with ∣x∣≥1|x| \ge 1 and ∣x∣≥2D∣an∣|x| \ge \dfrac{2D}{|a_n|}. Because ∣x∣≥1|x| \ge 1, every denominator obeys ∣x∣k≥∣x∣|x|^k \ge |x|, and the size of a sum is at most the sum of the sizes, so the whole tail after ana_n satisfies

∣an−1x+⋯+a0xn∣≤∣an−1∣+⋯+∣a0∣∣x∣=D∣x∣≤∣an∣2,\left| \frac{a_{n-1}}{x} + \cdots + \frac{a_0}{x^n} \right| \le \frac{|a_{n-1}| + \cdots + |a_0|}{|x|} = \frac{D}{|x|} \le \frac{|a_n|}{2},

the last step because ∣x∣≥2D∣an∣|x| \ge \dfrac{2D}{|a_n|}. The bracket therefore stays within ∣an∣2\dfrac{|a_n|}{2} of ana_n: it keeps the sign of ana_n and has size at least ∣an∣2\dfrac{|a_n|}{2}. Beyond the threshold, then, f(x)f(x) is xnx^n times a number that acts like ana_n, so two conclusions drop out. The sign of f(x)f(x) is the sign of ana_n times the sign of xnx^n, and the size obeys

∣f(x)∣≥∣an∣2 ∣x∣n≥∣an∣2 ∣x∣,|f(x)| \ge \frac{|a_n|}{2}\,|x|^n \ge \frac{|a_n|}{2}\,|x|,

which passes any bound you name once ∣x∣|x| is large enough. Finally, read off the sign of xnx^n. When nn is even, xn>0x^n > 0 on both far ends, so both ends carry the sign of ana_n. When nn is odd, xnx^n keeps the sign of xx itself, so the far right carries the sign of ana_n and the far left carries the opposite sign.

For the cubic in the table, D=100D = 100 and an=1a_n = 1, so the threshold is ∣x∣≥200|x| \ge 200. From there on, x3−100xx^3 - 100x is pinned between half the cube and the full cube, and indeed f(200)=7,980,000f(200) = 7{,}980{,}000, comfortably past half of 2003=8,000,000200^3 = 8{,}000{,}000.

A bit of history (optional)

The word polynomial is a hybrid: the Greek polys, many, welded onto the Latin nomen, name. It was coined on the pattern of the older word binomial, meaning two names.

Methods equivalent to solving a quadratic equation go back to ancient mathematics, long before anyone wrote them as the single symbolic formula used today. The cubic and the quartic resisted far longer. Gerolamo Cardano’s Ars Magna, printed in 1545, finally gave a method for the cubic, and a method for the quartic followed it in the same book. Everyone expected the fifth degree, the quintic, to fall next.

It never did. After more than two centuries of searching, Niels Henrik Abel proved in 1824 that no formula built from the ordinary arithmetic operations and radicals can solve every quintic equation. Evariste Galois soon explained exactly which equations do have such a formula.

That result shapes the chapter ahead. No such formula will hand you the roots of a general high-degree polynomial. What division gives you instead: test a candidate, and if it truly is a root, dividing out its factor lowers the degree by exactly one. Repeating that test-and-divide step, one candidate at a time, is how this chapter narrows a high-degree polynomial down to its roots.