Polynomials and their roots

The line and the parabola are the first two members of a family, and there is nothing in either that suggests the family should stop at degree two. Cubics and quartics are written the same way, arise from the same kinds of problem, and are handled by one set of tools.

This lesson builds those tools. It assumes the algebra of the earlier lessons and the graphs of the previous one, and it produces the central fact about polynomials: a root and a factor are the same thing seen twice, so finding one is finding the other.

Degree, and what a polynomial is not

A polynomial in $x$ is a finite sum of terms $a_{k} x^{k}$ with whole-number exponents and constant coefficients, such as $3 x^{4} - x^{3} + 7 x - 2$ . The largest exponent with a non-zero coefficient is the degree, and the coefficient sitting on it is the leading coefficient. Degree zero is a constant, one is linear, two quadratic, three cubic, four quartic.

The restriction to whole-number exponents is what makes the family well behaved, and it excludes a great deal. Neither $1 / x$ , which is $x^{- 1}$ , nor $\sqrt{x}$ , which is $x^{1 / 2}$ , is a polynomial, and neither is $2^{x}$ , where the variable is in the exponent rather than the base. Each of those three has its own lesson later in this course, and each behaves quite differently.

Adding two polynomials gives a polynomial whose degree is at most the larger of the two, and can be less when leading terms cancel. Multiplying gives one whose degree is exactly the sum, since the leading terms multiply and nothing else can reach that power. So degree behaves under multiplication the way exponents do, which is the first hint of the index laws that a later lesson makes general.

At large $lvert x rvert$ the leading term swamps everything else, and this settles the shape of the far ends of the graph. For $p (x) = x^{3} - 4 x^{2} + 2 x + 5$ at $x = 100$ , the leading term is $1 {,} 000 {,} 000$ while the rest contributes $- 40 {,} 000 + 200 + 5$ , four per cent of the total, and the fraction shrinks as $x$ grows. So an odd-degree polynomial with positive leading coefficient falls without limit to the left and rises without limit to the right, and an even-degree one goes the same way at both ends.

Division with a remainder

Whole numbers can be divided with a remainder smaller than the divisor: $17 = 5 \times 3 + 2$ . Polynomials do the same, with the degree of the remainder taking the place of the size: dividing $p (x)$ by a non-zero $d (x)$ gives a unique quotient $q (x)$ and remainder $r (x)$ with

p (x) = d (x) q (x) + r (x)

where the degree of $r$ is less than the degree of $d$ . Dividing by a linear $x - a$ therefore leaves a remainder of degree zero, a plain number.

The procedure is long division, arranged like the arithmetic version. Divide $x^{3} - 4 x^{2} + 2 x + 5$ by $x - 3$ . The leading term $x^{3}$ divided by $x$ gives $x^{2}$ ; multiply $x - 3$ by $x^{2}$ to get $x^{3} - 3 x^{2}$ and subtract, leaving $- x^{2} + 2 x + 5$ . Now $- x^{2}$ divided by $x$ gives $- x$ ; multiply and subtract, leaving $- x + 5$ . Finally $- x$ divided by $x$ gives $- 1$ ; multiply and subtract, leaving $2$ . So the quotient is $x^{2} - x - 1$ and the remainder is $2$ .

Multiplying back is the check, and it should always be done: $(x - 3) (x^{2} - x - 1) = x^{3} - x^{2} - x - 3 x^{2} + 3 x + 3 = x^{3} - 4 x^{2} + 2 x + 3$ , and adding the remainder $2$ recovers the original.

The remainder theorem, and why it is obvious in hindsight

There is a shortcut for the remainder that avoids the division entirely. Write the division statement for a linear divisor: $p (x) = (x - a) q (x) + r$ , with $r$ a constant. This holds for every $x$ , so it holds at $x = a$ , where the first term dies:

p (a) = 0 \times q (a) + r = r

The remainder on dividing by $x - a$ is simply $p (a)$ . Checking against the division just done, $p (3) = 27 - 36 + 6 + 5 = 2$ , which is the remainder found the long way.

The factor theorem is the case $r = 0$ , and it is the reason the whole apparatus matters. If $p (a) = 0$ then the remainder is zero, so $x - a$ divides $p$ exactly; and if $x - a$ divides $p$ exactly then $p (a) = 0$ . A number $a$ with $p (a) = 0$ is a root, and the theorem says roots and linear factors are two descriptions of one fact. Finding a root hands you a factor, and dividing by that factor leaves a polynomial of one lower degree, which is a smaller problem of the same kind.

Example. Show that $2$ is a root of $p (x) = 2 x^{3} + x^{2} - 13 x + 6$ , and factor it completely.

Evaluate: $p (2) = 16 + 4 - 26 + 6 = 0$ , so $x - 2$ is a factor. Dividing gives $2 x^{2} + 5 x - 3$ , and that quadratic factors as $(2 x - 1) (x + 3)$ , since $- 3 \times 2 = - 6$ and the middle term works out. So $p (x) = (x - 2) (2 x - 1) (x + 3)$ and the roots are $2$ , $1 / 2$ and $- 3$ . Check the middle one: $p (0.5) = 0.25 + 0.25 - 6.5 + 6 = 0$ .

Now you. Show that $1$ is a root of $q (x) = x^{3} - 7 x + 6$ and factor it completely.

Answer

$q (1) = 1 - 7 + 6 = 0$ , so $x - 1$ is a factor. Dividing gives $x^{2} + x - 6 = (x + 3) (x - 2)$ , so $q (x) = (x - 1) (x - 2) (x + 3)$ and the roots are $1$ , $2$ and $- 3$ . Check: $q (2) = 8 - 14 + 6 = 0$ and $q (- 3) = - 27 + 21 + 6 = 0$ .

Where to look for a root

The factor theorem is useless without a root to start from, and guessing is only tolerable if the guesses are few. The rational root theorem makes them few. If a polynomial with integer coefficients has a rational root $p / q$ in lowest terms, then $p$ divides the constant term and $q$ divides the leading coefficient.

For $2 x^{3} + x^{2} - 13 x + 6$ the constant is $6$ and the leading coefficient is $2$ , so the numerator is $\pm 1, \pm 2, \pm 3, \pm 6$ and the denominator is $1$ or $2$ . That is sixteen candidates rather than infinitely many, and the roots found above, $2$ , $1 / 2$ and $- 3$ , are all on the list as they must be.

The theorem is a genuine restriction and also a warning about how special rational roots are. Most polynomials with integer coefficients have none: $x^{3} - 2$ has only $\sqrt[3]{2}$ , which is irrational, and the candidates $\pm 1, \pm 2$ all fail on inspection. Textbook exercises are chosen to have rational roots, and this creates a false impression that finding roots is a matter of persistence. In the wild it is not, and numerical methods do the work.

Multiplicity, and what the graph does

A factor can repeat. In $p (x) = (x - 2)^{2} (x + 1)$ the root $2$ has multiplicity two, and the graph shows the difference plainly: at a root of odd multiplicity the curve crosses the axis, and at a root of even multiplicity it touches and turns back. The reason is a sign argument. Near $x = 2$ the factor $(x - 2)^{2}$ is positive on both sides, so the sign of the product is controlled by $(x + 1)$ alone and does not change; with a single factor $(x - 2)$ the sign flips as $x$ passes through.

Counting roots with their multiplicities makes the bookkeeping come out right. A polynomial of degree $n$ has at most $n$ roots, because each root contributes a linear factor and the degrees of the factors must add to $n$ . It may have fewer real roots: $x^{2} + 1$ has none, the gap left open in the first lesson, and $x^{3} - x$ has three while $x^{3} + x$ has one.

The fundamental theorem of algebra, proved by Gauss in his doctoral thesis of 1799 with several later and more rigorous proofs, says that over the complex numbers the count is always exact: every polynomial of degree $n \geq 1$ has exactly $n$ roots counted with multiplicity. This is the payoff of the extension the first lesson declined to make. Admitting the solutions of $x^{2} = - 1$ does not merely patch that one equation, it makes every polynomial equation solvable at once, and no further extension is ever needed. On the real line the statement is weaker and messier, and the honest version is that the roots come in a bounded number and some of them may be missing.

Example. Factor $x^{4} - 5 x^{2} + 4$ completely and describe its graph.

The expression is a quadratic in $x^{2}$ , so treat $u = x^{2}$ : $u^{2} - 5 u + 4 = (u - 1) (u - 4)$ . Substituting back, $(x^{2} - 1) (x^{2} - 4) = (x - 1) (x + 1) (x - 2) (x + 2)$ . Four distinct roots, all of multiplicity one, so the curve crosses the axis at $- 2, - 1, 1, 2$ . The degree is even with a positive leading coefficient, so both ends rise, and the curve must therefore dip below the axis between $- 2$ and $- 1$ , rise above between $- 1$ and $1$ , and dip again between $1$ and $2$ .

Now you. Factor $x^{3} - 3 x^{2} - 4 x + 12$ completely, given that $3$ is a root.

Answer

Dividing by $x - 3$ gives $x^{2} - 4$ , so the factorisation is $(x - 3) (x - 2) (x + 2)$ and the roots are $3$ , $2$ and $- 2$ . Check $x = - 2$ in the original: $- 8 - 12 + 8 + 12 = 0$ .

Building a polynomial from what it must do

The correspondence runs both ways, which is how polynomials get used as models. If a cubic must vanish at $- 1$ , $2$ and $5$ , it is $a (x + 1) (x - 2) (x - 5)$ for some constant $a$ , and one further condition fixes $a$ .

Example. Find the cubic with roots $- 1$ , $2$ , $5$ passing through $(0, 20)$ .

Write $p (x) = a (x + 1) (x - 2) (x - 5)$ . At $x = 0$ this is $a \times 1 \times (- 2) \times (- 5) = 10 a$ , and that must equal $20$ , so $a = 2$ . Expanding, $p (x) = 2 (x + 1) (x^{2} - 7 x + 10) = 2 (x^{3} - 6 x^{2} + 3 x + 10) = 2 x^{3} - 12 x^{2} + 6 x + 20$ . Check at $x = 2$ : $16 - 48 + 12 + 20 = 0$ .

Now you. Find the quadratic with roots $3$ and $- 4$ passing through $(1, - 12)$ .

Answer

$p (x) = a (x - 3) (x + 4)$ , and at $x = 1$ that is $a (- 2) (5) = - 10 a = - 12$ , so $a = 1.2$ and $p (x) = 1.2 (x - 3) (x + 4) = 1.2 x^{2} + 1.2 x - 14.4$ .

The limits of exact solution

Everything above is exact, and it depends on someone handing over a root. When no rational root exists, degree three and four can still be solved in radicals by the formulas of Cardano and Ferrari, published in the Ars Magna of 1545, though the cubic formula is unpleasant enough that almost nobody uses it.

Beyond that the road ends. Ruffini argued in 1799 and Abel proved in 1824 that the general fifth-degree equation has no solution by radicals, and Galois in 1832 explained exactly which equations do and which do not, in work that founded modern algebra. This is not a gap awaiting a cleverer formula. It is a proof that no such formula can exist, and it is one of the places where mathematics stops offering better methods and starts explaining why a method is impossible.

What is left is numerical work, which is enough for every practical purpose: bisection, which uses the fact that a sign change traps a root, and faster iterative schemes built on the derivative. Those belong to calculus.

Polynomials share one further limitation, and it is the one that opens the next lesson. No polynomial doubles at fixed intervals. A quantity that grows by a fixed factor for each unit of time, as money at compound interest does and as a bacterial culture does, cannot be described by any of these functions, however high the degree. Describing it needs the variable in the exponent, which is where the family ends and a new one begins.