Polynomials and their roots

What a polynomial is

A polynomial adds up whole-number powers of $x$, each multiplied by a fixed number:

The fixed numbers $a_n, \ldots, a_1, a_0$ are the coefficients. The highest power with a nonzero coefficient, $n$, is the degree. The last coefficient, $a_0$, is the constant term: it is the value at zero, $p(0) = a_0$.

The powers must be 0, 1, 2, 3 and so on. So $\frac{1}{x}$ and $\sqrt{x}$ are not polynomials, and neither is $2^x$, where $x$ sits in the power.

Evaluating a polynomial

To evaluate $p$ at a number, put that number in place of every $x$. The running example in this lesson is the cubic $p(x) = x^3 - 6x^2 + 11x - 6$. At $x = 4$ its terms are $64 - 96 + 44 - 6$, so $p(4) = 6$.

Here is $p$ at more inputs. At 0 it is $-6$, its constant term.

Roots and factors

A root of $p$ is an input $r$ with $p(r) = 0$. On the graph, the roots are where the curve meets the $x$-axis: the zeros of lesson 1.1.1. The table shows three roots of the cubic: 1, 2 and 3.

Those roots come from factors. Multiply out $(x - 1)(x - 2)(x - 3)$, two brackets at a time:

A product is zero exactly when one of its factors is zero. So the factored form shows the roots at a glance.

Finding the other roots from one

Once you know one root, the rest come from a polynomial of lower degree. Take $q(x) = x^3 - 7x + 6$.

When the coefficients are whole numbers, a whole-number root must divide the constant term, so try $\pm 1, \pm 2, \pm 3, \pm 6$. The first try works: $q(1) = 1 - 7 + 6 = 0$.

So $q(x) = (x - 1)(x^2 + bx + c)$ for some numbers $b$ and $c$. Multiply out the right side, then match each power with $x^3 + 0x^2 - 7x + 6$:

The $x^2$ terms give $b - 1 = 0$, so $b = 1$. The constants give $-c = 6$, so $c = -6$. The $x$ terms check the answer: $c - b = -7$, as it should be.

The quadratic factor is $x^2 + x - 6 = (x - 2)(x + 3)$, by factoring or by the quadratic formula of lesson 1.2.2. So $q(x) = (x - 1)(x - 2)(x + 3)$, with roots 1, 2 and $-3$.

Double roots

The quadratic $x^2 - 4x + 4$ is $(x - 2)^2$. Its discriminant is 0 (lesson 1.2.2), and its only root is 2. The factor $x - 2$ appears twice, so 2 is a double root.

At a double root the graph touches the axis and turns back without crossing. On both sides of 2 the value is positive: 0.01 at 1.9 and 0.01 at 2.1. The sign does not change.

A cubic can have one too: $(x - 1)^2(x - 3)$ multiplies out to $x^3 - 5x^2 + 7x - 3$. It is $-0.625$ at 0.5, 0 at 1 and $-0.375$ at 1.5, so it touches the axis at 1; at 3 it crosses.

In general, a factor repeated an even number of times touches the axis, and one repeated an odd number of times crosses it.

How many roots

The factor theorem limits the count. Each root takes a factor $(x - r)$ of degree 1, and a polynomial of degree $n$ has room for only $n$ of them.

The cubic $x^3 - 4x = x(x - 2)(x + 2)$ has the full three: $-2$, 0 and 2. Its constant term is 0, so 0 is a root. Fewer is common: $(x - 1)^2(x - 3)$ has two different roots, and $x^3 - 8$ has just one, 2.

An odd degree always gives at least one real root. Far from 0 the highest power outweighs the rest: at $x = 100$, $x^3$ is 1,000,000 while the other terms of $p$ add up to $-58{,}906$. So a cubic ends far below the axis on one side and far above it on the other, and must cross in between.

An even degree can miss the axis altogether. $x^2 + 1$ is never below 1, so it has no real root.

Sign changes

Between two neighbouring roots a polynomial keeps one sign, and at a single root the sign flips. The cubic $p$ shows it: $-1.875$ at 0.5, then 0.375 at 1.5, then $-0.375$ at 2.5.

The graph of a polynomial has no gaps or jumps. So it cannot get from below the axis to above it without crossing, and two values are enough to prove a root exists.

The test only works one way.

• A sign change means at least one root, not exactly one. $p(0) = -6$ and $p(4) = 6$, and all three roots lie between.
• No sign change does not mean no root. $p(0.5)$ and $p(2.5)$ are both negative, yet the roots 1 and 2 lie between them. A double root gives no sign change either.

A graph with no gaps or jumps is called continuous; lesson 2.1.2 makes the idea exact. Case study 1.3.5 meets a polynomial whose roots are interest rates.

Common mistakes

1. A root $r$ gives the factor $(x - r)$, not $(x + r)$. The root $-3$ gives $x + 3$.
2. The degree is the highest power, not the number of terms. $x^5 + 1$ has degree 5.
3. Give NumPy every power, highest first. $x^3 - 7x + 6$ is [1, 0, -7, 6]; [1, -7, 6] would mean $x^2 - 7x + 6$.
4. At most $n$ roots, not exactly $n$. The j answers of np.roots are not real roots, and a tiny j part on a real number is rounding.
5. No sign change does not mean no root. Two roots, or a double root, leave the sign unchanged.
6. A sign change means at least one root. There may be three, or any odd number.