Theory

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the polynomial story

In the first Precalculus course, we investigated polynomial functions from several viewpoints. Time to collect all of our thoughts and characterize all polynomial functions...as well as we can.

Polynomial Functions

A polynomial function is a function that CAN be represented by a formula of the form

\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \]

where the $a_i$ are real numbers and $a_n \ne 0$.

This is called standard form.

$n$ is called the degree of the polynomial

$\blacktriangleright $ Domain:
The natural domain of a polynomial function is all real numbers. Of course, a particular polynomial function may be defined with a restricted domain.

It seems weird to define polynomials and then quickly annouce that we don’t like the way we defined them. However, it just turns out that the standard form for a polynomial is just not that helpful. Our analysis prefers a different form.

Forms

Formulas for polynomial functions have two forms:

Standard Form: The standard form for a polynomial function is a sum. It looks like
\[ a_n x^n + a_{n-1} x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0 \]
Factored Form: The factored form for a polynomial function is a product. It looks like
\[ a(x - r_n)(x - r_{n-1}) \cdots (x - r_1) \]

We prefer the factored form for our function analysis. The standard form is just not that helpful for function analysis.

Note: A polynomial may have repeated zeros. For this reason, the factored form usually collects like factors:

\[ a (x - r_k)^{e_k} (x - r_{k-1})^{e_{k-1}} \cdots (x - r_1)^{e_1} \]

Here, the zeros are distinct: $r_i \ne r_j$ when $i \ne j$.

The exponents are called multiplicities of the zeros or the factors.

$\blacktriangleright $ Roots and Zeros:
Zeros of polynomials are also called roots of the polynomials. A polynomial function behaves in one of two ways around a root.

If the multiplicity is odd then the function changes sign over the root. The graph crosses over the horizontal axis at the corresponding intercept.
If the multiplicity is even then the function does not change sign over the root. The graph does not cross over the horizontal axis at the corresponding intercept. Instead, it bounces back in the direction from which it came.

Constant functions are polynomial functions. They don’t have zeros, unless the constant function just happens to be the zero constant function.

Linear functions are polynomial functions. They have one real root, unless it is a constant function. You can solve for this root.

Quadratic functions are polynomial functions. They have $0$, $1$, or $2$ real roots. You can solve for these real roots.

Higher Order Polynomials are polynomial functions with degree $3$ or greater. We do not have quick formulas to obtain their roots. Our usual strategy is to factor, if we can.

$\blacktriangleright $ Continuity:
Polynomials are nice functions. They are continuous everywhere. They have no discontinuities or singularities.

$\blacktriangleright $ Extrema:
Polynomial functions can have global and local maximums and/or minimums, which our algebra is unlikely to identify.

In Calculus, when we have a derivative, then we can attempt to locate exact values of critical numbers. Without the derivative, we turn to technology for some assistance in approximating. That’s the best we can do.

$\blacktriangleright $ Rate-of-Change:
The critical numbers partition the real line into intervals where the polynomial function increases or decreases. All of this depends on getting the critical numbers, which will require a derivative.

A random polynomial is almost impossible to accurately analyze. Our algebra is not strong enough.

Unless the polynomial is special (can be factored nicely), then approximating with a graph is usually unavoidable.

$\blacktriangleright $ Graphs:
Graphs of polynomials are nice. They are smooth. They do not have corners or spikes or breaks or asymptotes. Once we have the roots, we can plot the intercepts. Then we can smoothly connect them according to their multiplicities and have a pretty good sketch of the shape of the graph.

With a basic general shape, we can estimate critical numbers and types of extrema values.

Polynomial

Completely analyze $p(w) = -\frac {1}{5}(w+4)(w-3)(w-3)$

First, let’s collect like factors: $p(w) = -\frac {1}{5}(w+4)(w-3)^2$.

This formula matches our template for a polynomial in factored form.

Domain

Since, $p$ is a polynomial function, its natural domain is $(-\infty , \infty )$.

Zeros

$p$ is a polynomial of degree $3$ and we can see from its factorization that it has two roots, $-4$ and $3$.

Continuity

Since, $p$ is a polynomial function, it is continuous.

End-Behavior

The end-behavior of $p$ is dictated by the leading term and the degree, $-\frac {1}{5} w^3$.

An odd degree tells us that the two directions will have different signs for the end-behavior. The negative leading coefficient tells us that the polynomial is unbounded negatively in the positive direction.

\[ \lim \limits _{w \to \infty }p(w) = -\infty \]

\[ \lim \limits _{w \to -\infty }p(w) = \infty \]

Behavior

We can’t get the behavior exactly. We need the derivatvie for that. We can get an approximate idea.

Since $p(w)$ is a polynomial, these roots (zeros) are the only candidates for where $p(w)$ might change signs. The multiplicity will tell us if $p(w)$ changes sign at each root. If we can determine the sign of $p(w)$ anywhere, then we’ll know the sign of $p(w)$ everywhere.

$-4$ is a root of multiplicity $\answer {1}$. Since this multiplicity is odd, $p$ will change sign not change sign through $-4$ and the graph will cross at $(-4,0)$.
$3$ is a root of multiplicity $\answer {2}$. Since this multiplicity is even, $p$ will change sign not change sign through $3$ and the graph will not cross at $(3,0)$. The graph will touch and then bounce back.

Let’s collect our ideas graphically.

The graph is very suggestive that there is a local minimum somewhere around $-1$ and a local maximum at $3$.

$3$ is a zero, $p(3) = 0$. In addition, $p$ is negative around $3$. That makes $p(3) = 0$ a local maximum. That makes $3$ a critical number as well.

With some technology, we can approximate the other critical number to be $-1.67$ and the local minimum to be $-10.163$.

$p$ is increasing decreasing on $(-\infty , -1.67]$.
$p$ is increasing decreasing on $[-1.67, 3]$.
$p$ is increasing decreasing on $[3, \infty )$.

There is no global maximum or minimum, because $\lim \limits _{w \to -\infty }p(w) = \infty $ and $\lim \limits _{w \to \infty }p(w) = -\infty $.

That’s as close as we are going to get it.

with Calculus

With some Calculus, we could get the exact values of the critical numbers.

Calculus would give us the derivative, $p'(w) = -\frac {1}{5}(3w^2 - 4w - 15)$.

$p'(x)$ is another polynomial. The zeros of this derivative would be critical numbers of $p(x)$. $p'(w)$ is a quadratic. Therefore, we can obtain its zeros, the critical numbers of $p(x)$, via the quadratic formula.

\[ \frac {4 \pm \sqrt {(-4)^2 - 4 \cdot 3 \cdot (-15)}}{2 \cdot 3} = \frac {4 \pm \sqrt {196}}{6} = \frac {4 \pm 14}{6} \]

We get two real roots: $\frac {4 + 14}{6} = \frac {18}{6} = 3$ and $\frac {4 - 14}{6} = \frac {-10}{6} = \frac {-5}{3} \approx -1.67$

This allows us to factor the derivative

\[ p'(w) = -\frac {3}{5}(w - 3) \left ( w + \frac {5}{3} \right ) \]

Each of these roots or factors has a multiplicy of $1$, which means $p'(w)$ changes signs over $-\frac {5}{3}$ and $3$.

$p'(w)$ is a polynomial, which means it is continuous. It is certainly negative for very large negative values of $w$. Then it changes sign over $-\frac {5}{3}$. So, $p'(w)$ is positive between $-\frac {5}{3}$ and $3$. Then, it changes sign to negative at $3$.

$p$ decreases on $\left (-\infty , \frac {-5}{3}\right ]$.
$p$ increases on $\left [\frac {-5}{3}, 3\right ]$.
$p$ decreases on $[3, \infty )$.

Another approach to obtaining the roots of $p'(w) = \frac {1}{5}(3w^2 - 4w - 15)$ would be to factor.

\[ 3w^2 - 4w - 15 = (A \cdot w + B)(C \cdot w + D) \]

$A \cdot C = 3$ and $3$ is a prime number. Therefore, one of $A$ and $C$ is $3$ and the other is $1$.

\[ 3w^2 - 4w - 15 = (w + B)(3 \, w + D) \]

$B \cdot D = -15$. A first guess is that one of $B$ and $D$ is $-5$ and the other is $3$.

\[ 3w^2 - 4w - 15 = (w - 5)(3 \, w + 3) = 3 \, w^2 -12 \, w - 15 \]

Let’s try it the other way, $-3$ and the other is $5$.

\[ 3w^2 - 4w - 15 = (w - 3)(3 \, w + 5) = 3 \, w^2 -4 \, w - 15 \]

That worked.

$p'(w) = \frac {1}{5}(3w^2 - 4w - 15) = \frac {1}{5} (w - 3)(3 \, w + 5)$.

That tells us that $3$ and $-\frac {5}{3}$ are the critical numbers of $p(x)$.

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more examples can be found by following this link
More Examples of Polynomial Functions

2025-08-05 23:49:46