Polynomials

A quadratic function

f(x)=ax^2+bx+c

is also called a polynomial of degree 2. polynomials with larger degrees can be formed by adding terms of the form $x^3$ , $x^4$ and so on. The term with the largest power determines the degree of the polynomial.

Example 1

$f(x)=3x^8+4x^3-7x+1$ is a polynomial of degree $8$ .
$g(x)=-6.1 x^4-x^3+x^2-1$ is a polynomial of degree $4$
$h(x)=\frac{2}{7} x+1$ is a polynomial of degree $1$ (a linear function)
$k(x)=3$ for all $x$ is a polynomial of degree $0$ (a constant function). Why of degree $0$ ? Because we can write $k(x)=3x^0$ , since $x^0=1$ .
$l(x)=x^{10}$ is a polynomial of degree $10$ .
$l(x)=x^{10}+x$ is still a polynomial of degree $10$ .

The numbers in front of the $x$ -terms are called (as always) coefficients. They can assume all values. So, let's give a general definition of polynomials:

Definition 1

In general, a polynomial of degree $n$ (where $n$ can have the value $n=0,1,2,...$ ) has the following form:

\begin{array}{ll} f(x)&=&a_n x^n + a_{n-1}x^{n-1}+...+a_2 x^2+a_1 x^1 + a_0 x^0\\ &=&a_n x^n + a_{n-1}x^{n-1}+...+a_2 x^2+a_1 x^1 + a_0 \end{array}

where for the coefficient $a_n \neq 0$ must hold (otherwise the degree of the polynomial is not $n$ ).

A few more remarks:

We usually write down the polynomials sorted by descending power. So rather like this:
$f(x)=2x^3+1+3x^4$
we write
$f(x)=3x^4+2x^3+1$
All $x$ -terms occurring in the polynomial must have the form $x^0, x^1, x^2, x^3, ...$ . If several such $x$ -terms occur, they are combined. For example:
$f(x)=3x^4+2x^2+x+3x^2+1+5-4x$
is summarised to
$f(x)=3x^4+5x^2-3x+6$
If $x$ -terms with other powers occur, for example
$f(x)=x^4+2\sqrt{x}$
or
$f(x)=x^4-\frac{1}{x^2}+1$
then these are not polynomials.

It is often possible to transform a function into a polynomial.

Example 2

$f(x)=(x-1)^2=x^2-2x+1$
$g(x)=4\sqrt{x}\cdot x^{1.5}=4x^2$

Exercise 1

Can the following functions be converted into a polynomial? If so, what is the degree? Use the correct notation (sorted, same terms combined, see comments above).

$f(x)=x^6 +3 x^{1/2} -2$
$g(x)=(x-1) (x+1)$
$h(x)=\frac{\sqrt{x}}{x^{-7/2}}$
$i(x)=2x^2 (5x^2+x^3-2)$
$j(x)=(1-\sqrt{x})(1+\sqrt{x})$
$k(x)=\frac{4x^3-2x^7+0.5x^3}{2x^2}$
$l(x)=(x-2)(x+1)(x+2)$
$m(x)=-3x^4 -2x +1 +3x^4 -x^2$

Solution

no polynomial.
$g(x)=x^2-1$ , polynomial of degree $2$
$h(x)=x^{1/2} x^{7/2} = x^4$ polynomial of degree $4$
$i(x)=2x^5+10x^4-4x^2$ polynomial of degree $5$
$j(x)=-x+1$ , polynomial of degree $1$
$k(x)=2x-x^5+0.25x=-x^5+2.25x$ , polynomial of degree $5$
$l(x)=(x-2)(x^2+3x+2)=x^3+3x^2+2x-2x^2-6x-4=x^3+x^2-4x-4$ polynomial of degree $3$
$m(x)=-x^2-2x+1$ polynomial of degree $2$ .

The graph of polynomials of a degree $2$ or higher can often be compared to a landscape, with hills and valleys (see picture below).

We already know the graph of some polynomials:

constant function, e.g. $f(x)=3$ (horizontal line)
linear function, e.g. $f(x)=3x+1$ (a straight line)
quadratic function, e.g. $f(x)=2x^2+3x-1$ ( $\cap$ or $\cup$ form, a parabola).

With the exception of the constant function, the graphs of polynomials always tend towards $+\infty$ (upwards) or towards $-\infty$ (downwards). This is indicated in the picture above with arrows on the graphs. So if we run sufficiently far to the right or to the left on the $x$ -axis, the hills and valleys will eventually stop and the graph will only go up or only down.

Whether the graph of a polynomial finally tends upwards or downwards can be seen from the term with the highest power (the degree). This is so because for $x$ -values far away from the zero point (to the left or to the right) this term dominates all other terms. By this we mean that all other terms are so much smaller that they can be neglected. To illustrate this, consider the function

f(x)=3x^2+5x-1

For large $x$ -values (say $x=1000$ ) or for large negative $x$ -values (say $x=-1000$ ) the following applies

f(x)\approx 3x^2

that is, all other terms can be neglected. In fact we have

f(1000)=3\cdot 1000^2+5\cdot 1000-1=3\,004\,999

and

3\cdot 1000^2 = 3\,000\,000

thus

f(1000)\approx 3\cdot 1000^2

Using the function $3x^2$ it is easy to read the behaviour of the graph, hence whether it tends upwards or downwards (upwards). Indeed, the graph of the function $3x^2$ is a power function $x^2$ , scaled in $y$ -direction by $3$ .

In general, we have the following:

Theorem 1

For large positive or large negative $x$ -values, a polynomial of degree $n$ is basically a power function of degree $n$ :

a_n x^n + a_{n-1}x^{n-1}+...+a_2 x^2+a_1 x^1+a_0 \approx a_n x^n

Exercise 2

The following tasks are to be solved without calculator.

Q1

Determine whether for the following polynomials the graph tends upwards or downwards for large positive and negative $x$ values.

$g(x)=-3x^5+2x^4-1$
$g(x)=2(1-x)(x+1)$
$g(x)=-0.1(x-1)^3+1$

Q2

Is the power function $f(x)=2(x-1)^{10}+1$ a polynomial? If so, determine the degree, and the behaviour of the graph for large positive and negative $x$ values.

Solution

A1

$g(x)=-3x^5+2x^4-1\approx -3x^5$ , so for large positive $x$ values the graph goes down, and for large negative $x$ values it goes up.
We have
$\begin{array}{lll} g(x) &= &2(1-x)(x+1)\\ &=&2(x+1-x^2-x)=-2x^2+2 \\ &\approx& -2x^2 \end{array}$
so for large positive $x$ -values the graph goes down, and for large negative $x$ -values as well.
It is
$\begin{array}{lll} g(x)&=&-0.1(x-1)^3+1\\ &=&-0.1(x-1)(x-1)(x-1)+1\\ &=&-0.1(x-1)(x^2-2x+1)+1\\ &=&-0.1(x^3-2x^2+x-x^2+2x-1)+1\\ &=&-0.1 x^3+0.3 x^2-0.3 x+0.1+1\\ &\approx& -0.1\cdot x^3 \end{array}$
so for large positive $x$ values the graph goes down, and for large negative $x$ values it goes up.

A2

It is

\begin{array}{lll} f(x)&=& 2(x-1)^{10}+1\\ &=& 2\cdot \underbrace{(x-1)\cdot (x-1)\cdot (x-1)\cdot ...\cdot(x-1)}_{10 \text{ factors}}+1\\ &=& 2x^{10}+...+3 \end{array}

and thus it follows that $f$ is a polynomial of degree $10$ . Because

f(x)\approx 2x^{10}

it follows that the graph tends upwards for large positive $x$ -values as well as for large negative $x$ -values.