Precalculus: Lecture Notes

Section 2.4 Polynomials

Truth is much too complicated to allow anything but approximations. — John Von Neumann, Hungarian-American pure and applied mathematician, physicist, inventor, and polymath, 1903 – 1957

Why polynomials: The irrational number \(e\) is an important mathematical constant. It is approximately equal to \(2.718282846\text{.}\) A scientific calculator gives that

Problem: Is it possible to quickly find a reasonable approximation of the number \(\sqrt{e}\) by using only the four basic arithmetic operations: addition, subtraction, multiplication and division?

Try this:

• Let \(p_1(x)=1+x\text{.}\) Then \(\displaystyle p_1\left( \frac{1}{2}\right)=\)

  Solution

  \(\displaystyle p_1\left( \frac{1}{2}\right)=1+\frac{1}{2}=\frac{3}{2}=1.5\)

• Let \(\displaystyle p_2(x)=1+x+\frac{1}{2}\cdot x^2\text{.}\) Then \(\displaystyle p_2\left( \frac{1}{2}\right)=\)

  Solution

  \(\displaystyle p_2\left( \frac{1}{2}\right)=1+\frac{1}{2}+\frac{1}{8}=\frac{13}{8}=1.625\)

• Let \(\displaystyle p_3(x)=1+x+\frac{1}{2}\cdot x^2+\frac{1}{6}\cdot x^3\text{.}\) Then \(\displaystyle p_3\left( \frac{1}{2}\right)=\)

  Solution

  \(\displaystyle p_3\left( \frac{1}{2}\right)=1+\frac{1}{2}+\frac{1}{8}+\frac{1}{48}=\frac{79}{48}\approx 1.645\)

These three functions, \(p_1(x)=1+x\text{,}\) \(\displaystyle p_2(x)=1+x+\frac{1}{2}\cdot x^2\text{,}\) and \(\displaystyle p_3(x)=1+x+\frac{1}{2}\cdot x^2+\frac{1}{6}\cdot x^3\text{,}\) belong to an important family of functions called polynomials.

Polynomial: A polynomial is a function \(p\) such that

where \(n\) is a nonnegative integer and \(a_0,a_1,a_2,\ldots , a_n\) are given numbers with \(a_n\not=0\text{.}\)

Why polynomials: See Figure 2.36.

Vocabulary and Facts: Let

be a polynomial.

• The number \(n\) is called the degree of the polynomial \(p\) and it is denoted by \(\deg p=n\text{.}\)

  Example 2.4.1. Find thedegree.

  The degree of the polynomial

  \begin{equation*} p(x)=3x+4+5x^3-x^4-x^2. \end{equation*}

  is equal to:

  1. \(\deg p =1\text{?}\)

  2. \(\deg p =2\text{?}\)

  3. \(\deg p =3\text{?}\)

  4. \(\deg p =4\text{?}\)

  Solution

  Observe that

  \begin{equation*} p(x)=3x+4+5x^3-x^4-x^2=-x^4+5x^3-x^2+3x. \end{equation*}

  Hence, \(\deg p =4\text{.}\)

• If the number \(x_0\) is such that

  \begin{equation*} p(x_0)=0 \end{equation*}

  then \(x_0\) is called a zero of the polynomial \(p\text{.}\)

  Example 2.4.2. Find a zero.

  Which of the following numbers is a zero of of the polynomial

  \begin{equation*} p(x)=x^3-2x^2+x? \end{equation*}

  1. \(x =1\text{?}\)

  2. \(x =2\text{?}\)

  3. \(x =3\text{?}\)

  4. \(x =4\text{?}\)

  Solution

  From

  \begin{equation*} p(1)=1^3-2\cdot1^2+1=0, p(2)=2^3-2\cdot 2^2+1=1, p(3)=12, \mbox{ and } p(4)=33 \end{equation*}

  it follows that \(x=1\) is a zero of the the polynomial \(p\text{.}\)

• If the number \(x_0\) is a zero of the polynomial \(p\) then there is a polynomial \(q(x)\) such that

  \begin{equation*} p(x)=(x-x_0)\cdot q(x). \end{equation*}

  Example 2.4.3. Zeros and factors.

  Check that

  \begin{equation*} p(x)=x^3-2x^2+x=(x-1)(x^2-x). \end{equation*}

  Solution

  First observe that \(p(1)=0\) and conclude that \(x=1\) is a zero of the polynomial \(p\)

  Next we observe that, by the distributive law,

  \begin{equation*} (x-1)(x^2-x)=x(x^-x)-(x^2-x)=x^3-x^2-x^2+x=x^3-2x^2+x=p(x). \end{equation*}

• If the number \(x_0\) is a zero of the polynomial \(p\) and if there are a number \(m\) and a polynomial \(q(x)\) such that

  \begin{equation*} p(x)=(x-x_0)^m\cdot q(x) \end{equation*}

  with \(q(x_0)\not=0\) then we say that \(x_0\) is a zero of order \(m\text{.}\)

  Example 2.4.4. Degree of a zero.

  Check that \(x=1\) is a zero of order 2 of the polynomial

  \begin{equation*} p(x)=x^3-2x^2+x. \end{equation*}

  Solution

  First observe that \(p(1)=0\) and conclude that \(x=1\) is a zero of the polynomial \(p\text{.}\)

  Recall from the previous example that

  \begin{equation*} p(x)=x^3-2x^2+x=(x-1)(x^2-x)=(x-1)r(x), \end{equation*}

  where \(r(x)=x^2-x\text{.}\)

  Next, we note that \(r(1)=0\) and that \(r(x)=(x-1)x\text{.}\)

  It follows that \(p(x)=(x-1)r(x)=(x-1)^2x=(x-1)^2q(x)\text{,}\) where \(q(x)=x\text{.}\)

  Since \(q(1)=1\not=0\text{,}\) by definition, \(x=1\) is a zero of order 2 of the polynomial \(p\text{.}\)

  See Figure 2.37.

  

• A polynomial of degree \(n\) can have at most \(n\) zeros.

  Example 2.4.5. How many zeros?

  See Figure 2.37 and Figure 2.39.

  

  

All You Need to Know About Polynomials: Let

be a polynomial.

1. What is the domain of the function \(p\text{?}\)

   Solution

   Observe that to evaluate \(p(x)\) one needs only to use two algebraic operations, addition and multiplications. This implies that the domain of any polynomial is the set of all real numbers.

2. Determine the behaviour of the polynomial \(p\) if \(x\to\infty\) and \(x\to -\infty\text{.}\)

   The same … just little bit different: What can you say about the ratio

   \begin{equation*} \frac{p(x)}{a_nx^n}=\frac{a_0+a_1x+a_2x^2+\ldots+a_nx^n}{a_nx^n} \end{equation*}

   if \(x\) is such that \(|x|\) is very big?

   Solution

   1. We first recall that if a real number \(x\) is such that \(|x|\) is very big then its reciprocal \(\displaystyle \frac{1}{x}\) is very small.

      In other words,

      \begin{equation*} x\in \mathbb{R}\text{ is such that } |x| \text{ is very big } \Leftrightarrow \frac{1}{x}\approx 0. \end{equation*}

      For example, think about the reciprocal of \(x=10,000\) or \(x=-100,000\text{.}\)

   2. Next we observe that

      \begin{equation*} \frac{p(x)}{a_nx^n}= \frac{a_0+a_1x+a_2x^2+\ldots+a_nx^n}{a_nx^n} \end{equation*}
      \begin{equation*} = \frac{a_0}{a_nx^n}+\frac{a_1x}{a_nx^n}+\frac{a_2x^2}{a_nx^n}+\ldots+\frac{a_{n-1}x^{n-1}}{a_nx^n}+\frac{a_nx^n}{a_nx^n} \end{equation*}
      \begin{equation*} = \frac{a_0}{a_nx^n}+\frac{a_1}{a_nx^{n-1}}+\frac{a_2}{a_nx^{n-2}}+\ldots+\frac{a_{n-1}}{a_nx}+1 \end{equation*}
      \begin{equation*} = \frac{a_0}{a_n}\cdot\frac{1}{x^n}+\frac{a_1}{a_n}\cdot\frac{1}{x^{n-1}}+\frac{a_2}{a_n}\cdot\frac{1}{x^{n-2}}+\ldots+\frac{a_{n-1}}{a_n}\cdot\frac{1}{x}+1 \end{equation*}
      \begin{equation*} \approx\frac{a_0}{a_n}\cdot 0+\frac{a_1}{a_n}\cdot 0+\frac{a_2}{a_n}\cdot 0+\ldots+\frac{a_{n-1}}{a_n} \cdot 0+1 = 1. \end{equation*}

   3. Therefore, if \(x\) is such that \(|x|\) is very big then,

      \begin{equation*} \frac{p(x)}{a_nx^n}\approx 1 \Leftrightarrow p(x)\approx a_nx^n. \end{equation*}

      In other words, if \(x\to\infty\) or if \(x\to -\infty\) then the polynomial \(p\) behaves as its leading term \(a_nx^n\text{.}\)

   4. As an example consider the graphs of the polynomials

      \begin{equation*} p(x)=0.5x^3-2x+1 \text{ and }q(x)=0.5x^3. \end{equation*}

      See Figure 2.40.

      

3. What is the range of the function \(p\text{?}\)

   Solution

   See Figure 2.41 and Figure 2.39.

   

   Conclusion:

   \begin{equation*} \begin{array}{c|c|c} \text{Degree}\amp a_n\amp \text{Range}\\ \hline \text{Odd}\amp \amp (-\infty,\infty)\\ \hline \text{Even}\amp a_n\gt 0\amp [a,\infty)\\ \hline \text{Even}\amp a_n\lt 0\amp (-\infty,b]\\ \end{array} \end{equation*}

4. Find the zeros of the function \(p\text{,}\) i.e., solve the equation \(p(x)=0\text{.}\)

   1. If \(p(x)=ax+b\) is a polynomial of degree 1 then we solve the linear equation \(ax+b=0\) to obtain \(\displaystyle x=-\frac{b}{a}\text{.}\)

   2. If \(p(x)=ax^2+bx+c\) is a polynomial of degree 2 then we solve the quadratic equation \(ax^2+bx+c=0\) to using the quadratic formula \(\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\text{.}\)

   3. If \(p(x)\) is a polynomial of degree greater than 2 then we try to factor the given polynomial. Often we use the modern technology. See, for example, WolframAlpha.

      Reminder: A polynomial of degree \(n\) has at most \(n\) real zeros

   4. Fact: Polynomials are so–called continuous functions: This means if \(a\) and \(b\) are real numbers such that \(a\lt b\) and \(p(a)\) and \(p(b)\) are not of the same sign (i.e. one of them is positive and the other is negative) then if you wish to connect the points \((a,p(a))\) and \((b,p(b))\) without lifting your pen then you have to cross the \(x\)–axis. In other words: there is \(c\)between \(a\) and \(b\text{,}\) i.e. \(a\lt c\lt b\text{,}\) such that \(p(c)=0\text{.}\)

      See Figure 2.42.

      

Example 2.4.6. Putting everything together.

You know that \(p\) is a polynomial of an odd degree, with a positive leading coefficient, and that

\begin{equation*} p(-1)=1, p(0)=-1, p(1)=1, p(2)=0. \end{equation*}

See Figure 2.43.

1. What can you tell about the number of zeros of the polynomial \(p\text{?}\)

2. What can you tell about the degree of the polynomial \(p\text{?}\)

Solution

1. If you connect given elements of the graph of the mysterious polynomial you may get something like the graph shown in Figure 2.44.

   

2. Since there are at least four zeros and since the polynomial is of an odd degree, its degree is at least five.