A polynomial is a function \(f(x)\) that can be written as a finite sum of terms of the form \(a x^n\) where \(a\) is a constant and \(k\) is a non-negative integer (\(k=0, 1, 2, \ldots\)). Such terms are called monomials. So a polynomial is a finite sum of monomials.
\(f(x) = 3\) is a polynomial. There is exactly one term with \(a=3\) and \(k=0\).
\(g(x) = 2x-5\) is a polynomial. There are exactly two terms. The first term is \(2x\) with \(a=2\) and \(k=1\). The second term is \(-5\) with \(a=-5\) and \(k=0\).
\(h(x) = (x-5)(x+1)\) is a polynomial. To see this, we must first expand it so that \(h(x) = x^2-4x-5\). In this form, there are three terms. The term is \(x^2\) has \(a=1\) and \(k=2\). The term \(-4x\) has \(a=-4\) and \(k=1\). And the term \(-5\) has \(a=-5\) and \(k=0\).
\(k(x) = 1+x+x^2+x^3+ \cdots\) is not a polynomial. Although each term is the right type, there are infinitely many terms. This type of function is called a series and is a limit of polynomials. Such functions are studied in Calculus 2.
\(z(x) = 3x^{1/2} - 4x^{-3}\) is a not a polynomial. The powers for the terms are not non-negative integers.
For a given monomial \(ax^k\), we call the constant \(a\) the coefficient and the power \(k\) the degree. For a polynomial, the largest degree of all the terms is called the degree of the polynomial, and its corresponding coefficient is called the leading coefficient. For a given polynomial \(f(x)\), we write \(\deg(f)\) to represent the degree.
Once we know the degree \(n\) of the polynomial, we know that there are at most \(n+1\) monomials with possible degrees \(0, 1, \ldots, n\). A common way to describe the degree \(k\) monomial as \(a_k x^k\) so that we have distinct labels for the coefficients \(a_0, a_1, \ldots, a_n\). Then a general polynomial of degree \(n\) can therefore be written as \[ f(x) = a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n. \] Every polynomial can be uniquely described using the list of coefficients.
\(f(x) = 3\) has \(\deg(f)=0\) and coefficient list \((a_0)=(3)\).
\(g(x) = 2x-5\) has \(\deg(g)=1\) and coefficient list \((a_0,a_1)=(-5,2)\).
\(h(x) = (x-5)(x+1) = x^2-4x-5\) has \(\deg(h)=2\) and coefficient list \((a_0,a_1,a_2)=(-5, -4, 1)\).
Find the polynomial \(k(x)\) with coefficient list \((1, -3, 0, 2)\). We can read off the coefficients: \(a_0=1\), \(a_1=-3\), \(a_2=0\), and \(a_3=2\). So we know \(\deg(k)=3\). The \(x^2\) term has coefficient 0, so it will not appear. We find \[k(x) = 1 + -3x + 0x^2 + 2x^3 = 2x^3-3x+1.\]
Polynomials are a particularly simple family of functions. One of the reasons they are so nice is that they can be constructed using only the operations of addition and multiplication. Even the powers of \(x\) can be found by multiplication. And we don't even need division. Some of the most important results for polynomials are related to the ideas of factoring.
The division algorithm is similar to the idea of integer division. Given positive integers \(n\) and \(k\), we can find integers \(q\) (the quotient) and \(r\) (the remainder) so that \[n = k \cdot q + r.\] Using integer long division, we would divide \(n\) by \(k\) to get a quotient \(q\) and remainder \(r\). You probably learned this in the context of improper fractions and mixed fractions as \[ \frac{n}{k} = q + \frac{r}{k}. \]
Polynomial division can lead to similar representations, which we call rational functions. By the division algorithm, given polynomials \(f(x)\) and \(p(x)\), we can find the quotient and remainder polynomials \(q(x)\) and \(r(x)\), respectively. In the language of rational functions, this means \[ \frac{f(x)}{p(x)} = q(x) + \frac{r(x)}{p(x)}. \]
We will explore the ideas of rational functions and how to perform this division later. For us, our main interest is in understanding how the division algorithm relates to factoring. Consider any polynomial \(f(x)\) with \(\deg(f) \ge 1\). Now consider a divisor \(p(x)=x-c\) where \(c\) is some number. The division algorithm will then give us two polynomials \(q(x)\) with \(\deg(q) = \deg(f)-1\) and \(r(x)\) with \(deg(r) = 0\), meaning that \(r(x) = r\) is a constant, so that \[ f(x) = (x-c) \cdot g(x) + r. \]
What happens if we evaluate \(f(x)\) at the number \(x=c\)? \[ f(c) = (c-c) \cdot g(c) + r = r. \] That is, the remainder from the division algorithm is the value of \(f(c) = r\). This is the source of the Factor–Root Theorem of polynomials, which focuses on the case when there is no remainder,\(r=0\).
That is, if we are trying to factor a polynomial, then we can start by trying to identify the roots (or zeros).
Example: Factor \(f(x) = x^3 - 2x - 4\).
This polynomial is not a quadratic, so our usual methods do not work. However, creating a quick table (or looking at a graph) we find that \(f(2) = 0\). This means that \(x-2\) must be a factor by the Factor–Root Theorem. We can find the complementary factor \(q(x)\) (which must have \(\deg(q)=2\)) by choosing coefficients that properly expand to the polynomial \(f(x)\): \[ \begin{aligned} x^3 - 2x - 4 &= (x-2)(a_2x^2+a_1x+a_0) \\ &= a_2 x^3 + (a_1-2a_2) x^2 + (a_0 - 2a_1) x + -2a_0 \end{aligned} \] Terms must match, so we know \(a_2 = 1\), \(a_1-2a_2 = 0\), \(a_0-2a_1 = -2\) and \(-2a_0 = -4\). The first equation automatically tells us \(a_2 = 1\). The next equation can then allow us to solve for \(a_1 = 0 + 2a_2 = 2\). Now that we know \(a_1=2\), the third equation lets us solve for \(a_0 = -2 + 2a_1 = 2\), which is consistent with the last equation \(-2a_0 = -4\). Thus, our factorization becomes \[ f(x) = (x-2) \cdot (x^2+2x+2). \]
A method called synthetic division provides an efficient algorithm to accomplish this task.
Consider attempting to factor \(f(x) = 2x^3 - 3x^2 - 4x + 5\) for a potential root at \(x=c\). We do not know if there is a remainder or not. The division algorithm lets us write and then expand \[ \begin{aligned} 2x^3 - 3x^2 - 4x + 5 &= (x-c) \cdot (a_2 x^2 + a_1 x + a_0) + r \\ &= a_2 x^3 + (a_1 - c a_2) x^2 + (a_0- c a_1)x + (r-c a_0). \end{aligned} \] Again, we can match individual terms to give us equations for the coefficients. \[ \begin{aligned} a_2 &= 2 \\ a_1 - c a_2 &= -3 \\ a_0 - c a_1 &= -4 \\ r - c a_0 &= 5 \end{aligned} \qquad \Rightarrow \qquad \begin{aligned} a_2 &= 2 \\ a_1 &= -3 + c a_2 \\ a_0 &= -4 + c a_1 \\ r &= 5 + c a_0 \end{aligned} \]
This process is always the same when dealing with factors of the form \(x-c\) for a proposed root at \(x=c\). We can use a table with three rows to compute the values. The first row will contain the coefficients of the original function \(f(x)\) by descending degree, which for our example above are 2, -3, -4, and 5. (Be sure to include 0 for any missing monomials.) The third row will have the coefficients of the quotient, which will be \(a_2, a_1, a_0, r\). The middle row will contain the multiples that are added to get the next coefficient.
To learn the process, choose a value for \(c\) and then follow the steps that are presented one step at a time.
Choose where you think a root occurs. Then press Next to see the steps to fill the table.
c=
For each subsequent column, do the following two steps:
(a) Multiply the previous column's third row number by \(c\)
and put the result in the second row.
(b) Add the numbers in the first and second row to obtain the next number in the third row, which is the next coefficient of the quotient.
Repeat the previous sequence of steps for every column to obtain all of the coefficients.
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