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In this section, we review the vocabulary and arithmetic of polynomials. We start by defining what is meant by the word ‘polynomial’ in general. A more narrow definition of a ‘polynomial function’ will be given in Section GraphsofPolynomials. The general definition suffices for the purposes of this review.
Some examples of polynomials are \(x^2 + x\sqrt {3} + 4\), \(27x^2y + \frac {7x}{2}\) and \(6\). Things like \(3\sqrt {x}\), \(4x - \frac {2}{x+1}\) and \(13x^{2/3}y^{2}\) are not polynomials. In the box below we review some of the terminology associated with polynomials.
Classification by Number of Terms: A simplified polynomial is called a
For example, \(x^2 + x\sqrt {3} +4\) is a trinomial of degree \(2\). The coefficient of \(x^2\) is \(1\) and the constant term is \(4\). The polynomial \(27x^2y + \frac {7x}{2}\) is a binomial of degree \(3\) (\(x^2y = x^2 y^1\)) with constant term \(0\).
The concept of ‘like’ terms really amounts to finding terms which can be combined using the Distributive Property. For example, in the polynomial \(17x^2y - 3xy^2 + 7xy^2\), \(-3xy^2\) and \(7xy^2\) are like terms, since they have the same variables with the same corresponding exponents. This allows us to combine these two terms as follows:
Note that even though \(17x^2y\) and \(4xy^2\) have the same variables, they are not like terms since in the first term we have \(x^2\) and \(y = y^1\) but in the second we have \(x = x^1\) and \(y = y^2\) so the corresponding exponents aren’t the same. Hence, \(17x^2y + 4xy^2\) is the simplified form of the polynomial.
There are four basic operations we can perform with polynomials: addition, subtraction, multiplication and division. The first three of these follow directly from properties of real number arithmetic and will be discussed together. Division, on the other hand, is a bit more complicated and will be discussed separately.
Adding and subtracting polynomials comes down to identifying like terms and then adding or subtracting the coefficients of those like terms. Multiplying polynomials comes to us courtesy of the Generalized Distributive Property.
In particular, Theorem 1 says that, before combining like terms, a product of an \(n\)-term polynomial and an \(m\)-term polynomial will generate \((n \cdot m)\)-terms. For example, a binomial times a trinomial will produce six terms some of which may be like terms. Thus the simplified end result may have fewer than six terms but you will start with six terms.
A special case of Theorem 1 is the famous F.O.I.L., listed here:
That is, \((a+b)(c+d) = ac + ad + bc + bd\).
Theorem 1 is best proved using the technique known as Mathematical Induction which is covered in Section ??. The result is really nothing more than repeated applications of the Distributive Property so it seems reasonable and we’ll use it without proof for now. The other major piece of polynomial multiplication is one of the Power Rules of Exponents from page ?? in Section ??, namely \(a^n a^m = a^{n+m}\). The Commutative and Associative Properties of addition and multiplication are also used extensively. We put all of these properties to good use in the next example.
Perform the indicated operations and simplify.
We begin ‘distributing the negative’ as indicated on page ?? in Section ??, then we rearrange and combine like terms:
Following in our footsteps from the previous example, we first distribute the \(-3z\) through, then rearrange and combine like terms:
At last, we have a chance to use our F.O.I.L. technique:
We use the Generalized Distributive Property here, multiplying each term in the second quantity first by \(3y\), then by \(-\sqrt [3]{2}\):
Exponents do not distribute across powers so we know that \(\left (4w - \frac {1}{2} \right )^2 \neq (4w)^2 - \left (\frac {1}{2}\right )^2\). Instead, we proceed as follows:
Our (correct) final answer is \(16w^2 - 4w + \frac {1}{4}\).
Our last example has two levels of grouping symbols. We begin simplifying the quantity inside the brackets, expanding \((x+h)^2\) in the same way we expanded \((4w - \frac {1}{2})^{2}\) in our previous example:
We conclude our discussion of polynomial multiplication by showcasing two special products which happen often enough they should be committed to memory.
The formulas in Theorem 3 can be verified by working through the multiplication.
We now turn our attention to polynomial long division. Dividing two polynomials follows the same algorithm, in principle, as dividing two natural numbers so we review that process first. Suppose we wished to divide \(2585\) by \(79\). The standard division tableau is given below.
In this case, \(79\) is called the divisor, \(2585\) is called the dividend, \(32\) is called the quotient and \(57\) is called the remainder. We can check our answer by showing:
or in this case, \(2585 = (79)(32) + 57 \checkmark \). We hope that the long division tableau evokes warm, fuzzy memories of your formative years as opposed to feelings of hopelessness and frustration. If you experience the latter, keep in mind that the Division Algorithm essentially is a two-step process, iterated over and over again. First, we guess the number of times the divisor goes into the dividend and then we subtract off our guess. We repeat those steps with what’s left over until what’s left over (the remainder) is less than what we started with (the divisor). That’s all there is to it!
The division algorithm for polynomials has the same basic two steps but when we subtract polynomials, we must take care to subtract like terms only. As a transition to polynomial division, let’s write out our previous division tableau in expanded form.
Written this way, we see that when we line up the digits we are really lining up the coefficients of the corresponding powers of \(10\) - much like how we’ll have to keep the powers of \(x\) lined up in the same columns. The big difference between polynomial division and the division of natural numbers is that the value of \(x\) is an unknown quantity. So unlike using the known value of \(10\), when we subtract there can be no regrouping of coefficients as in our previous example. (The subtraction \(215 - 158\) requires us to ‘regroup’ or ‘borrow’ from the tens digit, then the hundreds digit.) This actually makes polynomial division easier. Before we dive into examples, we first state a theorem telling us when we can divide two polynomials, and what to expect when we do so.
Essentially, Theorem 4 tells us that we can divide polynomials whenever the degree of the divisor is less than or equal to the degree of the dividend. We know we’re done with the division when the polynomial left over (the remainder) has a degree strictly less than the divisor. It’s time to walk through a few examples to refresh your memory.
To begin \(\left (x^3 + 4x^2 - 5x - 14\right ) \div (x-2)\), we divide the first term in the dividend, namely \(x^3\), by the first term in the divisor, namely \(x\), and get \(\frac {x^3}{x} = x^2\). This then becomes the first term in the quotient. We proceed as in regular long division at this point: we multiply the entire divisor, \(x-2\), by this first term in the quotient to get \(x^{2}(x - 2) = x^3 - 2x^2\). We then subtract this result from the dividend.
Now we ‘bring down’ the next term of the quotient, namely \(-5x\), and repeat the process. We divide \(\frac {6x^2}{x} = 6x\), and add this to the quotient polynomial, multiply it by the divisor (which yields \(6x(x - 2) = 6x^{2} - 12x\)) and subtract.
Finally, we ‘bring down’ the last term of the dividend, namely \(-14\), and repeat the process. We divide \(\frac {7x}{x} = 7\), add this to the quotient, multiply it by the divisor (which yields \(7(x - 2) = 7x - 14\)) and subtract.
In this case, we get a quotient of \(x^2 + 6x + 7\) with a remainder of \(0\). To check our answer, we compute
To compute \(\left (2t + 7\right ) \div \left (3t - 4\right )\), we start as before. We find \(\frac {2t}{3t} = \frac {2}{3}\), so that becomes the first (and only) term in the quotient. We multiply the divisor \((3t-4)\) by \(\frac {2}{3}\) and get \(2t - \frac {8}{3}\). We subtract this from the divided and get \(\frac {29}{3}\).
Our answer is \(\frac {2}{3}\) with a remainder of \(\frac {29}{3}\). To check our answer, we compute
When we set-up the tableau for \(\left (6y^2 - 1 \right ) \div \left (2y + 5\right )\), we must first issue a ‘placeholder’ for the ‘missing’ \(y\)-term in the dividend, \(6y^2 -1 = 6y^2 + 0y - 1\). We then proceed as before. Since \(\frac {6y^2}{2y} = 3y\), \(3y\) is the first term in our quotient. We multiply \((2y+5)\) times \(3y\) and subtract it from the dividend. We bring down the \(-1\), and repeat.
For our last example, we need ‘placeholders’ for both the divisor \(w^2 - \sqrt {2} = w^2 + 0w -\sqrt {2}\) and the dividend \(w^3 = w^3 + 0w^2 + 0w + 0\). The first term in the quotient is \(\frac {w^3}{w^2} = w\), and when we multiply and subtract this from the dividend, we’re left with just \(0w^2 + w\sqrt {2} + 0 = w\sqrt {2}\).
Since the degree of \(w\sqrt {2}\) (which is \(1\)) is less than the degree of the divisor (which is \(2\)), we are done. Our answer is \(w\) with a remainder of \(w \sqrt {2}\). To check, we compute:
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