This section covers:
What is a Polynomial?
About the time when you learn exponents in variables, you’ll also learn how to add, subtract, multiply, and then divide (or factor) what we call polynomials. A polynomial (meaning “many” “nomial’s”, or many terms) is basically just a collection of numbers (coefficients) multiplied by letters (variables) that can have exponents, possibly added to or subtracted from other numbers without letters (constants). Polynomials can have fractions as coefficients, but have no variables in denominators – these are called rationals, and we will work with them later in the Rational Functions section. Note that in order for a function to be a polynomial, it’s domain must be all real numbers!
Note: Later, we’ll talk about Graphing and Finding Roots of Polynomial Functions here.
A polynomial with one term is called a monomial; two terms is a binomial; three terms is a trinomial, and four and more terms is typically just called a polynomial. There can be one or more variables in the polynomials, or no variables in polynomials (for example, if there is just a number, or constant).
The degree of the polynomial is the highest exponent of one of the terms (add exponents if there are more than one variable in that term); for example, the degree of \({{x}^{3}}y\) is \(3+1=4\).
A polynomials having a degree higher than four is typically just called a polynomial, or a polynomial of degree \(n\), the highest degree of five is called a quintic, the highest degree of four is called a quartic, three is a cubic, two is a quadratic (you’ll hear a lot more about these!), one a linear, and zero (just a number) is a constant. So make sure you understand how we named the polynomials below:
Polynomial 
Degree 
Number of Terms 
Name 
\(10{{x}^{3}}+4{{x}^{2}}+x4\) 
3 (from the \({{x}^{3}}\)) 
4  Cubic Polynomial 
\(t\left( {{{t}^{3}}+t} \right)={{t}^{4}}+{{t}^{2}}\) 
4 (from the \({{t}^{4}}\)) 
2  Quartic Binomial 
\(8\) 
0 (no variables) 
1  Constant Monomial 
\(\displaystyle \frac{{\left( {x+4} \right)}}{2}+\frac{{xy}}{{\sqrt{3}}}+3\) 
2 (from the \(xy\)) 
3  Quadratic Trinomial 
\(4{{x}^{3}}{{y}^{4}}+2{{x}^{2}}y+xy+3xy+x+y4\) 
7 (from the \({{x}^{3}}{{y}^{4}}\)) 
6 (since we can combine the \(xy\) and \(3xy\))  Polynomial of Degree 7 
\(x{{\left( {x+4} \right)}^{2}}{{\left( {x3} \right)}^{5}}\) 
8 (add up the exponents: \(1+2+5=8\)). 
(Difficult to say unless multiply out)  (Difficult to say unless multiply out) 
These are not polynomials (notice the variable in the denominator or a root of a variable):
Expression 
Why it is not a Polynomial 
\(\displaystyle 3{{x}^{2}}+\frac{1}{x}+1\)  There can’t be a variable in the denominator. 
\(\displaystyle 4+2{{t}^{{3}}}=4+\frac{2}{{{{t}^{3}}}}\)  There can’t a variable with a negative exponent (variable in the denominator). 
\({{x}^{2}}\sqrt{x}\)  There can’t be any variables that are radicals (roots). 
\(\displaystyle \frac{{x+3}}{{x2}}\)  There can’t be any variable in the denominator, even if there’s one on the top. 
When adding and subtracting polynomials, you just put the terms with the same variables and exponents together. These are called “like terms”. So \(3{{x}^{2}}y\) and \({{x}^{2}}y\) are like terms (adding them would be \(2{{x}^{2}}y\)), \(4xy\) and \(yx\) are like terms (adding them would be \(5xy\)), but \(4{{x}^{2}}{{y}^{2}}\) and \(4x{{y}^{2}}\) are not.
We’re actually using the distributive property (sort of backwards) when we put together like terms. We saw this with linear functions, but it applies to other functions as well. Notice that we have invisible “1”’s before variables with no numbers (coefficients). For example:
\(\begin{array}{c}\color{#800000}{{3{{x}^{2}}+4{{x}^{2}}+3xx}}=\left( {3+4} \right){{x}^{2}}+\left( {31} \right)x=7{{x}^{2}}+2x\\\color{#800000}{{{{t}^{3}}4{{t}^{3}}}}=\left( {14} \right){{t}^{3}}=3{{t}^{3}}\\\color{#800000}{{7{{x}^{2}}y+3x{{y}^{2}}+4y+11{{x}^{2}}y2y2x{{y}^{2}}+4}}=\left( {7+11} \right){{x}^{2}}y+(32)x{{y}^{2}}+\left( {42} \right)y+4=18{{x}^{2}}y+x{{y}^{2}}+2y+4\end{array}\)
Note on the last example of a polynomial above, we had two different letters (variables) in the same term; they each represent a different amount. We’ll talk more about working with more than one variable later.
Multiplying Polynomials: FOILING, and “Pushing Through”
Multiplying binomials (also sometimes called FOILING) is done frequently in your algebra classes. There are two different ways to do this: the FOIL (First Outer Inner Last) method or the more generic “pushing through” (distributing) method or just doing “long multiplication”. Later, we’ll learn how to undo the multiplication of the polynomials (or factor or “UNFOIL”) when we want to turn them back into factors.
Let’s first show multiplying binomials with a multiplication box. Let’s multiply \(\left( {x+3} \right)\left( {x2} \right)\).
We split up the first binomial on the top, and the second down the left side. (We have to watch the signs, and turn the minus 2 in the second binomial into a negative 2). Then we multiply across and down to fill in all the boxes, and then add up all the boxes:
Since we won’t want to draw a box every time we multiply binomials, we have a several methods to help us.
FOILing
FOIL stands for First Outer Inner Last. Just remember that we multiply the First Outer Inner and then Last terms and put plus signs between them (unless the product is negative). Here are some examples. Note that FOILing only works if you multiply binomials – each factor has two terms. In other cases, we’ll do the pure “distributing” method.
Note the last two examples are “special cases” that you’ll see a lot: difference of two squares, and perfect square trinomials; note the shortcuts with these cases.
Multiplying Binomials  Notes 
\(\displaystyle \begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,F\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, O \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L\\\color{#800000}{{\left( {x+1} \right)\left( {x+2} \right)}}\,\,&=\left( {x\cdot x} \right)+\left( {x\cdot 2} \right)+\left( {1\cdot x} \right)+\left( {1\cdot 2} \right)\\&={{x}^{2}}+2x+1x+2\,\,\,\\&=\,\,\,{{x}^{2}}+3x+2\end{align}\)  In most cases, when the terms are the same in each binomial, you’ll end up combining the “O” and “I” of FOIL (First, Outer, Inner, Last). These are called the middle terms. 
\(\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, O \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L\\\color{#800000}{{\left( {4x7} \right)\left( {2x2} \right)}}&=\left( {4x\cdot 2x} \right)+\left( {4x\cdot 2} \right)+\left( {7\cdot 2x} \right)+\left( {7\cdot 2} \right)\\&=8{{x}^{2}}8x14x+\left( {–14} \right)\\&=8{{x}^{2}}22x+14\end{align}\)  Watch the negatives. Remember that \(+ \) is the same as \(\), and \(– \) is the same as \(+\). 
\(\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,F\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, O \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L\\\color{#800000}{{\left( {x7y} \right)\left( {x+3y} \right)}}&=\left( {x\cdot x} \right)+\left( {x\cdot 3y} \right)+\left( {7y\cdot x} \right)+\left( {7y\cdot 3y} \right)\\&={{x}^{2}}+3xy7xy\left( {21{{y}^{2}}} \right)\\&={{x}^{2}}4xy21{{y}^{2}}\end{align}\)  Sometimes you have more than one variable in the binomials. Notice in this case that the middle terms could still be combined, since the variables are the same in the binomials. 
\(\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,F\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, O \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L\\\color{#800000}{{\left( {ab} \right)\left( {c+d} \right)}}&=\left( {a\cdot c} \right)+\left( {a\cdot d} \right)+\left( {b\cdot c} \right)+\left( {b\cdot d} \right)\\\,&=ac+adbcbd\end{align}\)  When the two binomials contain different variables, you can’t combine the middle terms. 
\(\begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,F\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, O \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L\\\color{#800000}{{\left( {x+3} \right)\left( {x3} \right)}}&=\left( {x\cdot x} \right)+\left( {3\cdot x} \right)+\left( {3\cdot x} \right)+\left( {3\cdot 3} \right)\\&={{x}^{2}}+3x+3x9\\&={{x}^{2}}9\text{ (notice difference of squares)}\end{align}\) 
Difference of Two Squares This is a special case when you add something and subtract the same thing. The middle terms cancel out and you just have the square of the first term minus the square of the second term. 
\(\displaystyle \begin{align}&\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,F\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, O \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, I\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, L\\\color{#800000}{{{{{\left( {x+3} \right)}}^{2}}}}&=\left( {x+3} \right)\left( {x+3} \right)=\left( {x\cdot x} \right)+\left( {x\cdot 3} \right)+\left( {3\cdot x} \right)+\left( {3\cdot 3} \right)\\&={{x}^{2}}+3x+3x+9\\&={{x}^{2}}+6x+9\end{align}\) 
Perfect Square Trinomial This is a special case where the binomials are exactly the same. We end up with the sum of the square of the first term, added to twice the product of the two terms, added to the square of the last term. 
“Pushing Through” or Distributing Terms of Polynomials
Really, what we are doing when we are FOILing is using the distributive method to make sure every term (variable or number) is multiplied by every other one and then you add them all up. We can also think of this as “pushing through” the terms to every other term.
Also called “double distributing”, this way of multiplying binomials is more popular now, since it can be used for any polynomial.
Let’s do a couple of the problems above and also see how “pushing through” can be used with polynomial products when we don’t have two binomials:
Distributing Polynomials  Notes 
We start with the first \(\color{blue}{x}\) and “push it through” to both the \(\color{red}{x}\) and the \(\color{#66ff00}{2}\) in the second binominal.
Then we take the \(\color{#D982b5}{1}\) and “push it through” to both the \(\color{red}{x}\) and the \(\color{#66ff00}{2}\) in the second binomial. Remember to put “\(+\)” signs in between all the terms, unless we have to deal with negatives. 

Sometimes it’s easier to “push through” the terms on two (or more if necessary) separate rows so you can line up any “like” terms to add them.  
In fact, if you put the polynomial with the most terms first, you can actually multiply them like long multiplication and line up the terms.
Multiply the \({{x}^{2}}x+6\) by 8 first, and then by \(x\). 

\(\displaystyle \begin{array}{l}\color{#800000}{{\left( {x+2} \right)\left( {x1} \right)\left( {2x9} \right)}}\\=\left( {\left( {x+2} \right)\left( {x1} \right)} \right)\left( {2x9} \right)\\=\left( {{{x}^{2}}x+2x2} \right)\left( {2x9} \right)\\=2{{x}^{3}}9{{x}^{2}}2{{x}^{2}}+9x+4{{x}^{2}}18x4x+18\\=2{{x}^{3}}7{{x}^{2}}13x+18\end{array}\)  If you have three binomials, you can multiply any two first and then multiply this product by the other. See how I “pushed through” the \({{x}^{2}}\), and then the \(x\), and then the \(2x\), and then the \(2\) to the \(2x\) and then the \(9\) Then combine the middle terms. 
Much more multiplying and factoring polynomials will be in the Introduction to Quadratics and Graphing and Solving Polynomials sections.
Learn these rules, and practice, practice, practice!
For Practice: Use the Mathway widget below to try a Polynomial Multiplication problem. Click on Submit (the blue arrow to the right of the problem) and click on Multiply to see the answer.
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On to Introduction to Quadratics – you are ready!