Introduction to Multiplying Polynomials

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 an expression that includes variables, constants, and exponents (not radicals) that are combined by addition, subtraction, and multiplication. They can also be combined by division, but no variables can be in the denominator; these are called rational functions, and we will work with them later in the Rational Functions, Equations, and Inequalities section. Note that in order for a function to be a polynomial, it’s domain must be all real numbers!

(Note that more multiplying (and factoring) polynomials will be in the Introduction to Quadratics and the Graphing and Finding Roots of Polynomial Functions sections.)

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 polynomial having a degree higher than five is typically just called a polynomial, or a polynomial of degree $ n$. Otherwise, a degree is five is called a quintic, a degree of four a quartic, three a cubic, two a quadratic (you’ll hear a lot more about these!), one a linear, and zero (just a number) a constant. Here are some examples:

Polynomial

Degree

Number of Terms

Name

$ 10{{x}^{3}}+4{{x}^{2}}+x-4$

$ 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+y-4$

$ 7$ (from the $ {{x}^{3}}{{y}^{4}}$)

$ 6$, since we can combine the $ xy$ and $ 3xy$ Polynomial of Degree $ \boldsymbol {7}$
 $ x{{\left( {x+4} \right)}^{2}}{{\left( {x-3} \right)}^{5}}$

$ 8$

(add up the exponents: $ 1+2+5=8$).

(Difficult to say unless multiply out) Polynomial of Degree $ \boldsymbol {8}$

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}}{{x-2}}$ There can’t be any variable in the denominator, even if there’s one on the top.

When adding and subtracting polynomials, put the terms with the same variables and exponents together. These are called “like terms”. For example, $ 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 when Solving Algebraic (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:

$ \displaystyle \begin{align}\color{#800000}{{3{{x}^{2}}+4{{x}^{2}}+3x-x}}&=\left( {3+4} \right){{x}^{2}}+\left( {3-1} \right)x=7{{x}^{2}}+2x\\\color{#800000}{{{{t}^{3}}-4{{t}^{3}}}}&=\left( {1-4} \right){{t}^{3}}=-3{{t}^{3}}\\\color{#800000}{{7{{x}^{2}}y+3x{{y}^{2}}+4y+11{{x}^{2}}y-2y-2x{{y}^{2}}+4}}&=\left( {7+11} \right){{x}^{2}}y+(3-2)x{{y}^{2}}+\left( {4-2} \right)y+4\\&=18{{x}^{2}}y+x{{y}^{2}}+2y+4\end{align}$

Multiplying Polynomials: FOILING, and “Pushing Through”

Multiplying binomials (also sometimes called FOILING) is done frequently in your algebra classes. There are 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 (factoring) when we want to turn them back into factors.

You can use a multiplication box to multiply polynomials; let’s multiply $ \left( {x+3} \right)\left( {x-2} \right)$. Split up the first binomial on the top, and the second down the left side. (Watch the signs, and turn the minus $ 2$ in the second binomial into a $ -2$). Multiply across and down to fill in all the boxes, and then add up all the boxes:

$ \boldsymbol {x}$ $ \boldsymbol {3}$

When we add up everything in the boxes, we get:

 

$ \begin{array}{l}{{x}^{2}}+3x-2x+-6\\\,\,\,={{x}^{2}}+x-6\end{array}$

$ \boldsymbol {x}$ $ {{x}^{2}}$ $ 3x$
$ \boldsymbol {-2}$ $ -2x$ $ -6$

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. Multiply together the First, Outer, Inner, and then Last terms, and put plus signs between them (or a minus sign if  the product is negative). 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; there are shortcuts for these cases.

Notice how the middle term is always the sum of the products of the inside and outside terms; soon this will become second nature and you won’t think about it!

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( {4x-7} \right)\left( {2x-2} \right)}}&=\left( {4x\cdot 2x} \right)+\left( {4x\cdot -2} \right)+\left( {-7\cdot 2x} \right)+\left( {-7\cdot -2} \right)\\&=8{{x}^{2}}-8x-14x+\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( {x-7y} \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}}+3xy-7xy-\left( {21{{y}^{2}}} \right)\\&={{x}^{2}}-4xy-21{{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( {a-b} \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+ad-bc-bd\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( {x-3} \right)}}&=\left( {x\cdot x} \right)+\left( {-3\cdot x} \right)+\left( {3\cdot x} \right)+\left( {3\cdot -3} \right)\\&={{x}^{2}}+-3x+3x-9\\&={{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

When we are FOILing, we are actually using the Distributive Property 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.

Here are some examples; notice how this way is much more generic and can be used with polynomials with any number of terms.

Distributing Polynomials Notes
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 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 a product is negative.

Sometimes it’s easier to “push through” the terms on two (or more if necessary) separate rows so you can line up 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 $ {{y}^{2}}-y+6$ by $ 8$ first, and then by $ y$.

$ \displaystyle \begin{array}{l}\color{#800000}{{\left( {x+2} \right)\left( {x-1} \right)\left( {2x-9} \right)}}\\\,\,=\left[ {\left( {x+2} \right)\left( {x-1} \right)} \right]\left( {2x-9} \right)\\\,=\left( {{{x}^{2}}+x-2} \right)\left( {2x-9} \right)\\\,=2{{x}^{3}}-9{{x}^{2}}+2{{x}^{2}}-9x+-4x+18\\\,=2{{x}^{3}}-7{{x}^{2}}-13x+18\end{array}$ If you have three binomials, multiply any two first and then multiply this product by the other. See how I first multiplied out the first two terms, and then multiplied this through to the last one?

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.

You can also type in your own problem, or click on the three dots in the upper right hand corner and click on “Examples” to drill down by topic.

If you click on “Tap to view steps”, or Click Here, you can register at Mathway for a free trial, and then upgrade to a paid subscription at any time (to get any type of math problem solved!).

On to Introduction to Quadratics – you are ready!