# Multiplying Polynomials and FFFT!

Most folks learn to multiply increasingly complex quantities gradually over time, starting with constants in elementary school, and eventually continuing on to polynomials in high school.  As the quantities become more complex, students master “the distributive property”, “collecting like terms”, and perhaps even procedures made more memorable with acronyms like “FOIL”.

When learning to multiply binomials and polynomials, people often focus more on the process than the reasoning behind it – which can makes things feel complex. Once you understand the reasoning, multiplying polynomials will hopefully become straightforward. And with a small measure of melodrama, I will describe “FFFT!”, a trivial technique with a silly name that can help make multiplying polynomials easy.

Way back in elementary school, perhaps in first grade, you were probably taught to multiply two integers:

$8\cdot 7\\*~\\*=56$

From there, you were probably asked to learn your multiplication tables. It is extremely useful, particularly when studying algebra, to know your multiplication tables by rote. This “reflex knowledge” will help you work faster, rely on a calculator less, and verify that solutions are correct with greater speed and confidence. No matter how old you may be, no matter whether you are still in school or not, I recommend mastering any gaps your multiplication tables.

Having said that, I confess that I have a few gaps in my multiplication tables (memorizing has never been a talent of mine).  Despite the gaps, I am confident that I can figure out a value I need quickly. For example, if I do not recall the answer to 8 x 7, but I do recall 8 x 8, I can subtract one eight from the product I recall to get to the one I need:

$8\cdot 7=\\*~\\*8\cdot (8-1)=\\*~\\*8\cdot 8-8\cdot 1=\\*~\\*64-8=\\*~\\*56$

or, if I recall 8 x 6, add one eight to the result:

$8\cdot 7=\\*~\\*8\cdot (6+1)=\\*~\\*8\cdot 6+8\cdot 1=\\*~\\*48+8=\\*~\\*56$

### The Distributive Property

As you can see above, I rely on the distributive property of multiplication over addition or subtraction to fill in my gaps. Understanding why the distributive property of multiplication works as it does helps make polynomial multiplication easy… so let’s explore the distributive property a bit.

What does the expression (3)(5) represent? Five scaled by a factor of three (using the scaling model of multiplication), or three fives added together (using the repeated addition model of multiplication). Either way, the result is 15.

What happens to the above if we substitute an expression that equals five for the number five in the above: (3)(4+1)? Clearly, we are still being asked to evaluate three times five… but five is now expressed as a sum instead of as a single quantity.

If we use the scaling model of multiplication, the above expression asks us to scale what is inside the parentheses, everything inside the parenthesesby a factor of three:

$3\cdot (4+1)=\\*~\\*3\cdot 4+3\cdot 1=\\*~\\*12+3=\\*~\\*15$

If we use the repeated addition model of multiplication, the expression asks us to duplicate “the parentheses”, along with everything inside of them, three times and add the results:

$3\cdot (4+1)=\\*~\\*(4+1)+(4+1)+(4+1)=\\*~\\*5+5+5=\\*~\\*15$

In addition to verifying the above two approaches produce the same result (15), we can show that the scaling and repeated addition models must be algebraically equivalent (when multiplying by a whole number) by using the commutative and associative properties of addition:

$3\cdot (4+1)=\\*~\\*(4+1)+(4+1)+(4+1)=\\*~\\*4+1+4+1+4+1=\\*~\\*4+4+4+1+1+1=\\*~\\*(4+4+4)+(1+1+1)=\\*~\\*3\cdot 4+3\cdot 1=\\*~\\*3\cdot (4+1)$

From here on I will use the scaling model of multiplication because it is a bit easier on the eyes (as you may have noticed above).

Even if that first factor is in parentheses, the process remains the same:

$(3)\cdot (3+2)=\\*~\\*(3)\cdot 3+(3)\cdot 2=\\*~\\*9+6=\\*~\\*15$

or in a slightly more abstract form to help highlight the pattern:

$(~~)\cdot (a+b)=\\*~\\*(~~)\cdot a+(~~)\cdot b$

And now for the final step… what if the number/blank in first set of parentheses above happens to be a sum or a difference, or in other words, has more than one term?  The process of distributing does not change. The entire quantity in the first set of parentheses is multiplied by each of the terms in the second set of parentheses.

$(1+2)\cdot (3+4)=\\*~\\*(1+2)\cdot 3+(1+2)\cdot 4$

Note that the above example will require us to use the distributive property two more times before we can get rid of all parentheses:

$(1+2)\cdot (3+4)=\\*~\\*(1+2)\cdot 3+(1+2)\cdot 4=\\*~\\*1\cdot 3+2\cdot 3+1\cdot 4+2\cdot 4=\\*~\\*~\\*3+6+4+8=\\*~\\*21=\\*~\\*3\cdot 7$

Once we have exhausted our opportunities to use the distributive property above, if we look carefully at the result (please review the line before the gap in the work) we see the following pattern:

Every term in the first set of parentheses has been
multiplied by every term in the second set of parentheses.

### Geometric Interpretation of the Distributive Property

The distributive property of multiplication over addition has a geometric interpretation as well. Suppose I seek to multiply two quantities, each of which is a sum:

$(a+b+c)\cdot (d+e+f)$

This corresponds to stretching a line of length (a+b+c) through a distance (d+e+f) in a direction perpendicular to its length – creating a rectangle as shown below.  The total area of this rectangle is equal to the product of the two sums:

Yet the total area of the rectangle can also be calculated by adding the areas of the nine smaller rectangles that piece together to create the large one.  Each smaller rectangle has an area that equals its length times its width (as shown in its center).  So,

$ad+ae+af\\*~\\*+bd+be+bf\\*~\\*+cd+ce+cf\\*~\\*=(a+b+c)\cdot (d+e+f)$

Multiplying the two polynomials using the approach described above produces the exact same result as calculating the area of the large rectangle by adding together all of its component rectangles.

### Where have you seen this before?

Do you remember multiplying multi-digit numbers in elementary school?  For example:

$123\cdot 45=5,535$

Doing this by hand involves multiplying every digit in the first number by every digit in the second one… but in a way that also keeps track of the position of each number (ones, tens, hundreds, etc.).  You can also do long multiplication algebraically, if you write each digit in a way that also makes its place value explicit:

$123=1\cdot 10^2+2\cdot 10^1+3\cdot 10^0\\*~\\*45=4\cdot 10^1+5\cdot 10^0$

so the product of these two numbers would be:

$(1\cdot 10^2+2\cdot 10^1+3\cdot 10^0)\cdot (4\cdot 10^1+5\cdot 10^0)$

Which provides a wonderful excuse to introduce a quick, easy, and reliable way of multiplying expressions that each have more than one term.

### FFFT!

FFFT, pronounced “fffffffT”, is an acronym for the “Famous Ford Finger Technique.” At this point, a disclaimer is in order: this technique is not famous, and is not something I invented… but the use of a finger is required.

Why does FFFT! work? Because it provides a physical aid (your finger) to help you ensure that every term in one set of parentheses is indeed multiplied by all terms in the other set of parentheses. If someone next to you sneezes loudly as you are in the middle of the problem, your finger helps you recall where you were in the problem. If one particular product took so much concentration that you are not sure quite where it came from once you finish calculating it, your finger helps you quickly find the first factor that led to it. So, without further obfuscation, here is the procedure:

For right-handed writers, when multiplying two polynomials do the following:

1. Place your left index finger under the left-most term in the left set of parentheses.
2. Here’s the hard part: Don’t move your finger until step 4.
3. Multiply the term above your finger by every term in the next set of parentheses, and write each product down with the appropriate sign.
4. Now that the term above your finger has been multiplied by all terms in the second set of parentheses, make a loud tearing sound as you laboriously move your finger to rest under the next term to the right.
5. Repeat steps 3-5 until you have reached the right side of the left set of parentheses
6. Count the number of products you have written down… it had better be equal to the number of terms in the first set of parentheses times the number of terms in the second set of parentheses.

For left-handed writers, when multiplying two polynomials do the following:

1. Place your right index finger under the right-most term in the right set of parentheses.
2. Here’s the hard part: Don’t move your finger until step 4.
3. Multiply the term above your finger by every term in the left set of parentheses, and write each product down with the appropriate sign.
4. Now that the term above your finger has been multiplied by all terms in the first set of parentheses, make a loud tearing sound as you laboriously move your finger to rest under the next term to the left.
5. Repeat steps 3-5 until you have reached the left side of the right set of parentheses
6. Count the number of products you have written down… it had better be equal to the number of terms in the first set of parentheses times the number of terms in the second set of parentheses.

### Practice

Now, practice using FFFT! on the following:

1. Multiply a monomial by a binomial (easy):

$(2)\cdot (3x-5)=$

Your initial answer should have had 1 x 2 = 2 terms, and your simplified answer should be equal to:  $6x-10$

2. Multiply two binomials (medium):

$(x+1)\cdot (2y-3)=$

Your initial answer should have had 2 x 2 = 4 terms, and your simplified answer should be equal to : $2y+2xy-3x-3$

3. And, for something a bit more complex to test your complete mastery of FFFT!, multiply two trinomials (ugly):

$(x+y+2)\cdot (2x-y+3)=$

Your initial answer should have had 3 x 3 = 9 terms, and your simplified answer should be equal to:  $2x^2+7x+xy+y-y^2+6$

Problems like this last one illustrate why it can be very helpful to organize your work. As the number of terms you have to keep track of grows, it is much easier to verify you have the correct total number of terms, and avoid overlooking a term as you collect like terms, if you:
– work neatly
– organized your work in some way (such as like terms each in their own column)
– left enough white space between terms that they don’t run together
– put a check mark over each term after you have “collected” it with like terms
– etc.

4. And finally, try the multiplication problem described above:

$(1\cdot 10^2+2\cdot 10^1+3\cdot 10^0)\cdot (4\cdot 10^1+5\cdot 10^0)=$

Once you have finished multiplying, add the terms together and verify you get the right answer: 5,535.

At this point, you may wish to make up a few of your own polynomial multiplication problems, then solve them, to give yourself a bit more practice and ensure you have both mastered the process and understand why it works.

Published
Categorized as Concepts

## By Whit Ford

Math tutor since 1992. Former math teacher, product manager, software developer, research analyst, etc.

1. This is exactly how I teach polynomial multiplication, but I never had a name for it. I, though, have students cover up the terms their not multiplying with yet, so I have them cover up the 5 in (x+5)(x+6) say, and distribute the x, then I have them cover up the x and distribute the 5. I’m glad I’ll have a name for it. If you get a chance, I recently through together an activity I hope helps them understand the geometric interpretation of multiplying binomials like this intuitively that I put up on my blog. I would love any advice on improving it. I’m still not sure it’s the best way to introduce the topic, but I’ve tried giving the distributive property argument you have above and the students got pretty lost with it.

1. I have had much more luck explaining the “distributive property” approach to individual students while tutoring. I agree that it does not always work well when attempting to explain it to an entire class. However, if you were to turn it into an activity for student groups – that might work more smoothly… it could look something like:

In your groups, first show each other how to use the distributive property to expand:
(3x)(2x + 5)
Once you have agreed upon how to complete that multiplication problem and are certain that you have arrived at a correct answer, discuss how the distributive property would be applied to a situation like the following, where the contents of the first set of parentheses are not known yet:
( )(2x + 5)
When you have agreed upon how this situation should be handled, use the process you described to multiply the following:
(3x + 1)(2x + 5)
(Hint: you will need to use the distributive property a total of three times before you will have finished expanding this product.)

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