Polynomial Distribution
A reasonable mastery of distribution of polynomials will facilitate the understanding of certain kinds of mathematics, certain later proofs, and higher scores on popular standardized tests. We will discuss the “factoring” of quadratics and polynomials in a later section.
1. Polynomials: An Introduction
Polynomials are built of many (“poly”) terms (“nomials”); each of these singular terms is called a “monomial” (“mono” meaning “one” or “single”). Below is a polynomial with four monomials or terms:
$$x^2{y^4}+2xy^3-5x^2+7y$$
One might call this a “quadrinomial” (the prefix “quad” meaning “four”); however, we generally do not classify polynomials into special varieties like this. Generally, any polynomial with three or more terms is simply called a “polynomial,” though it is good to be aware that sometimes they are referred to with prefixes that specify the number of monomials in the polynomial. Below are some relatively common examples:
- Binomials : Two monomials
- Trinomials : Three monomials
- Quadrinomials: Four monomials
- Quintinomial : Five monomials
Rather than classify polynomials in this cumbersome fashion, we often simply refer to the degree of the polynomial, which is determined by the highest exponent on a monomial within the polynomial; the degree of the polynomials has no necessary correlation to the number of monomials in the polynomial. For example, the below polynomial is a “seventh” degree polynomial, though there are only four monomials in it:
$$5a^3+4a^5-a^2+a^7$$
Notice that the above polynomial has been given in a jumbled order according to the monomials’ exponents. Generally, polynomials are and should be written with descending (or sometimes ascending) exponents. Thus, the above degree-four polynomial is best written as:
$$a^7+4a^5+5a^3-a^2$$
Keep in mind that $ax^0$ is also a monomial; these are usually called constant terms, since they are simply numbers without a variable component. Even fractional and negative exponents can find their ways into complex polynomials.
2. Performing Operations on Polynomials: Monomial Addition and Monomial Division
2a. Combining Like Terms: Monomial Addition/Subtraction
Before we start doing more complex work with polynomials, the first step is to simplify as much as possible by “combining like terms” and sorting these monomials in descending order of exponent. There are a few considerations for the combination of terms in a polynomial.
- You can only combine terms that have the same exponent on its variable(s)
$$x^2-xy+2x-2y+2xy-y^2\rightarrow{x^2+xy+2x+2y-y^2}$$
- If there is a common factor on all your values, it is fine and likely advisable to simply factor it out of all monomials:
$$2x^3-4x^2+8x-16\rightarrow{2(x^3-2x^2+4x-8)}$$
Perhaps the biggest hurdle here is the realization of what like terms are. Like terms are variable expressions that may be validly added together to yield a mathematical identity. In the expression x + y + 5, there are no like terms. We cannot simplify this expression further. Similarly, in the expression $x^2 + x + 1$ the expression cannot be further simplified.
However, in expressions like
$4x + x^2 + x + 2x^2 + 3 = 4x + x + x^2 + 2x^2 + 3 = 5x + 3x^2 + 3 = 3x^2 + 5x + 3$
The terms with same exponent can all be added and thus simplified. Think of terms with different exponents like different kinds of fruit. You can certain say 10 oranges + 8 apples + 2 oranges is the same as 12 oranges and 8 apples. However, it is certainly DIFFERENT than 20 oranges or 20 apples.
Consider this example:
If $10x-3xy+x^2$ is subtracted from 8xy-2x^2+5x, what is the resulting expression?
First let us write the expression in mathematical terms:
$$(8xy-2x^2+5x)-(10x-3xy+x^2)$$
We should now be sure to distribute the negative that is in front of the second polynomial:
$$(8xy-2x^2+5x)+(-1)(10x-3xy+x^2)$$
$$(8xy-2x^2+5x)+(-10x+3xy-x^2)$$
Now we can combine all like terms. To do this, we might sort all the monomials such that the like terms are adjacent::
$$8xy-2x^2+5x-10x-3xy+x^2$$
$$8xy-3xy-2x^2+x^2+5x-10x$$
$$(8xy-3xy)+(-2x^2+x^2)+(5x-10x)$$
$$5xy-x^2-5x$$
2b. Monomial Division
Dividing polynomials by polynomials is a rather complex topic for us right now; polynomial division is part of a topic called “rational functions,” so we will return to this in more detail in later sections. However, right now let us consider the case of dividing a polynomial by only a monomial:
- First, be sure to simplify the polynomial as much as possible through the combination of like terms and listing by descending exponents.
- Divide all terms of the polynomial by the monomial.
Consider the following examples:
Example with a constant monomial::
$$\frac{6x^2+3y-12}{3}$$
Since we are dividing the whole polynomial by 3, we can consider this like multiplying the whole polynomial by $\frac{1}{3}$. When we do this, we must distribute the $\frac{1}{3}$ to each member of the polynomial by the Distributive Property.
$$\frac{6x^2+3y-12}{3}\rightarrow{\frac{1}{3}(6x^2+3y-12)}$$
$$\frac{6x^2}{3}+\frac{3y}{3}-\frac{12}{3}$$
$$2x^2+y-4$$
Example with a variable monomial:
$$\frac{12x^3+9x^2-12x+3}{3x}$$
Since we are dividing the whole polynomial by $3x$, we can consider this like multiplying the whole polynomial by $\frac{1}{3x}$. When we do this, we must distribute the $\frac{1}{3x}$ to each member of the polynomial by the Distributive Property.
$$\frac{12x^3+9x^2-12x+3}{3x}\rightarrow{\frac{1}{3x}(12x^3+9x^2-12x+3)}$$
$$\frac{12x^3}{3x}+\frac{9x^2}{3x}-\frac{12x}{3x}+\frac{3}{3x}$$
Using the Quotient of Powers Rule, we can simplify these fractions:
$$\frac{12}{3}x^{3-1}+\frac{9}{3}x^{2-1}-\frac{12}{3}x^{1-1}+\frac{3}{3}x^{0-1}$$
$$4x^{2}+3x^{1}-4x^{0}+1x^{-1}$$
$$4x^{2}+3x-4+x^{-1}$$
3. Multiplying Polynomials: Polynomial Distribution, “FOILing,” and “The Grid Method”
Multiplying a polynomial is an extension of the Distributive Property. Recall that the distributive property of multiplication states that:
$$a(b+c)=ab+ac$$
3a. FOIL and Binomial Distribution
Most frequently in introductory algebra courses you will work with multiplying two binomials; this process is often called “FOILing,” which is the denominalization of the acronym FOIL (First, Outside. Inside, Last). FOIL tells you simply an order for the distribution of the elements in the multiplication of two binomials.
The FOIL method can be simply proved according to the distributive property either by definition or through algebraic process:
Proof of the FOIL Method:
Given $(a+b)(h+k)$ where $a$, $b$, $h$, and $k$ are non-zero number, the FOIL Method would state that
$$(a+b)(h+k)=ah+ak+bh+bk$$
Let us prove this
Given $(a+b)(h+k)$ where $a$, $b$, $h$, and $k$ are non-zero number, assume $(a+b)=c$ since $a$ and $b$ can be summed as two numbers.
Now, we have $c(h+k)$, which can be simply distributed according to the Distributive Property:
$$c(h+k)=ch+ck$$
Now, let us substitute back the value for $c$ in both cases:
$$ch+ck=(a+b)h+(a+b)k$$
We can now distribute the values of $h$ and $k$ according to the Distributive Property:
$$(a+b)h+(a+b)k=ah+bh+ak+bk$$
Since addition is commutative, we can reorder this summation:
$$ah+bh+ak+bk=ah+ak+bh+bk$$
This result is the same as the FOIL method. However, let us continue a bit further to demonstrate that this can be undone via a different path to return the original statement. Given the grouping of the monomials, we could factor out an $a$ and $b$ as follows:
$$ah+ak+bh+bk=a(h+k)+b(h+k)$$
Let $(h+k)=g$ for some value g, since $h$ and $k$ being numbers can sum to some quantity. Thus,
$$a(h+k)+b(h+k)=ag+bg$$
We can now factor out $g$:
$$ag+bg=g(a+b)$$
Substituting back in the value for $g$, we return to the original statement with the binomials swapped; however, since multiplication is commutative, we can reverse this order and return the original statement precisely.
$$g(a+b)=(h+k)(a+b)\rightarrow{(a+b)(h+k)}$$
3b. The Grid Method and Polynomial Distribution
Sometimes FOILing is taught via the “Grid Method,” which is merely a visual representation of the distribution of polynomials. To use the Grid Method, you simply make a chart with one polynomial forming the vertical side and one forming the horizontal side. You then multiply at the intersections of the rows and columns of this chart.
General Example of Grid Method:
Given $(a+b)(h+k)$, we make a 3×3 grid with one binomal across the top and the other along the side. We will then fill in the grid by taking the product of the row’s and column’s element respectively.
Thus, $(a+b)(h+k)=ah+ak+bh+bk$
Keep in mind that the grid you make, while frequently will be 3×3, can be of any dimension. Consider the following example:
Specific Example of Grid Method:
Given $(x^3+2x^2+3x+4)(3y^2+2y)$, we make the grid of dimension 5×3 and take the products along the columns and rows respectively.
Thus,
$$(x^3+2x^2+3x+4)(3y^2+2y+1)=$$
$$3x^3{y^2}+2x^3{y}+6x^2{y^2}+4x^2{y}+$$
$$9xy^2+6xy+12y^2+8y$$
While FOIL works very well to help keep track of binomial distributions like $(x+y)(2x+3y)$, FOIL does not help with polynomial distribution in more complex situations. Keep this in mind. However, ultimately you can distribute polynomials by merely following the definition of distribution, which says to multiply each element of one polynomial onto each element of the other polynomial.
The Grid Method can be simply proved according to the distributive property either by definition or through algebraic process. Using the same methodology from the Proof of the FOIL Method, we can derive a general proof for the distribution of polynomials of any size or degree; however, we will need to make our mathematical language a bit more sophisticated. This will be good practice for us in starting to think about mathematical proof and generalizing statements.
General Proof of the Distribution of Polynomials
Given two polynomials, $P(x)$ and $P(y)$ of degrees $n$ and $m$ respectively, where
$$P(x)=a_1{x^n}+a_2{x^{n-1}}+…+a_{n-2}{x^2}+a_{n-1}{x}+a_{n}$$
and
$$P(y)=b_1{y^m}+b_2{y^{m-1}}+b_3{y^{m-2}}+…+b_{m-2}{y^2}+b_{m-1}{y}+b_{m}$$
where all $a$, $b$, $n$, and $m$ Real Numbers, assume the sum of all elements of $P(x)$ is $c$ and the sum of all elements of $P(y)$ is $g$:
$$P(x)=c\quad\text{and}\quad{P(y)=g}$$
Then $P(x)P(y)=cP(y)$, and by the Distributive Property we may distribute the value of $c$ to all elements of $P(y)$:
$$cP(y)=cb_1{y^m}+cb_2{y^{m-1}}+cb_3{y^{m-2}}+…+cb_{m-2}{y^2}+cb_{m-1}{y}+cb_{m}$$
Now, let us substitute back the value for $c$ in both cases:
$$P(x)b_1{y^m}+P(x)b_2{y^{m-1}}+P(x)b_3{y^{m-2}}+…+P(x)b_{m-2}{y^2}+P(x)b_{m-1}{y}+P(x)b_{m}$$
Thus, the distribution of polynomials is the distribution of each element of one monomial onto each element of the other, which accords to the definition of distribution.