Polynomial Operations

This article discusses the basics of a polynomial and how it can be manipulated for easier use in further investigations.

Polynomials

A polynomial is a collection of terms linked together with addition or subtraction. A polynomial must have no variable exponent, no negative exponent, and no fractional exponent. Examples of polynomials are 3x2+2x-3, 5, 2x2-4xy+2y2, and a14-abc12+b7c4+1. However, expressions such as 5x+3, x1/3-y3, and xy-2 are not polynomials.

Polynomials have three main components. One is the amount of variables. The more variables an equation has, the more dimensions it can handle. For example, a polynomial of 3x2+5x-1 is less sophisticated than a polynomial of 2a2+3ab-6b2. Polynomials with more variables have more things to take into consideration and can prove useful when modeling some very vague data. Another component is constants. The constant of a polynomial is the part with no variable. This shows a limit where no matter what the variables are, this value stays the same. It is like the anchor of the polynomial. In the expression 2x3-27, the constant is -27. The final component is the degree of the polynomial. The degree of a polynomial is the greatest sum of the exponents of the variables in a term relative to the other terms. So, the polynomial 2x2-4x+2 has a degree of 2 since the largest degree is 2 in 2x2. The degree of the polynomial a2b2-a3+2ab2 is 4 since the sum of the exponents in the first term is 4 which is greater than the other terms exponent sums. Finally, the degree of a constant (say, 14) is 0 since you can imagine a variable with the exponent 0 (or in other words, a one.) The degree shows what shape the graph will take.

Polynomials are typically used for identifying trends and giving them a model. Linear models tend to be sufficient, but there are times where a larger degree polynomial is needed. Take the stock market for example. It goes up and down all the time, but it seems to increase over a general period of time. We could model it with a linear equation, but it does not tell the whole story. If we were to use a polynomial of a degree 5, then we can add curves and bumps that match the graph a little more. Unfortunately, polynomials of larger degrees can get pretty complex, so we need to find out what they do and how we can manipulate them so that in the future, polynomials can be easier.

Graphs

Polynomials with degrees greater than 1 are curves. Ones with even degrees tend to shape like U's and odd degrees look kind of like S's. The amount of the degree is the maximum amount of times that graph can intersect a horizontal axis (typically the x-axis, but also refers to other lines parallel to the x-axis). This means, that a polynomial equation like y=2x4-x3-4x2+2x+4 is capable of crossing the x-axis four times giving it a W type shape. This is what a graph for a third and fourth degree polynomial might look like.

coordinate5.png

Here, the red is a third degree equation while the green is a fourth degree equation. As you can see, these equations can get really complicated. Finding models and graphing them can be quite hard, but with the help of our elementary operations, we can see how these polynomials behave and thus help us in the future.

Operations

Polynomials are able to be altered using the elementary operations. However, you can't just go in and do it like you can with pure constants, but you have to follow a few rules since we are dealing with variables. The reason we sometimes need to use these operations is to change them into forms that can be used for graphing, modeling, and solving. We will discuss each operation as well as how we use those to put a polynomial in better more usable forms.

Adding

In adding polynomials, you can only add if the terms are alike. This means that the terms you are adding must have the same variables with the same exponents on each one. For example, 4x+5x is 9x. However, 2x+3x2 is simply 3x2+2x. They cannot be combined because the terms don't have the same exponential value. When adding polynomials with more than one term, gather your terms into an arrangement so that like terms are aligned. Here is an example.

(1)
\begin{align} \left( 2x+4x^2-6-x^3 \right) \\ + \left( 2x^3-3x+5 \right) \end{align}

We must add these polynomials. First, you need to align the terms of each polynomial do that alike terms are in the same column. If a term is missing or there is none in the other polynomial, consider that term 0 so nothing changes. First, align this expression to its like terms.

(2)
\begin{align} \left( -x^3+4x^2+2x-6 \right) \\ + \left( 2x^3+0x^2-3x+5 \right) \end{align}

Now, add vertically. -x3+2x3 is x3. We used a 0x2 to represent what we are adding to the 4x2. Adding the rest of the terms, you get x3+4x2-x-1. So, to add polynomials, line up the like terms and combine those vertically.

Be careful. Sometimes, something like this occurs.
Add 2a+3b2 and 4ab-2a2.
In this case, there are no like terms. So, your answer is simply -2a2+2a+4ab+3b2.

Subtracting

Subtracting polynomials is very similar to adding them. Subtraction is the inverse of addition. Pretty much, the only difference is the change in signs. When you are subtracting a polynomial from the other, you simply reverse the signs of every term of the second polynomial and add. So, say you had the problem (2x2-4x+5)-(x2+3x+3). Reverse the signs of the second polynomial and add. (2x2-4x+5)+(-x2-3x-3). When you align the like terms, you end up with

(3)
\begin{align} \left( 2x^2-4x+5 \right) \\ + \left( \underline{-x^2-3x-3} \right) \\ x^2-7x+2 \; \end{align}

That is all you do with subtraction: reverse the signs and add.

Multiplying

Multiplying polynomials can be done no matter what the terms are. Multiplication can be very time consuming after you get a lot of terms. In it, all you are doing is multiplying your constants/coefficients, and adding the exponents of the alike variables. When multiplying terms, you must consider what you are multiplying. Consider the term 3xy2. This term is bounded together by multiplication, as are all terms. So, that term actually can translate to 3 * x * y2.

Let's review a fundamental rule. If at anytime you are multiplying two numbers or variables of the same base, then you add the exponential values of each number. So, if you have x2 * x3, you can simplify that to x5. If you have y3 * y-3, then you get y0, which is one. Similarly, if you get zz * zz, then you get z2z. However, you cannot do that with, say, x4 * y2. That simplifies to x4y2 because x and y are not identical bases. So, the fundamental rule for multiplying is xa * xb = xa+b.

When you multiply terms, you are multiplying all the like components of each terms. For example, let's multiply 3xy2z and 6x2yz3. All you have to do is multiply the numbers and add the exponent values of the like variables together. When you multiply these, what you are doing is splitting each term into individual components and multiplying like terms. Like so: 3 * x * y2 * z * 6 * x2 * y * z3. You can align your look-alikes to get this: 3 * 6 * x * x2 * y2 * y * z * z3. Now, it can be seen how this is simplified. Combine each by multiplying to get 18x3y3z4. That is how terms are multiplied.

Multiplication of polynomials is a fairly easy concept if you look at it right. There are two main types of multiplication: a monomial against a polynomial, and a polynomial against a polynomial.

Distribution

Distribution refers to monomial against polynomial multiplication. This refers to multiplication like 4x(3x+2). All you are doing is multiplying the outside term by every single term in the parenthesis and finding the sum or difference. So, 4x(3x+2) = 4x * 3x + 4x * 2. This can be simplified to 12x2+8x. This can be done anytime you find a term being multiplied by multiple terms bound by addition or subtraction. This works for expressions such as

  • 3x(4-2x) = 3x * -2x + 3x * 4 = -6x2+12x
  • 2xy(x+y) = 2xy * x + 2xy * y = 2x2y+2xy2
  • 4x2(3y+2z) = 4x2 * 3y + 4x2 * 2z = 12x2y+8x2z
  • 6x2y4z3(2x-3xz+4yz2-xyz) = 6x2y4z3 * 2x + 6x2y4z3 * -3xz + 6x2y4z3 * 4yz2 + 6x2y4z3 * -xyz = 12x3y4z3-18x3y4z4+24x2y5z5-6x3y5z4

No matter how many terms are in the parenthesis, you must multiply the outside term by all of them. Also note something like this: 2x(6+2). This is 12x+4x, which can actually be further simplified to 16x. Distribution cannot occur when two or more parenthetical groups are multiplied, like (3x+3)(2x-5). That is discussed in the next section.

Binomial by Binomial

The other type of multiplication is polynomial by polynomial, or two groups of parentheses being multiplied. The most basic of these is the binomial times a binomial, like (a+b)(a+b). In this case, you have to multiply every term in the first set with every term in the second set. So, you would multiply a * a + a * b, and then b * a + b * b. This gives a final answer of a2+2ab+b2. It happens like in this diagram with arrows representing multiplication.

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Multiplying in this fashion is known as the FOIL method. FOIL means first, outer, inner, last. In other words, it is telling you how to go about simplifying binomial multiplication. Multiply the first term of each set, then multiply the inner two terms, then the outer two terms, and finally the last terms.

So, simplify this:
(2x+3)(x-3)
We can use the FOIL method. Multiply your first terms.
2x2
Multiply the outer terms.
-6x
Multiply the inner terms.
3x
Multiply the last terms.
-9
Now, combine all the terms on one line.
2x2-6x+3x-9
Finally, combine all your like terms.
2x2-3x-9 is the final answer.
That is how the FOIL method works. It can only be done for binomial by binomial multiplication, but the way it works can be applied to other longer multiplication expressions.

General Scope

The general idea behind multiplication is that you multiply every term in one set of parenthesis by every term in the other set it is multiplied by. No term is counted out. Knowing this, we can simplify any polynomial multiplication problem, including the monster (2x3+3x2-5x+1)(3x2+6x-4)
Although this is rather long, it isn't that much different from the other multiplication problems. You would just proceed by multiplying 2x3 by each term in the second set, then 3x2, then 5x and so on. That is all multiplication is: adding exponents and multiplying every term by each term in their multiplicative parts.

Division

Just as you can add, subtract, and multiply polynomials, you can also divide them. Division of polynomials is a little tougher, but it is very systematic. Division is the inverse of multiplication. In this, you have your numerator and denominator. Division of polynomials is a lot like division of integers where you get a quotient and remainder, and the method by which to use is very similar.

Before you can start, you must first know about dividing monomials and variables. Remember that when you multiply same bases, you add the exponents? Well, when you divide identical bases, you subtract the exponents. So, x4 divided by x2 is x4-2, which is x2. When you divide terms, like here: $4x^3y^6 \over 2xy^4$, you simply divide each part: the constants and the variable bases. So, that expression simplified is 2x2y2. So remember this rule: ${x^a \over x^b} = x^{a-b}$.

Another thing is that an expression is simplified when it has no negative exponents. When you have a negative exponent, move that part to the denominator and change the sign of the exponent. Why? Well, consider a-1. When something is taken to the negative first power, it is said to be inversed. Or, what we are really doing is finding its reciprocal. And remember, the reciprocal is simply $1 \over a$. The a in the denominator has an exponent of one which is the opposite of -1. Just as well, this applies to higher exponents. The expression a-3 is equal to $1 \over a^3$. When this occurs in an expression like 6a2b-2c3, all you do is move the negative exponent's base to the denominator. So that is equivalent to $6a^2c^3 \over b^2$. That is how terms are divided. When you divide a polynomial by a monomial, then all you have to do is divide every term in the numerator by the term of the denominator.

Long Division

Some division problems are a little more complicated. Long division is used when you are dividing a polynomial by a polynomial of more than one term and lower degree. Long division is the process of dividing an expression into another by smaller parts. For example, consider an elementary problem 1475/13. How do you solve this? Remember, you ask yourself, "How many times does 13 go into 14?" The answer is 1. Then, you multiply 1 by 13 and subtract that from 14, and then carry down everything else. This is exactly how you divide polynomials. You pretend each term is a "digit". So, let us attempt to solve this problem: (4x3-2x2-1)/(x+1)

(4)
\begin{align} x+1\overline{\vert 4x^3 - 2x^2 + 0x - 1} \end{align}

The first thing you must do is write it in this form. This is the dividing house type thing where the dividend/numerator goes under the house and the divisor/denominator goes outside. The quotient ends up on top. This will make the long division much more organized and easy to understand. Also notice that we added a +0x term. This acts as a place holder and must be included for the operation to work correctly. The reason is because the 0x identifies how many x's there are. This is like when we are considering the difference between 129 and 1209. Notice that the 0 is there because it is indicating how many 10 units there are. So, always be sure to include the 0x? whenever you see a term is "missing".

(5)
\begin{matrix} \quad 4x^2\\ \qquad\qquad\quad x+1\overline{\vert 4x^3 - 2x^2 + 0x - 1}\\ \qquad\;\;\;\;\; \underline{4x^3 + 4x^2}\\ \qquad\qquad\qquad\qquad\quad\;\; -6x^2+0x-1 \end{matrix}

So, the first step of polynomial division is asking yourself, "How many times does the leading term of the divisor go into the first term of the dividend?" So, how many times does x go into 4x3? Most of polynomial division consists of finding how many times the leading term of the divisor goes into the next term of the dividend. So, to find out the answer to this, you need to find 4x3 divided by x. That yields 4x2, which is the first term, or "digit" of the quotient. Write that up top.

Now, you need to multiply that part of the solution by the entire divisor, which means you are multiplying that term by every term in the divisor. So, for this example, you must multiply the 4x2 by x and 1. Then, you take that part and place it under the dividend aligned with the appropriate terms. So, we would get 4x3+4x2. Align that with the terms of the same degree of x. Then, subtract that part from the dividend. So, what you end up with is (4x3-2x2)-(4x3+4x2). This simplifies to -6x2. All this is done in the picture above. Finally, you would take down the rest of the equation and prepare for the next step.

(6)
\begin{matrix} \qquad\quad\; 4x^2-6x\\ \qquad\qquad\quad x+1\overline{\vert 4x^3 - 2x^2 + 0x - 1}\\ \qquad\;\;\;\;\; \underline{4x^3 + 4x^2}\\ \qquad\qquad\qquad\qquad\quad\;\; -6x^2+0x-1\\ \qquad\qquad\qquad\qquad \underline{-6x^2-6x}\\ \qquad\qquad\qquad\qquad\qquad\qquad\quad\; 6x-1 \end{matrix}

So, like you would on normal division problems, you must bring down the rest of the problem. Now, you practically repeat the previous step. You must find how many times the first term of the divisor, x, goes into the difference of the subtraction set you just completed. So, you are finding how many times x goes into -6x2. That ends up being -6x times. So, you would write that after the previous quotient as shown above. Then, you multiply your new part by the entire divisor yielding -6x2-6x. You must subtract that polynomial from what you brought down remembering to align your like terms. When you do that, you get 6x.

(7)
\begin{matrix} \qquad\qquad\quad\; 4x^2-6x+6\\ \qquad\qquad\quad x+1\overline{\vert 4x^3 - 2x^2 + 0x - 1}\\ \qquad\;\;\;\;\; \underline{4x^3 + 4x^2}\\ \qquad\qquad\qquad\qquad\quad\;\; -6x^2+0x-1\\ \qquad\qquad\qquad\qquad \underline{-6x^2-6x}\\ \qquad\qquad\qquad\qquad\qquad\qquad\quad\; 6x-1\\ \qquad\qquad\qquad\qquad\qquad\qquad\quad\; \underline{6x+6}\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad -7 \end{matrix}

Bring down the rest of the problem. Now, you must repeat the step once again. The core of long division is that you repeat the same step multiple times making it long. Again, how many times does the x go into your previous difference? It divides to be 6, so you would write +6 up top. Multiply that by the divisor to get 6x+6 and subtract that from the current polynomial. After that, you are left with a lone -7. Notice that there is nothing more to bring down. When you run out of numbers to bring down, all you have to do is stop. There is no decimal parts or anything ugly like that. Now, all you have to do is write the answer, but do remember the -7.

(8)
\begin{align} {4x^3-2x^2-1 \over x+1} = 4x^2-6x+6-{7 \over x+1} \end{align}

The above end up being your answer. Write down the quotient, or what is on top of the house. Then, consider the -7. That number ends up being your remainder. When you get a remainder, whether it be a constant or even another polynomial with variables, you put it over the divisor and tag it at the end of the original quotient.

The long division method works for any polynomial division problem as long as the dividend's degree is greater than the divisor's degree. The only hazard is that the method is very long and can be time consuming sometimes. However, there is another method which is considerably faster.

Synthetic Division

Synthetic division is the other way to solve polynomial division. What it does is it ignores the variables and concentrates only on the coefficients and constants. This makes it much faster and a lot easier, but there are limitations. First of all, the divisor must be first degree, and secondly, the divisor's leading coefficient plays a major factor in the final quotient. Knowing that, let's jump in.
Let us divide this problem.
(4x3-6x+2)(x+2)-1 (Please note that this is another way to write division.)
The first thing you need to do is set up the problem.
First, extract the coefficient of every term in the dividend and put them in the order of descending powers. So, the x3 term would come first, then the x2, etc. Make sure to identify the zeroes! In our example, we would have 4, 0, -6, and 2. Then, take the divisor and set it equal to 0. Solve for x and take that number. So, ours would be x+2=0 which makes x=-2. Now we have the numbers needed set up the chart-like apparatus. Organize the numbers in a chart like this one:

(9)
\begin{matrix} \underline{-2}| & | & 4 & 0 & -6 & 2\\ & | & & & & \\ & | & & & & \end{matrix}

Notice where the numbers went. The coefficients of the dividend go on the right side of the vertical line in order and the number we obtained by solving for x in the divisor goes on the left side. The process of synthetic division really isn't that hard to follow, and since adding is involved, it can be a little easier. Essentially, the process of long division is actually taking place, but it is happening in a way that is easily understood and less messy. Let us discuss the course of action.

(10)
\begin{matrix} \underline{-2}| & | & 4 & 0 & -6 & 2\\ & | & & -8 & & \\ & | & 4 & & & \end{matrix}

The first step is always to bring down the first number to the bottom row. You always do that. The next step is to multiply that number on the bottom row by the number on the left side of the vertical line. So, for this, we would multiply the 4 by -2 to get -8. The product then goes under the number next in the sequence. The illustration shows this.

(11)
\begin{matrix} \underline{-2}| & | & 4 & 0 & -6 & 2\\ & | & & -8 & 16 & \\ & | & 4 & -8 & & \end{matrix}

Now, you add the two numbers in that column. See how we added -8 and 0? The answer is -8, and we put the sum below the column. The next thing would be to multiply that sum by the outside number and to place the product under the next number in the sequence. So, -8 and -2 make 16, and that goes under -6. See a pattern?

(12)
\begin{matrix} \underline{-2}| & | & 4 & 0 & -6 & 2\\ & | & & -8 & 16 & -20 \\ & | & 4 & -8 & 10 & \end{matrix}

So, add the two numbers in that column and place the sum under the previous product. -6+16 is equal to 10, so observe where that went. Multiply the sum by the outside number (-2) and place the product under the next number. Pretty much, you are doing the same thing over and over again until you reach the end: add, multiply, add, multiply,…. -2 times 10 is -20 which goes under 2.

(13)
\begin{matrix} \underline{-2}| & | & 4 & 0 & -6 & 2\\ & | & & -8 & 16 & -20 \\ & | & 4 & -8 & 10 & |\overline{-18} \end{matrix}

When you reach the end, you simply add the numbers and box off that last result. It is the remainder. So, 2+(-20) is 2-20 which is -18. You put that under the -20 and box it off. The next step is to divide everything on the bottom row except the remainder by the leading coefficient of the divisor. Since this problem's leading coefficient was one, it keeps the same. However, if it were 2 or some other number, you would have to divide by that number or else the quotient is entirely wrong. Now, the final step is to put back the variables.

(14)
\begin{align} {4x^3-6x+2 \over x+2} = 4x^2-8x+10-{18 \over x+2} \end{align}

When putting back the variables, simply look at the original problem. The degree of the quotient is one less than the degree of the dividend. So, since the degree of 4x3-6x+2 is three, the degree of the quotient is 2. Look at the chart. The numbers in the last row are the coefficients of the quotient in descending degree. So, the first number, when reading left to right, is the x2 coefficient. The next is the x coefficient. The remainder is simply that number over the divisor. When we apply this to our problem, we get the above result.

The core of synthetic division is adding and multiplying over and over again and then writing the final answer in the correct form. This method will not work with divisor degrees higher than 1 or multi-variable division problems. If you encounter problems such as:
(5x5-1)/(x2-1)
(x2+5xy-6y2)(x+y)-1
then use the long division method because it will always work if you do it correctly.

Expanding

Expanding is the same thing as simplifying a polynomial. This is the process of adding, subtracting, multiplying, or dividing in order to get the standard form:

(15)
\begin{equation} ax^n+bx^{n-1}+cx^{n-2}+...+dx^2+ex+f \end{equation}

What you are doing is putting whatever you are given into descending degree order with no replications or random floats. Also take to mind that when applying division with remainders, those are not polynomials, and negative exponents arise.
So, say you are given the expression (3x+4)(x-3)+(5x+4)2. You must put this into the above form. Remembering PEMDAS, just start doing your operations.

The first thing is to do the exponents (since all the parentheses cannot be simplified further). When looking at (5x+4)2, that is actually (5x+4)(5x+4). Now, we can do multiplication. Since this is a binomial multiplied by a binomial, we can use the FOIL method discussed above to simplify it. You get 25x2, 20x, 20x, and 16 all added together. However, you can't just place that there. You must put that whole product in a set of parenthesis (this is really just an organizational matter). So, now you have (3x+4)(x-3)+(25x2+40x+16). Notice that there is still some multiplication on the left side of the addition. So now, we need to multiply those two binomials, and we can use FOIL. Multiplying (3x+4)(x-3), you get 3x2-9x+4x-12. This simplifies to 3x2-5x-12. Now, you get (3x2-5x-12)+(25x2+40x+16). Referring to what you know about addition of polynomials, combine your like terms. This yields a final result of 28x2+35x+4 which is in the form above.

This is how you would do any sort of simplifying polynomials. The purpose for putting a polynomial in expanded form is that it immediately shows the degree which reveals some properties the graph of the polynomial. The coefficients of the terms shows how much influence they have on the final product.

Factoring

Factoring is the reverse process of expanding. Rather than trying to get a polynomial in its standard form, you are trying to get a polynomial in standard form, or similar thereof, into a form that identifies all the roots. That is, you are putting in the parentheses and such. This is a lot like "unsimplifying" a polynomial. Unfortunately, there is no set way to do this, and a lot of logic and guessing is needed to put a polynomial in factored form.

Say we had the polynomial 2x2-5x-12. We must factor this polynomial.
Since this is a quadratic polynomial, we can assume that the FOIL method was used to end up with the result. So, the factored form will have a binomial multiplied by a binomial.
First, you would want to find all the factors of 12 and 2. The factors of 12 are {1,2,3,4,6,12}. The factors of 2 are {1,2}. So, say that you had the factored form as (ax+b)(cx+d). b and d must be one of the paired factors of 12, and a and c must be 1 and 2. Knowing this, we can begin to fill our template: (x+b)(2x+d).
Now, you need to fill in b and d with the factor pairs to see which pair will yield a result of -5x when subtracted since the sign of the 12 is negative. If it was a positive twelve, the numbers would add to be -5. So, begin the various checkings by making estimated guesses. Let us try these to see which one might work:
(x+2)(2x-6) = 2x2-2x-12
(x-6)(2x+2) = 2x2-10x-12
(x+3)(2x-4) = 2x2+2x-12
(x+4)(2x-3) = 2x2+5x-12
The last answer seems very close. Let's try flipping the pluses and minuses.
(x-4)(2x+3) = 2x2-5x-12
So, the factored form of 2x2-5x-12 is (x-4)(2x+3).

Factoring is used to help graph polynomials by hand and find solutions. Unfortunately, not all polynomials can be factored which makes them prime. Since there are so many little instances for factoring shortcuts, it is a very large topic. The article Factoring describes it in better detail.

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