Determining polynomials, basic operations, most important rules

Polynomials are expressions consisted of variables and coefficients. Those variables can have non-negative exponents.

A polynomial is made out of one or more terms. Term is a smaller expression consisting of variables and coefficients bound with multiplication. In polynomial terms can only be bound by subtraction and addition, and variables within terms with multiplication and positive exponents.

For example:

$\ x^2 + 2x + 4$ is a polynomial.

$\ x^4 + 2x^3 + x + 4$ is a polynomial.

$\ x^{-2} + 2x + 4$ is not a polynomial because one variable has negative exponent.

$\frac{x^{-2} + 2x + 4}{4x}$ is not a polynomial because the terms here are bound with division.

They can also have more than one variable. Same rule applies to them as well.
$\ x^2 y^4 + 2xy + 4$ is a polynomial with two variables. They can also be written in a standard form, that means that unknowns are sorted by the value of their exponent, starting from the largest to the smallest, and in non-standard form, where they don’t have to be in order.

Polynomials also have a name based on their properties. There are two parts in each of their name. First part represents the highest exponent and second one how many terms that polynomial has. The highest exponent is called the leading exponent.

If that leading exponent is equal to 0, that polynomial is a constant, 1 linear, 2 quadratic and 3 cubic and so on. If that number has one term it is called a monomial, 2 binomial, 3 trinomial and if it consists of more, it is called just polynomial of n terms.

What is the easiest way to determine name of given polynomial? First, it doesn’t matter if the polynomial is in standard or non-standard form, you find your biggest exponent, for real numbers like 1, 2 or 3 that exponent is 0 which means that this is a constant, for any variable unknown x, y, z…that exponent is equal to 1 which means that this is a linear polynomial, and so on. Second you find the other part of their name, for example if you have $\ x^2 + 4$, this polynomial has two terms which means that this is a binomial, if you have $\ x^3 + x^2 + 1$ it has three terms which implies that this is a trinomial.

And you have to put those parts together. So:

$\ x^2 + x$ is a quadratic polynomial.

$\ x$ is a linear monomial.

$\ x^3 + x^2 + x + 1$ is a cubic polynomial of 4 terms.

Addition and subtraction

Adding and subtracting polynomials is very similar to adding and subtracting exponents and radicals. You have to make sure you’re adding or subtracting terms that have the same exponents on matching variables.

Example 1: $ ( 7x + 2) + (3y – 4x) = ?$

(They are bound only by addition so we can just erase braces, and then we add what we can)
$\ (7x + 2) + (3y – 4x) = 7x + 2 + 3y – 4x = (7x – 4x) + 3y + 2 = 3x + 3y + 2$

Example 2: $\ (7x^2 + 4y) – (4x + 4y) = ?$

(They are bound by subtraction so every term in other polynomial will change its sign, and then we add what we can)

$\ (7x^2 + 4y) – (4x + 4y) = 7x^2 + 4y – 4x -4y = 7x^2 – 4x$

Example 3: When things are getting a bit more complicated it is easier to just group the similar terms with same variables.

$\ (7x^2 + 7xy^2 + y^3) – (4x^2 – 3y^2x + y) = 7x^2 + 7xy^2 + y^3 – 4x^2 + 3y^2x – y = (7x^2 – 4x^2) + (7xy^2 + 3y^2x) + y^3 – y = 3x^2 + (7xy^2 + 3y^2x) + y^3 – y$

Now what to do with that $\ 7xy^2 + 3y^2x$ ? well as you can notice you have two variables, in first term x to the power of one, and y to the power of two, and in the second term y to the power of two and x to the power of one. Those are the same variables with same exponents so nothing is stopping us to add them together.

Why is that? Because of the commutative property of multiplication of real numbers $\ a * b = b * a$. So our polynomial can look like this:

$ 3x^2 + (7xy^2 + 3y^2x) + y^3 – y = 3x^2 + (7xy^2 + 3xy^2) + y^3 – y = 3x^2 + 10xy^2 + y^3 – y$

It doesn’t matter if constants that multiply your unknown are a decimal or a fraction, the procedure is the same. The only two things you have to pay attention to are that the things you are adding or subtracting are of the same kind- terms with same variables with matching exponents and minuses.

Example 4.:

$(0.2x + 4 + \frac{1}{2}y^2x) + (\frac{1}{5}x^2 + 7xy^2) = 0.2x + 4 + \frac{1}{2}y^2x + \frac{1}{5}x^2 + 7xy^2 = \frac{1}{5}x + (7xy^2 + \frac{1}{2}y^2x) + \frac{1}{5}x^2 = \frac{1}{5}x + \frac{15}{2}y^2x + \frac{1}{5}x^2$

Multiplication

Multiplying polynomials is a bit more complicated, because you have more than two factors which contain more than one term.

These are the most important rules for multiplication of polynomials:
1. Square of sum

2. Square of difference

3. Difference of squares

4. Sum of cubes

5. Difference of cubes

6. Cube of sum

7. Cube of difference

These formulas cannot be simplified and you just have to learn them by heart.

The easy way to remember is this one: When you have square of sum or difference, or cube of sum or difference, you have a pattern. Let’s see on (x+y)^2 = x^2 + 2xy + y^2. First exponent of x is 2, and of y is 0, in second term exponent of x is 1 and exponent of y is also 1, and in third term the exponent of y is 2 and x is 0. So their exponents always add up to 2, which is exactly the power which you are exponenting your polynomial. The exponent of the first variable is always giving away one exponent to the other variable until he gives it all away.

The same is happening with cube:

$\ (x – y)^3 = x^3 – 3x^2y + 3xy^2 + y^3$

$\ x^3y^0$ => $\ x^2y^1$ => $\ x^1y^2$ => $\ x^0y^3$

And the minuses in differences are alternating with pluses.
With practice you will see that they are more than helpful, and can turn something really complicated into something simple.

Let’s say you have two polynomials that you have to multiply. And you have few terms on the left and few terms on the right. How do you multiply them?

First you take one term from the left polynomial and multiply it with every term from the right polynomial. And then you do that for every term from the left polynomial.

Example 1: $\ (a + b) \cdot (c + d)$. Like we said: first you take a, and multiply it with c and d, and then you take b and multiply it with c and d.

You have to know this very well because this is basic for multiplication, and if you understand this you are able to understand everything from this part.

Example 2: $\ (x + 2) \cdot (x + 3) = x \cdot x + 3 \cdot x + 2 \cdot x + 3 \cdot 2 = x^2 + 3x + 2x + 6$

Example 3: $\ (\frac{1}{2}x^2 + 4x + y)(7x – y)$. Even when there are more than two terms in one polynomial that is multiplying the other, you solve it just the same:

Example 4: $\ (0.2 – y)^2 (x + 2) = ?$

When you have a polynomial that is bound by exponents, first you have to get rid of them and then multiply with another polynomial.

Here you can use those formulas you remembered:

$\ (x – y)^2 = x^2 – 2xy + y^2$

$\ (0.2 – y)^2 (x + 2) = (\frac{1}{5} – y)^2 (x + 2) = (\frac{1}{25} – \frac{2}{5}y + y^2)(x + 2) = \frac{1}{25}x + \frac{2}{25} – \frac{2}{5}xy – \frac{4}{5}y + y^2x + 2y^2$

Example 5:

$\ (x – \frac{1}{2})(x – 0.5)^2 = ?$

$\ (x – \frac{1}{2})(x – \frac{1}{2})^2 = (x – \frac{1}{2})^3 = x^3 – \frac{3}{2}x^2 + \frac{3}{4}x – \frac{1}{8}$

Division

Dividing polynomials is similar to dividing real numbers, but with some few changes.

The most important thing is to watch out for exponents, always put them in order from greatest exponent to lowest. This will stop you from making mistakes, and you don’t have to think much about what is dividing what.

Example 1:
Divide: $\ (x^2 – 4) : (x + 2)$ First you take the variables with greatest exponents on both the dividend and divisor. In these cases they are x^2 and x. First you divide them x^2 : x = x. And we write x on the left side. That is similar to dividing real numbers.

$\ (x^2 – 4) : (x + 2) = x$ and then you multiply x to your divisor, and subtract what you got from divider, and the rest is all the same:

dividing polynomials procedure

You can always check your solution by multiplying your solution with divisor:

$\ (x + 2) (x – 2) = x^2 – 4$

Example 2.

dividing polynomials example

Since 6 has a variable with exponent 0, and our divisor has exponent 1, we can’t divide anymore. The 6 is a remainder.

How do you check up your solution now? Just like in normal dividing, your solution multiplied to the divisor and then added remainder must be equal to dividend.

$\ (x + 1)(x^3 + 1) + 6 = x^4 + x^3 + x + 7$

Example 3.

example divide polynomials