## 17Calculus Precalculus - Polynomials

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Polynomials are a special class of equations that are used so much in calculus that a lot of attention is given to them in precalculus to prepare you for calculus. Polynomials have properties that are well-known and they are easy to work with which makes them quite useful. We can even approximate other, more complicated functions using polynomials. So you need to know how they work and how to use them, as well as be very familiar with their properties.

What Are Polynomials?

Polynomials are expressions that consist of a coefficient, a variable and an exponent. The exponent must be a non-negative integer. Here are some examples of polynomials and expressions that are not polynomials.

 Polynomials NOT Polynomials $$2x^3$$ $$t$$ $$17$$ $$2x^{-3}$$ $$t^{1/2}$$

Notes
1. For our discussion (and as usual in precalculus), we require that the coefficient be a real number (not complex).
2. In the above list, the expression $$2x^{-3}$$ is not a polynomial because the exponent is a negative integer. The $$-3$$ violates the rule that the exponent must be greater than or equal to zero. A similar argument holds for $$t^{1/2}$$ since $$1/2$$ is not an integer.
3. It may seem strange that we say that the number $$17$$ is a polynomial. This is true since we allow the exponent to be zero and since all numbers raised to the zero power are defined to be one, we can write $$17=17x^0$$. The expression $$17x^0$$ fits the definition of a polynomial, $$17$$ is considered a polynomial.
4. We said that all numbers raised to the zero power are defined to be one. However, this is true only as long as we don't have $$0^0$$. We will ignore this complication for now by replacing $$x^0$$ with 1 and you will get to deal with it in calculus. It's not hard but it does require some special handling.

Combining Polynomials

Okay, so now that we know what a polynomial looks like, we can combine them just like we did with regular numbers using basic arithmetic, addition, subtraction, and multiplication (division of polynomials produce rational expressions). So we can add $$2x^3$$ and $$17$$ to get $$2x^3+17$$. We also call this last expression a polynomial.

We usually simplify polynomials by factoring and combining common variable terms. For example, $$2x^3+5x^3 = (2+5)x^3 = 7x^3$$. In this equation, we factored the $$x^3$$ term and combined the coefficients.

Equations vs Functions

Before we go on, there is some terminology that you need to get used to. The terms equation, expression and function all refer to different things. Many instructors use them interchangeably but most textbooks do not. So now is a good time to get them straight in your head. We will show examples as we define them.

 expression $$3t^5+4t-1$$

An expression is the basic building block for the other two. There is really no way to solve this and it doesn't really say much.

 equation $$3t^5+4t-1=0$$

If you take an expression and assign a value (not necessarily zero) to it using an equal sign, you get an equation. This, in essence, restricts the expression to a set of values for the variable that satisfy the equation.

 function $$f(t)=3t^5+4t-1$$

When we take an expression and give it a name, we produce what we usually call a function.
Important - - All polynomials are functions, i.e. they all pass the vertical line test.

Zeroes

The zeroes of a polynomial are where the graph of the polynomial crosses the x-axis, called the x intercepts. Zeroes are often called roots and sometimes called poles (especially in electrical engineering). We will use the terms zeroes, roots and x-intercepts interchangeably. On graphs, zeroes are x-values where the graph crosses the x-axis, i.e. where $$y=0$$. These points have special meaning in many contexts.

One of the nice things about polynomials is that there are exactly the same number of zeroes as the highest power in the polynomial. So, if we have a polynomial $$g(x)=x^4+3x^3-1$$, the highest power is 4 and so we know that we have 4 zeroes.
Special Notes
1. Notice that we did not say that the zeroes are all real values. From the context, it is natural to assume this but that is not the case. For example, the graph of the polynomial $$y=x^2+1$$ never crosses the x-axis but it does have two roots, both complex.
2. Just because the zeroes exist, does not mean we can find them.

For more details on zeroes and how to find them, read the discussion the polynomial zeroes page.

Next

Okay, so that is a quick introduction to polynomials. On other pages, we discuss the properties of polynomials, finding zeroes and other details. Unless otherwise instructed, we recommend polynomial zeroes as the next topic.

Really UNDERSTAND Precalculus

### Topics You Need To Understand For This Page

 functions vertical line test

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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Differential Equations

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