]> Chapter 03 Notes (Sullivan & Sullivan 3e)

Chapter 03: Polynomial and Rational Functions

§3.1: Quadratic Functions and Models

The Three Forms of a Quadratic Function

The various forms make different questions easier to answer…so you should know all three forms! You should also know how to convert a quadratic equation to and from each of these forms.

General Form

$y = a x^{2} + b x + c$ is called the General Form. This form is useful for solving via the Quadratic Formula, and it’s the easiest version for finding the y-intercept. The value of a indicates how squeezed/tall the parabola is; the value of b is difficult to interpret. The hidden meaning of c should not be difficult to determine…

Vertex Form

Also called standard form… $y = a {(x - h)}^{2} + k$ . This form is most useful for graphing, or just locating the vertex (hence the name). The vertex is at $(h, k)$ . This is equivalent to my transformation template for quadratics…I’d write it as $y = a {(x - c)}^{2} + d$ for that reason. Once again, the value of a indicates how squeezed/tall the parabola is.

One must usually complete the square in order to convert into this form.

Factored Form

$y = a (x - m) (x - n)$ is what I call the Factored Form. This form is most useful for determining the zeros of the function, and hence the x-intercepts. The x-intercepts are at $(m, 0)$ and $(n, 0)$ . Once again, the value of a indicates how squeezed/tall the parabola is.

One must usually factor in order to convert into this form (hence the name).

It’s good to know that the x-coordinate of the vertex is at the average of the x-intercepts!

Examples

[1.] Sketch the graph of $y = 3 - {(x - 2)}^{2}$ .

This parabola has been flipped across the x-axis, shifted two right and up three.

An Image

[2.] Locate the vertex and axis of symmetry of $y = 2 (x - 1) (x - 3)$ .

There are a couple of ways of doing this…I’ll use the shortcut: the vertex is halfway between the x-intercepts, at $x = 2$ . This is the axis of symmetry.

Substitute the two into the equation to get the y-coordinate of the vertex: $y = - 2$ . The vertex is at $(2, - 2)$ .

[3.] Determine the equation of the parabola pictured below:

An Image

The vertex tells me that the equation looks like $y = a {(x + 2)}^{2} - 8$ ; I just need to figure out what a is. To do that, I’ll use one of the other points: $0 = a {(0 + 2)}^{2} - 8 \Rightarrow 0 = 4 a - 8 \Rightarrow 4 a = 8 \Rightarrow a = 2$ .

The equation is $y = 2 {(x + 2)}^{2} - 8$ .

(you could have used the same method but with the factored form of the parabola…)

§3.2: Power Functions and Models

Definition

A Power Function is an equation of the form $y = x^{n}$ , where $n \in ℤ^{+}$ . Notice that the simplest quadratic function is an example of a power function.

Properties

Different power functions share some traits, based on the magnitude of the exponent and whether the exponent is even or odd.

End Behavior

End Behavior refers to the direction that the graph turns on the far left and right ends. If you wanted to be really nerdy, you could say that the left end behavior was $\lim_{x \to - \infty} f (x)$ and that the right end behavior was $\lim_{x \to \infty} f (x)$ . Of course, you probably don’t know what some of those symbols mean…yet.

Since raising a negative number to an even power results in a positive number, power functions with even exponents will have both ends turned upwards. Thus, we could say “as x approaches negative infinity, y approaches positive infinity,” or we could write “ $x \to - \infty \Rightarrow y \to + \infty$ .” The single arrow is read as “approaches” and the double struck arrow is read as “implies.” That’s just the behavior on the left end, though—you’ve still got one more side to describe! You could write another one of those statements like I just used, or you could describe both sides with a little notation: $(\infty, \infty)$ . This looks like an interval, but in the context of end behavior, most people would understand that you meant both side of the function turn upwards.

Since raising a negative number to an odd exponent results in a negative number, the left side of power functions with an odd exponent will be turned down, and the right side will be turned up. Once again, you have a choice of how you want to communicate that—I won’t list all three possibilities here.

Extrema

It turns out that the number of possible extrema (or turning points) is determined by the magnitude of the exponent in a power function. I won’t show the series of graphs here that I use in class to show you this…and without those graphs, I can’t really explain one crucial part of this fact…so you’ll just have to take this one plain, without any condiments.

The maximum number of extrema is one less than the exponent. The possible number of extrema is an even number less than this maximum value—in other words, extema come and go in pairs. This is related to something else that comes and goes in pairs…more on that later in the chapter (and next year in Calculus).

Graphing with Transformations

Again, something is lost without seeing a page full of graphs here…you’ll have to make due.

You should know what the basic power functions look like— $x^{2}$ , $x^{3}$ , $x^{4}$ , etc.

Once you know what these parent functions look like, then you can graph simple transformations of them. Specifically, you can graph $y = a \cdot x^{b (x - c)} + d$ . This is my transformation template (for those who haven’t seen it before, or who—gasp!—have forgotten it). In this template, a multiplies the y-values (sometimes called the stretch factor by some authors); b divides the x-values (omitted by most authors until Trigonometry—a bad choice, if you ask me); c move the graph horizontally (many authors use an h here); d moves the graph vertically (and many authors use a k here).

You should be able to sketch a transformation; you should be able to describe a transformation in words; you should be able to read a verbal description of a transformation and then write the resulting equation.

Examples

[4.] Sketch the graph of $y = 2 {(x + 1)}^{3}$ .

Start with the graph of the parent $y = x^{3}$ :

An Image

Multiply the y-values by two, then shift the whole thing left one unit.

An Image

[5.] Describe a sequence of transformations that changes the graph of $y = x^{4}$ into the graph of $y = {(2 x + 1)}^{4} - 5$ .

First of all, let’s rearrange that to fit my template: $y = {(2 x + 1)}^{4} - 5 = {(2 (x + \frac{1}{2}))}^{4} - 5$ . The 2 will divide every x-value by two (pulling the graph closer to the y-axis); the $\frac{1}{2}$ will shift the graph left one half unit; the 5 will move the graph down five units.

[6.] A parabola has been flipped across the x-axis, moved three units right and two units up. Write the equation of the new parabola.

Flipping across the x-axis makes $a = - 1$ ; moving three units right makes $c = 3$ ; moving two units up makes $d = 2$ . Thus, the equation is $y = - {(x - 3)}^{2} + 2$ .

§3.3: Polynomial Functions and Models

Definition

A polynomial (with real coefficients) is an equation of the form $y = a_{n} x^{n} + a_{n - 1} x^{n - 1} + a_{n - 2} x^{n - 2} + ... + a_{2} x^{2} + a_{1} x + a_{0}$ , where each $a_{i} \in ℝ$ , $n \in ℤ^{+}$ and $a_{n} \neq 0$ . This can be written more compactly as $y = \sum_{i = 0}^{n} (a_{i} x^{i})$ , but that will probably cause some of you to fall over with seizures. Take your meds; I’ll be here waiting.

OK—so it typically bothers quite a few of you that every coefficient appears to be the letter a. However, if you look carefully, you’ll notice that each one has a subscript—that means that they are really lots of different values. Math people prefer writing it that way, rather than $y = a x^{n} + b x^{n - 1} + c x^{n - 2} + ... + v x + w$ or something like that. The bonus is that we can then compact the notation into the summation notation that sent some of you scurrying away, screaming. We’ll be dealing with that summation notation later!

The first term $a_{n} x^{n}$ is called the leading term of the polynomial. The first coefficient $a_{n}$ is called the leading coefficient. The value n is called the degree of the polynomial. The term $a_{0}$ is called the constant term.

The long form definition I used is called the expanded (or general) form. Much of the point of this chapter is to write polynomials in factored form: $y = a {(x - r_{1})}^{m_{1}} {(x - r_{2})}^{m_{2}} {(x - r_{3})}^{m_{3}} ... {(x - r_{j})}^{m_{j}}$ . It is not hard to see that a polynomial written in this form could be expanded and written in the long form!

Multiplicity

When written in factored form, each binomial piece $(x - r_{i})$ is a factor of the polynomial. The exponent $m_{i}$ is called the multiplicity of the factor.

The Graphical Effect of Multiplicity

First of all, note that when $x = r_{i}$ , the value of the polynomial must be zero—thus, the numbers $r_{i}$ are the locations of x-intercepts on the graph of the polynomial. Thus, each factor locates an x-intercept on the graph.

A factor that has an odd multiplicity will have an x-intercept where the graph pierces (passes through) the axis. A factor that has an even multiplicity will have an x-intercept where the graph grazes (bounces off of) the axis.

Examples

[7.] Write the equation of a polynomial with the following zeros and multiplicities:

Zero	1	-2	4
Multiplicity	2	1	3

How about $y = {(x - 1)}^{2} (x + 2) {(x - 4)}^{3}$ ?

[8.] Determine the zeros (and multiplicities) of $y = 4 x {(x + 1)}^{5} {(x - 3)}^{2} (x - 5)$ .

They are…

Zero	0	-1	3	5
Multiplicity	1	5	2	1

[9.] Determine the x-intercepts of $y = x^{3} + 2 x^{2} + x$ . At each x-intercept, does the graph cross or touch the axis?

First, some factoring: $y = x^{3} + 2 x^{2} + x = x (x^{2} + 2 x + 1) = x {(x + 1)}^{2}$ .

The x-intercepts are at $x = 0$ and $x = - 1$ . At zero, the graph crosses the axis. At negative one, the graph touches the axis.

§3.4: Rational Functions, Part 1

Disclaimer

I would not split this into two sections. How I write these notes here is NOT how I will present this information in class! Be warned.

Definition

A rational function is a function of the form $\frac{n (x)}{d (x)}$ , where both $n (x)$ and $d (x)$ are polynomials.

The Reciprocal Function

The simplest rational function—the parent function—is the reciprocal function, $y = \frac{1}{x}$ . We mentioned this function earlier as a basic one which everyone should know how to graph.

Graphing with Transformations

Thus, we can graph simple transformations of the reciprocal function: $y = a \frac{1}{b (x - c)} + d$ or $y = \frac{a}{b (x - c)} + d$ . The parameters a through d do the same thing that they did back in Chapter 2 (and that they will continue to do!).

Notice that the b is not really necessary, since $y = \frac{a}{b (x - c)} + d = (\frac{a}{b}) (\frac{1}{x - c}) + d$ . The b effect can be isolated, or it can be lumped together with the effect from a. This is true for most functions that you are going to encounter, and that’s why most authors neglect it.

Silly authors.

Asymptotes

An asymptote is a value that a function approaches as either x or y (or maybe both) moves away from zero. Asymptotes are lines (usually) that a graph tends to look like when you look really far away from the origin. Some asymptotes can be crossed; others cannot.

Vertical

A vertical asymptote will (usually; see the next section) occur at those values of x that cause the denominator of the rational function to be zero.

Since vertical asymptotes are caused by something related to the domain (you can’t divide by zero), they cannot ever be crossed.

Horizontal

A horizontal asymptote is a type of end behavior—it is a horizontal line which the graph tends to look like as the value of x gets really, really far from zero.

Since horizontal asymptotes are not related to the domain, they can be crossed.

Notice that I didn’t mention what causes them—more on that in the next section.

Oblique

An oblique asymptote is another type of end behavior—it is an oblique (neither horizontal nor vertical) line which the graph tends to look like when x gets really far from zero.

Oblique asymptotes can be crossed.

Other Asymptotes

The previous three types of asymptotes were all lines—but life is much more rich and interesting than that. There are some that claim asymptotes can only be lines…I’m not one of them. You can approach any type of graph when x gets really far from the origin! More on this in the…you guessed it; next section.

§3.5: Rational Functions, Part 2

Holloman’s Method for Rational Functions

So—here’s my (nearly) foolproof five step method for analyzing and graphing rational functions.

1. Factor the top and bottom

The key is factoring the numerator and denominator. If you can’t (are you sure that you can’t?), then much of the rest of this will not matter; the answers (and characteristics) of the function will be fairly obvious.

2. Look for Domain Issues

Since a rational function is composed of polynomials, there is only one kind of domain issue that can occur—division by zero. However, there are two ways that this division can occur: zero over zero, or non-zero over zero. The graphic effects of these two are different.

Holes

A hole is caused by the zero over zero condition. The only way that this can occur is to have a common factor between the numerator and denominator. A hole is a single value of x where, suddenly, the graph ceases to exist. Everything’s trucking along just fine, then all of a sudden there’s a tiny little gap in the graph.

The value of x that causes this common factor to be zero is the x-coordinate of the hole.

The key to knowing the y-coordinate of the hole is to realize that everywhere except for the location of the hole, the factor that caused the hole is reducing to one (you’ll get a detailed explanation of this in class). Thus, once you’ve determined the location of the hole, you should cancel the common factors and answer all of the remaining questions using this new, reduced version of the rational function.

Vertical Asymptotes

Any domain issues remaining after locating holes will be vertical asymptotes. These will be factors that occur in the denominator, but not in the numerator. Thus, vertical asymptotes will occur at the zeros of the (newly reduced) denominator function.

3. Find Axis Intercepts

y-intercept

A y-intercept is a place on the graph where $x = 0$ . Thus, to find/locate it, you should let $x = 0$ (if possible).

How many y-intercepts can a function have? Think!

x-intercepts

An x-intercept is a place on the graph where $y = 0$ …in other words, the function value is zero. The only way that a fraction can equal zero is when the numerator equals zero—thus, x-intercepts will come from the zeros of the (reduced) numerator function.

4. Determine the End Behavior Model

The Real Way

The End Behavior Model (my terminology) is an equation which the rational function approaches at the extreme left and right ends of the graph (when x is very, very far from zero). To determine the equation of the End Behavior Model, divide the denominator into the numerator. The quotient of this division is the equation of the End Behavior Model (just put $y =$ in front of it).

The Short Cut for Linear Asymptotes

If you’re only concerned about the three linear types of asymptotes, there is a short cut. For this short cut, let n be the degree of the numerator and let d be the degree of the denominator. There are four possibilities:

$d > n$ : There is a horizontal asymptote at $y = 0$ .
$d = n$ : There is a horizontal asymptote at the ratio of leading coefficients.
$d = n - 1$ : There is an oblique asymptote. You must divide to find its equation.
$d < n - 1$ : There is no linear asymptote.

5. Graph

Plot what you’ve found

Plot the holes, intercepts and asymptotes. This is the framework around which the graph must flow.

THINK

First, make sure there isn’t anything silly on your paper. You can’t have an intercept where there is a vertical asymptote! You can’t have more than one y-intercept! You can’t have a hole where a vertical asymptote exists!

Graphs are always drawn from left to right! Put your pencil on the end behavior on the left side of the graph. Now—where must the graph go in order to get to the next spot you’ve plotted?

THINK!

Remember that the graph can only cross the x-axis at the locations of the x-intercepts you have plotted. Remember that the graph can only cross the y-axis at the location of the y-intercept. Remember that the graph won’t stop until you’ve reached the far right hand side—and when you get there, your pencil must end on the end behavior.

Examples

[10.] Sketch $y = \frac{x^{2} + 3 x - 4}{x^{2} - x - 6}$ .

First of all, factor: $y = \frac{x^{2} + 3 x - 4}{x^{2} - x - 6} = \frac{(x + 4) (x - 1)}{(x - 3) (x + 2)}$ . Since there are no factors in common between the numerator and denominator, there are no holes. That means that the zeros of the denominator are the locations of vertical asymptotes—specifically, $x = 3$ and $x = - 2$ .

Letting $x = 0$ determines that the y-intercept is at $(0, \frac{2}{3})$ .

The zeros of the numerator locate the x-intercepts: $(- 4, 0)$ and $(1, 0)$ .

Since the numerator and denominator are of the same degree, the end behavior is a horizontal asymptote at the ratio of leading coefficients: $y = 1$ .

Plot all of that and you get this:

An Image

Start at the end behavior on the left. Which way must the graph go? Remember, it has to cross the axis at that x-intercept, and you can’t cross that first vertical asymptote.

An Image

What happens between those two vertical asymptotes? Remember that you must cross through both points, and you can’t cross an axis unless you’ve already got a point marked there.

An Image

Notice that we crossed the horizontal asymptote. That’s fine! You can cross the horizontal ones; you can’t cross the vertical ones.

Finally, what happens on the right? There aren’t any x-intercepts, and you’ve got to approach those asymptotes…

An Image

[11.] Sketch $y = \frac{x + 2}{2 x^{2} + 3 x - 2}$ .

Factor: $y = \frac{x + 2}{(x + 2) (2 x - 1)}$ . The common factor indicates that there is a hole at $x = - 2$ . The reduced function is $y = \frac{1}{2 x - 1}$ ; I’ll use that to answer all of the other questions…starting with the y-coordinate of the hole: $- \frac{1}{5}$ . There is still a domain issue indicating a vertical asymptote at $x = \frac{1}{2}$ .

Letting $x = 0$ gives the y-intercept at $(0, - 1)$ .

There are no values of x that will cause the numerator to equal zero; thus, there are no x-intercepts.

Since the denominator is of higher degree, the end behavior is $y = 0$ .

Here’s the graph:

An Image

[12.] Sketch $y = \frac{2 x^{3} + 5 x^{2} - 12 x}{x^{2} - 2 x}$ .

Factor: $y = \frac{x (2 x - 3) (x + 4)}{x (x - 2)}$ . There is a common factor, indicating a hole at $x = 0$ . Use the reduced function $y = \frac{(2 x - 3) (x + 4)}{x - 2}$ to find the y-coordinate of the hole: $y = 6$ . There is still a domain issue, indicating a vertical asymptote at $x = 2$ .

I can’t let $x = 0$ since there is a hole there, so there is no y-intercept.

There are two x-intercepts: $(\frac{3}{2}, 0)$ and $(- 4, 0)$ .

Since the degree of the numerator is one higher than the denominator, the end behavior will be an oblique asymptote—I’ll use synthetic division to find the equation.

$\begin{array}{r} 2 | & 2 & 5 & - 12 \\ 4 & 18 \\ \overline{2} & \overline{9} & \overline{| 6} \end{array}$

The end behavior is $y = 2 x + 9$ .

Plot all of that and you get this:

An Image

§3.6: Polynomial and Rational Inequalities

Solving Polynomial Inequalities

To solve a polynomial inequality, rearrange things so that there is a zero on one side. Factor the non-zero side. Now create a sign chart of the factors. I probably showed you how to do this back in Chapter 2; I will demonstrate it again in class.

Solving Rational Inequalities

To solve a rational inequality, you still begin by rearranging things so that there is a zero on one side. However, once you’ve done that, you now need to force that side to become a single fraction—a common denominator is a definite possibility! Once you’ve got a single fraction on one side, continue as if it were a polynomial inequality. Be careful with your answer from the sign chart—remember that the denominator function cannot equal zero!

Examples

[13.] Solve $6 x^{2} + 2 < - 7 x$ .

Get the zero: $6 x^{2} + 7 x + 2 < 0$ . Factor: $(3 x + 2) (2 x + 1) < 0$ . Make the sign chart:

An Image

We’re looking for where the result is less than zero; thus, the solution is $x \in (- \frac{2}{3}, - \frac{1}{2})$ .

[14.] Solve $x \geq \frac{4 x + 2}{x + 3}$ .

Get the zero: $x - \frac{4 x + 2}{x + 3} \geq 0$ . Now we need to make that a single fraction…using a common denominator: $x \frac{x + 3}{x + 3} - \frac{4 x + 2}{x + 3} \geq 0 \Rightarrow \frac{x^{2} + 3 x - 4 x - 2}{x + 3} \geq 0 \Rightarrow \frac{x^{2} - x - 2}{x + 3} \geq 0$ . Now, factor: $\frac{(x + 1) (x - 2)}{x + 3} \geq 0$ . Now for the sign chart:

An Image

We’re looking for where that is positive—but don’t forget that we can’t let $x = - 3$ ! The solution is $x \in (- 3, - 1] \cup [2, \infty)$ .

§3.7: The Real Zeros of a Polynomial Function

The Division Algorithm

For any two polynomials $p (x)$ and $q (x)$ , there exist unique polynomials $d (x)$ and $r (x)$ so that $p (x) = q (x) \cdot d (x) + r (x)$ and the degree of $r (x)$ is smaller than the degree of $d (x)$ .

In short, you can always divide two polynomials! In this, $p (x)$ is the dividend (polynomial being divided), $d (x)$ is the divisor (what you’re dividing by), $q (x)$ is the quotient (how many “times” the divisor “goes into” the dividend), and $r (x)$ is the remainder.

You may have some choice in how you actually divide two polynomials. It is always the case that you can use Polynomial Long Division—which is almost exactly like good old long division from elementary school. If, however, the divisor is of the form $(x - c)$ , you can opt to use Synthetic Division (also known as Ruffini’s Algorithm). I’ll demonstrate both of these in class.

The Remainder Theorem

The result from the last little bit of that comes when the divisor is of degree one—in particular, that will force the remainder to be a constant (degree zero).

Let’s just say that the divisor is the binomial $(x - c)$ . The division algorithm then says that $p (x) = q (x) \cdot (x - c) + r$ , where r must be a constant. Here’s the trick: use that equation I just wrote to evaluate $p (c)$ : $p (c) = q (c) \cdot (c - c) + r$ . Look at that again—can’t you simplify it a bit? $(c - c) = 0$ , so that equation becomes $p (c) = q (c) \cdot 0 + r \Rightarrow p (c) = r$ .

A-ha! So when $p (x)$ is divided by $(x - c)$ , the remainder is $p (c)$ . This is a useful way (sometimes) of finding the value of $p (c)$ without actually plugging c into $p (x)$ . It is also a quick way to find the remainder of a division without actually dividing…

The Factor Theorem

The binomial $(x - c)$ is a factor of the polynomial $p (x)$ if (and only if) $p (c) = 0$ .

…because if $p (c) = 0$ , then $p (x)$ divided by $(x - c)$ has a remainder of zero…if you divide two things and have a zero remainder, we say that one is a factor of the other!

This is the final theoretical piece that connects the roots of a function to the x-intercepts of the graph and the (real) factorization of a polynomial.

The Rational Root Theorem

It turns out that the polynomial itself will give you some hints as to which rational numbers might be roots…in particular, for a polynomial with leading coefficient q and constant term p, any rational root of the polynomial must be of the form $\frac{a factor of p}{a factor of q}$ .

The Procedure

If you’ve got a calculator, finding all of the real roots of a polynomial is almost trivial.

If you want to find all real roots of a polynomial by hand, you need somewhere to start—in particular, you should start by testing the possible rational roots. Once you find a root (let’s call it c), then factor out $(x - c)$ from the polynomial, either by using Long Division or Synthetic Division. Continue working with the quotient of that division—don’t go back to working on the original, longer polynomial!

A Reminder (Vocabulary)

All of the following are equivalent:

c is a real zero of $f (x)$
c is a real root (or solution) of $f (x) = 0$
$(x - c)$ is a factor of $f (x)$
$(c, 0)$ is an x-intercept on the graph of $f (x)$

Examples

[15.] Find the remainder of $\frac{6 x^{2047} - 5 x^{28} + 2 x^{4} + 2}{x - 1}$ .

Recall that the remainder of $\frac{f (x)}{x - c}$ is $f (c)$ ; hence, the remainder of $\frac{6 x^{2047} - 5 x^{28} + 2 x^{4} + 2}{x - 1} = \frac{f (x)}{x - 1}$ will be $f (1) = 6 {(1)}^{2047} - 5 {(1)}^{28} + 2 {(1)}^{4} + 2 = 6 - 5 + 2 + 2 = 5$ .

[16.] Determine the possible rational roots of $6 x^{4} - 5 x^{3} - 16 x^{2} + 10 x + 8 = 0$ .

The possible rational roots will be a factor of 8 divided by a factor of 6. Every one can be either positive or negative. I won’t list those separately; I’ll just a plus-or-minus sign out in front of the list: $\pm {1, 2, 4, 8, \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{4}{3}, \frac{8}{3}, \frac{1}{6}}$

[17.] Find all real zeros of $g (x) = 2 x^{4} - x^{3} - 7 x^{2} + 2 x + 6$ .

If you’re using a calculator, then this is easy. I’ll demonstrate it without a calculator.

First of all, the possible rational roots are $\pm {1, 2, 3, 6, \frac{1}{2}, \frac{3}{2}, \frac{1}{6}}$ . I won’t show the scratch work that allowed me to see that -1 works, but I will show how I factor that out:

$\begin{array}{r} - 1 | & 2 & - 1 & - 7 & 2 & 6 \\ - 2 & 3 & 4 & - 6 \\ \overline{2} & \overline{- 3} & \overline{- 4} & \overline{6} & \overline{| 0} \end{array}$

This factors out to $g (x) = (x + 1) (2 x^{3} - 3 x^{2} - 4 x + 6)$ ; I’ll continue working with the right hand factor. Immediately I notice that I can factor this piece directly, without testing possible rational roots!

$2 x^{3} - 3 x^{2} - 4 x + 6 = x^{2} (2 x - 3) - 2 (2 x - 3) = (2 x - 3) (x^{2} - 2) = (2 x - 3) (x - \sqrt{2}) (x + \sqrt{2})$

I now see four real roots—I must be done. The real zeros are $- 1$ , $\frac{3}{2}$ and $\pm \sqrt{2}$ .

§3.8: Complex Zeros; Fundamental Theorem of Algebra

The Fundamental Theorem of Algebra

All of this discussion of roots of polynomials continued from the early 1600’s up until the 1800’s. It had been noticed that the number of real roots could be equal to the degree of the polynomial, and it had also been noted that some polynomials had non-real roots, but it took a long time (until 1806) before someone could actually prove that the number of complex roots of a polynomial is equal to the degree of the polynomial.

The Conjugate Pairs Theorem

If your polynomial has real coefficients (which all of ours do), then a consequence of this is that non-real roots must come in conjugate pairs.

Remember that the conjugate of the complex number $a + b i$ is $a - b i$ . Notice that the product of a complex number with its conjugate is $(a + b i) (a - b i) = a^{2} + b^{2}$ , which is real number—perhaps you can see that it is this property which forces all of the coefficients to be real when the polynomial is converted from factored to expanded form.

The Procedure

If you are looking for all complex roots of a polynomial, then you can start by finding the real roots—either with a calculator or with the rational root theorem. As you find them, factor them out…eventually you will be left with a polynomial that won’t factor in the reals. Note that this polynomial must be of even degree! For us, this leftover will usually be degree two—which means that you can use the quadratic formula to find the non-real roots.

A Reminder (Vocabulary)

All of the following are equivalent:

c is a complex zero of $f (x)$
c is a complex root (or solution) of $f (x) = 0$
$(x - c)$ is a factor of $f (x)$

Examples

[18.] Find all complex roots of $x^{4} + 2 x^{2} - 15 = 0$ .

This looks a lot like a good old quadratic, doesn’t it? It just has all of its exponents doubled…which is a clue of how to factor it without a calculator.

Let $u = x^{2}$ , which will make the equation $u^{2} + 2 u - 15 = 0$ . Do you think you can factor that? I hope so! $u^{2} + 2 u - 15 = 0 \Rightarrow (u + 5) (u - 3) = 0$ , which gives us two equations: $x^{2} + 5 = 0$ and $x^{2} - 3 = 0$ . The solutions to the first equation are $x = \pm i \sqrt{5}$ ; the solutions to the second equation are $x = \pm \sqrt{3}$ . The polynomial was degree four and we found four roots, so that’s it!

[19.] Find all complex zeros of $h (x) = x^{3} - 6 x^{2} + 21 x - 26$ .

While that constant term looks daunting, it doesn’t have all that many factors, so don’t go running for the calculator! The possible rational roots are $\pm {1, 2, 13, 26}$ , and it doesn’t take much work to see that 2 is a real zero.

$\begin{array}{r} 2 | & 1 & - 6 & 21 & - 26 \\ 2 & - 8 & 26 \\ \overline{1} & \overline{- 4} & \overline{13} & \overline{| 0} \end{array}$

Now I have to factor $x^{2} - 4 x + 13$ …and it shouldn’t take long for you to decide that the quadratic formula is the way to go.

$x = \frac{4 \pm \sqrt{16 - 4 (1) (13)}}{2 (1)} = \frac{4 \pm \sqrt{16 - 52}}{2} = \frac{4 \pm \sqrt{- 36}}{2} = \frac{4 \pm 6 i}{2} = 2 \pm 3 i$ .

The complex zeros are 2 and $2 \pm 3 i$ .

Page last validated 2010-10-10