is a polynomial. Polynomial functions are sums of terms consisting of a numerical coefficient multiplied by a unique power of the independent variable. We generally write these terms in decreasing order of the power of the variable, from left to right*. Here is a summary of the structure and nomenclature of a polynomial function:
*Note: There is another school that writes the terms in order of increasing order of the power of x. This has some appeal because we write power series that way. You'll have to choose which works for you.
Here are some examples of polynomial functions and the language we use to describe them:
|f(x) = 3x - 2||Linear polynomial (linear function)|
|f(x) = x2 - 4x + 1||Quadratic polynomial|
|f(x) = -3x3 + x - 6||Cubic polynomial with no quadratic term|
|f(x) = (x - 3)2(2x - 1)||Cubic polynomial (convince yourself that the largest power will be three when expanded)|
|f(x) = 2x4 - 5x3 + 2x2 - x + 17||Quartic polynomial|
|f(x) = 18x5 - 7||Quintic polynomial with only the 5th degree and constant terms.|
It is important that you become adept at sketching the graphs of polynomial functions and finding their zeros (roots), and that you become familiar with the shapes and other characteristics of their graphs.
Each algebraic feature of a polynomial equation has a consequence for the graph of the function. Here is a table of those algebraic features, such as single and double roots, and how they are reflected in the graph of f(x).
|Single root||A solution of f(x) = 0 where the graph crosses the x-axis. For example, the quadratic function f(x) = (x+2)(x-4) has single roots at x = -2 and x = 4.|
|Double root||A solution of f(x) = 0 where the graph just touches the x-axis and turns around (creating a maximum or minimum - see below). For example, the cubic function f(x) = (x-2)2(x+5) has a double root at x = 2 and a single root at x = -5.|
|Triple root||A solution of f(x) = 0 where the graph crosses the x-axis and the curvature changes sign. See the graph below. For example, the cubic function f(x) = x3 has a triple root at x = 0.|
|Inflection point||The name of the point that is a triple root of a polynomial function. The curvature of the graph changes sign at an inflection point between concave-upward and concave-downward. Not all inflection points are located at triple roots (or even at roots at all), but all triple roots are inflection points located on the x-axis.|
|y-intercept||The solution to f(0); the point where a graph crosses the y-axis, usually a convenient (and very easy-to-find) point to plot when sketching a graph.|
|Local maximum/minimum||When a graph turns around (up to down or down to up), a maximum or minimum value is created. Local maxima or minima are not the highest or lowest points on a graph.|
|Global maximum/minimum||The parabola f(x) = x2 has a global minimum at x = 0, but no global maximum (it increases without bound). The parabola f(x) = -x2 has a global maximum, but no global minimum. The graph below has a global maximum at x = 1. The highest/lowest point on a graph (one may not exist).|
|End behavior||When x is large, either positive or negative, we are concerned with whether the function increases or decreases without bound (it will do one or the other).|
Here are the graphs of two cubic polynomials. they differ only in the sign of the leading coefficient.
The leading term of any polynomial function dominates its behavior. When that term has an odd power of the independent variable (x), negative values of x will yield (for large enough |x|) a negative function value, and positive x a positive value.
That is, for large enough | x |,
f(x > 0) > 0
f(x < 0) < 0
The opposite is true when the coefficient of the leading power of x is negative.
Note also in these figures and the ones below that a cubic polynomial (degree = 3) can have two turning points, points where the slope of the curve turns from positive to negative, or negative to positive. The quartic polynomial (below) has three turning points.
In general, we say that the graph of an nth degree polynomial has (at most) n-1 turning points. It may have fewer, however.
When the degree of a polynomial is even, negative and positive values of the independent variable will yield a positive leading term, unless its coefficient is negative. Negative numbers raised to an even power multiply to a positive result:
(-2)(-2) = 4
(-2)(-2)(-2) = -8
(-2)(-2)(-2)(-2) = 16, and so on.
The result for the graphs of polynomial functions of even degree is that their ends point in the same direction for large | x |:
up when the coefficient of the leading term is positive,
down when the coefficient is negative.
Notice that these quartic functions (left) have up to three turning points. A quartic function need not have all three, however. The graph of f(x) = x4 is U-shaped (not a parabola!), with only one turning point and one global minimum.
The table below summarizes some of these properties of polynomial graphs.
Use the sliders on the graph of y = Ax5 + Bx4 + Cx3 + Dx2 + Ex + F below to adjust the constants A, B, C, D and E, and watch the effect on the graph of the function. Pay attention to how the graph behaves at the left and right ends and to how many wiggles it has. Notice also the relative sizes of the effects of changing A-E: Changing A, the coefficient of x5, alters the look of the graph dramatically, while changing the linear parameter E has only a small effect.
4x4 - 3x3 + 6x2 = x + 12
Such an equation can always be rearranged by moving all of the terms to the left side, leaving zero on the right side:
4x4 - 3x3 + 6x2 - x - 12 = 0
Now the solutions to this equation are just the roots or zeros of the polynomial function f(x) = 4x4 - 3x3 + 6x2 - x - 12. They are the points at which the graph of f(x) crosses (or touches) the x-axis. Our task now is to explore how to solve polynomial functions with degree greater than two. We already know how to solve quadratic functions of all kinds. First, a little bit of formalism:
Further, when a polynomial function does have a complex root with an imaginary part, it always has a partner, its complex conjugate.
When a polynomial function has a complex root of the form a + bi, a - bi is also a root. Complex roots with imaginary parts always come in complex-conjugate pairs, a ± bi. When the imaginary part of a complex root is zero (b = 0), the root is a real root.
When faced with finding roots of a polynomial function, the first thing to check is whether there is something that can be factored away from all of its terms. Here are some examples:
x4 - 3x3 + x2 - 7x = 0
x(x3 - 3x2 + x - 7) = 0
|Factor out an x, which appears in all terms. We automatically know that x = 0 is a zero of the equation because when we set x = 0, the whole thing zeros out. This leaves us with finding the zeros of a simpler polynomial.|
3x3 - 27x + 81 = 0
x3 - 9x + 27 = 0
|All terms are divisible by three, so get rid of it. Note that the zero on the right makes this very convenient ... the 3 just "disappears". We haven't simplified our polynomial in degree, but it's nice not to carry around large coefficients.|
11x3 - 121x2 + 3x - 2 = 0
|Tantalizing when you look at the x's, and the 11 and 121, but there is no GCF here.|
Always look for a greatest common factor first when working with any polynomial function. Finding one can make things a lot easier.
Find all roots of these polynomial functions by finding the greatest common factor (GCF).
The example below shows how grouping works. First find common factors of subsets of the full polynomial, say two or three terms, and move that out as a common factor. If what's been left behind is common to all of the groups you started with, it can also be factored away, leaving a product of binomials that are simpler and easier to solve for roots.
The trickiest part of this for students to understand is the second factoring. Look at the example. Between the second and thrid steps. The binomial (x + 3) is just treated as any other number or variable. It appears in both added terms of the second step, therefore it can be factored out.
Find all roots of these polynomial functions by factoring by grouping.
f(x) = Ax2 + Bx + C.
In this form, there is a constant term, and the first term has twice the degree as the middle term. Now consider equations of the form
Notice that each of those equations has the same pattern. All have three terms, the highest power is twice that of the middle term, and each has a constant term (if it didn't, we'd be able to find a GCF). They have the same general form as a quadratic. Here's how to solve them:
Once you get the hang of this, you won't have to use the substitution trick, but it does help to keep things straight. This kind of substitution is also used in all kinds of other situations in algebra, so it's a good idea to learn it.
While this method of finding roots isn't used all that often, it's a huge timesaver when it can be used. You don't have to memorize these formulae (you can always look them up), but use them in situations where your polynomial equation is a sum or difference of cubes, such as
You should confirm these formulae for yourself by multiplying and simplifying the right sides.
Use the sum/difference of perfect cubes formulae (above) to find all of the roots (zeros) of these functions:
The important thing to keep in mind about the rational root theorem is that any given polynomial may not even have any rational roots. In those cases, we have to resort to estimating roots using a computer, using methods you will learn in calculus.
Consider a polynomial equation of the form
Axn + Bxn-1 + . . . + Z = 0,
where A is the coefficient of the leading term and Z is the constant term. Now let p = the set of all possible integer factors of Z, and their negatives, and let q = the set of all possible integer factors of A, and their negatives. The rational root theorem says that if there are any rational roots of the equation (there may not be), then they will have the form p/q. That is, any rational root of the equation will be one of the p's divided by one of the q's.
We will see how the rational root theorem works in a bit, but first, a little detour into ...
Synthetic substitution is simply a method for substituting a value into the independent variable of a polynomial function. It's quicker and easier than doing it directly, and kind of magic, really. Here's an example:
Sometimes (erroneously) called synthetic division, this procedure is illustrated by example on the left. It's a quick and easy method to test whether a value of the independent variable is a root. Remember that a root will make the function value zero.
The method starts with writing the coefficients of the polynomial in order of the power of x that they multiply, left to right. It's important to include a zero if a power of x is missing. In the example, if there were no linear term, we'd put 0 in instead of a 1 in the first step.
The number to be substituted for x is written in the square bracket on the left, and the first coefficient is written below a line drawn under everything (second step). That's the setup. Now it's just a matter of doing the same thing to the end.
The number in the bracket is multiplied by the first number below the line. The result becomes the next number in the second row. The numbers now aligned in the first and second row are added to become the next number under the line. Repeat until you're finished. The last number is the result.
In our example, -1 is a root because it makes the function zero. The binomial (x+1) must then be a factor of f(x).
The rational root theorem gives us possibilities of rational roots, if any exist. Now synthetic substitution gives us a quick method to check whether those possibilities are actually roots. Using the rational root theorem is a trial-and-error procedure, and it's important to remember that any given polynomial function may not actually have any rational roots. Its roots might be irrational (repeating decimals) or imaginary.
We begin by identifying the p's and q's. For this function it's pretty easy. The constant term is 3, so its integer factors are p = 1, 3. The coefficient of the highest degree term (x4), is one, so its only integer factor is q = 1. Therefore our candidates for rational roots are:
Now we test to see if any of these is a root. For work in math class, here's a hint: always try the smallest integer candidates first. This is just a matter of practicality; some of these problems can take a while and I wouldn't want you to spend an inordinate amount of time on any one, so I'll usually make at least the first root a pretty easy one. Here we try one and see that it's a root because the value of the function is zero. Notice that the coefficients of the new polynomial, with the degree dropped from 4 to 3, are right there in the bottom row of the synthetic substitution grid.
Now we don't want to try another positive root because the coefficients of the new cubic polynomial are all positive. There's no way that a positive value for x will ever make the function equal zero. We'll try the next-easiest candidate, x = -1:
That worked, and now we're left with a quadratic function multiplied by our two factors. That's good news because we know how to deal with quadratics. This one is easily factorable:
So that's the whole problem. The complete factorization is:
and the roots are
Find all of the roots (zeros) of these polynomial functions:
Sometimes you won't find a GCF, grouping won't work, it's not a sum or difference of cubes and it doesn't look like a quadratic, . . . and it doesn't have any rational roots. What to do? Well, you're stuck, and you'll have to resort to numerical methods to find the roots of your function.
That means graphing the function on a calculator and estimating x-axis crossings or using a numerical root-finding algorithm. Some calculators and many computer programs can do this. You'll also learn about Newton's method of finding roots in calculus.
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