Okay—you know what a variable is. A term is a product of constants and powers of variables (for example: ; ). For now, let's restrict our thinking to terms with real coefficients, one variable, and positive integer exponents.
A polynomial is the sum of some number of terms (with the restrictions I just mentioned). For example: ; .
In general, a polynomial looks like this: .
The highest power of the variable (n) is called the degree of the polynomial. The term where that power occurs is called the leading term ( ; it is typically listed first when writing the polynomial). The coefficient on the leading term is called the leading coefficient (). Of course, this is all familiar, since I told you about it in the last chapter…
If you write the polynomial so that the exponents on the variable are decreasing, then the polynomial is written in standard form.
Polynomials can be classified in two ways: by degree, and by the number of terms. When classified by degree, they are called linear, quadratic, cubic, quartic, etc. When classified by number of terms, they are called monomials, binomials, trinomials…and I've never heard of a term for more than three!
If you are given some points on a polynomial, and asked to find the model for that graph, there are two ways to handle it: algebraically (by hand) or with the calculator.
We saw how to handle it algebraically (by hand) in the last chapter. To determine the model of a quadratic, we plugged in the given points and were left with three equations and three variables (the three coefficients in the standard form of a quadratic). To do this with higher degree polynomials, plug in the points (you must have a number of points one greater than the degree of the polynomial) and solve the resulting system of equations.
To get help from the calculator, use the various regression models (quadratic, cubic, quartic) that your calculator gives you!
[1.] Write in standard form, and classify it by degree and number of terms.
Add the like terms (same variable, same exponent) and write in descending order of exponents: . This is a third-degree polynomial with 4 terms.
[2.] Find a cubic model for the following points: , , , .
Drop this into your calculator and let cubic regression do its work!
The model is .
In precisely the same way that real numbers can be factored into a unique product of primes, so too can polynomials be factored into a unique product of linear factors.
(You can also add, subtract, multiply and divide polynomials…you can even define a zero polynomial and a one polynomial! I wonder if polynomials form their own "number" system, like matrices do…)
A linear factor is a binomial of the form , where c is a real number. Note that there's a minus in there—that's important!
One of the interesting features on the graph of a polynomial function is the x-intercept. We have seen (in the previous chapter) how the factored form of a polynomial (a quadratic at the time) can be used along with the zero-product property to determine the zeros of the function. Now we're going to do the same thing with higher degree polynomials!
Let's make this clear: the zeros of the polynomial, the linear factors of the polynomial, and the x-intercepts of the polynomial are all really the same thing. Many books have a list of equivalent terms that sums this up.
All of the following are equivalent:
One of the things that we are going to spend some time doing is taking a polynomial and figuring out (by hand) what its factors are. This has the side benefit of telling us where the x-intercepts of the graph are.
Another thing that we will do is take a list of x-intercepts (or zeros), and from them, write the polynomial in standard form. This is easy!
Just as prime factors can show up multiple times (expressed as a power of the prime), so too can linear factors show up multiple times—take as an example. The number of times that the linear factor shows up (the exponent on that linear factor) is called the multiplicity of that factor. One really nice thing to notice is that factors with even multiplicity will have zeros (x-intercepts) that just barely touch/brush the x-axis, while factors with odd multiplicity will have zeros that pierce the axis.
In addition to the x-intercepts, other points of interest are the y-intercept, any relative maxima (hilltops) and relative minima (valleys). Finding the y-intercept is easy (you do remember how to do it, don't you?). Finding the extrema (maxima and minima) will require your calculator…at least, until you learn calculus…
[3.] Write in standard form.
When you multiply, remember to work every possible pair of terms from the left and right!
[4.] Locate the axis intercepts and extrema of .
We're going to have to use the calculator—put the function in and take a look.
Now use the built-in functions to find the x-intercepts at , and . Note that is also the y-intercept. The maximum is at , and the minimum is at .
[5.] Finds the zeros of , and state the multiplicity (even or odd) of each.
You don't yet have a way to find the zeros of this function algebraically, so you'll have to graph it.
The first zero (even multiplicity) is -2; the second (odd multiplicity) is -0.5; the final zero (odd multiplicity) is 1.
I alluded to this before—now it's time to do it. Dividing polynomials is a lot like the division algorithm that you (probably) learned a few years ago. There are a few differences, so pay attention!
First, some terminology: the numerator (the polynomial being divided) is called the dividend. The denominator (the polynomial doing the dividing) is called the divisor. The result of the division has two parts: the quotient, and the remainder.
Before you begin, make sure everything is lined up—the divisor and dividend should both be in standard form. Also, any missing powers of the variable should have 0 put in place—for example, should be written as .
Step one—look at the leading terms of the divisor and the dividend. What must you multiply the leading term of the divisor by so that it becomes the leading term of the dividend? Put this above the leading term (just above that horizontal line in the long division symbol…by the way, that line is called a virgule)—this is a term of the quotient.
Step two: multiply that new term of the quotient times the divisor. Place the result underneath the dividend (make sure to line up your powers of the variable!).
Step three: subtract—top minus bottom. The result of this becomes the new dividend.
Now repeat steps one through three until the new dividend is of a degree smaller than the divisor. The final dividend is the remainder.
Look at the example following this, and be sure to pay attention to the examples in class!
At some point in history, a really smart fellow (Paolo Ruffini; 1809) realized that the only thing that really mattered in certain forms of polynomial long division was the coefficients; the variables were just there for the ride. Thus, he devised a way to do division with just the coefficients. We call this synthetic division. Alas, it only works if the divisor is linear…
For this method, write the zero of the linear divisor on the left (where you would normally write the divisor), and write the coefficients of the dividend on the right (where you would normally put the dividend!). Place the bar (virgule) a short distance (not directly underneath) the dividend. Now: pull down the first number in the dividend—that's the first number of the quotient. Next, multiply the divisor times the newest part of the quotient, and place the result underneath the dividend in the next column. Add, and place the result in the quotient. Continue this process until you've worked every column. The last number on the bottom row isn't part of the quotient—it's the remainder.
Again, watch the examples closely!
A really nice fact about division of a polynomial by a linear binomial: the remainder of is .
[6.] Divide using long division: .
Here goes!
The quotient is and the remainder is .
[7.] Divide using synthetic division: .
A little shorter:
The quotient is and the remainder is -5.
[8.] Use synthetic division and the Remainder Theorem to find if .
Remember that the remainder of is . Since we want to find , we need to divide .
Since the remainder of is 2, .
I shouldn't have to say anything about solving polynomial equations graphically…they should be sufficiently easy on their own.
Solving polynomial equations by factoring is the same as solving quadratic equations by factoring—which we did in the previous chapter. Just remember that the other side of the equation must be zero in order for this to work!
There is no equivalent to the quadratic formula for higher-degree polynomials…in fact, Evariste Galois (French; 1811-1832) proved that for polynomials of degree 5 or higher, there can be no such formula!
Sometimes you can simplify the problem with a change of variable…you can make a higher-degree polynomial look like one of a smaller degree. See one of the following examples for more information.
[9.] Solve .
I would like to point out that you can factor this!
This means that either , or .
You can also graph this to solve it…
[10.] Factor .
In this case, you don't have any choice but to start by looking at the graph.
Now, find those zeros—they tell you what some of the factors are (-1.12, -1, 1.12, 2). Factor them (the integers) out!
That gives us .
That puts us at .
Now to work with what's left… note that this really is a difference of squares. The final factorization is .
[11.] Solve by factoring: .
Believe it or not, this is really a quadratic polynomial! Let …that makes the equation . This factors to —put it back in terms of x: . Now we can solve: , and . The solutions are and .
More information that may be useful in factoring polynomials…
The Rational Root Theorem says that all rational roots of polynomials are of the form , where p is a factor of the constant term, and q is a factor of the leading coefficient. This only provides a way to find possible rational roots—there is no guarantee that any of them actually work!
If you find a root (zero) of the polynomial, remember that you also now know one of the factors of that polynomial—so factor it out, and continue working with the quotient.
Irrational roots are harder to find. The only help here is that certain irrational roots must come in conjugate pairs. The conjugate of the irrational number is (sound familiar?).
Non-real roots can only be found through the quadratic formula. Note that non-real roots also come in conjugate pairs.
Remember: knowing the roots of a polynomial is the same as knowing the factors! Thus, if you are given a list of roots, you should be able to write a product of factors that (when multiplied out) produce the required polynomial.
[12.] List the possible rational roots of .
The possible values of p are 1, 2, or 4. The possible values of q are 1, 2, 3, or 6. That makes the possible rational roots , , , , (which we already listed; there is no need to repeat it!), (same comment), , , (again), , (again), .
The possible rational roots are .
[13.] A certain 4th degree polynomial has the following roots: and . Determine the other roots of the polynomial.
Another irrational root must be , and another non-real root must be .
[14.] Find a 3 rd degree polynomial (in standard form) with the following roots: 3, -1, and .
One answer is . Another possible answer is (note that this is just three times the previous answer).
The Fundamental Theorem of Algebra (which, it turns out, is really a theorem about something else—analysis) says that a polynomial (of degree at least 1—the other polynomials are boring) has at least one complex root. Our textbook recognizes Gauss (Karl Friedrich; 1777-1855) as the first to prove this theorem; however, by today's (rigorous) standards, his initial proof (in 1799) isn't good enough—it took him until 1816 to come up with a complete and correct proof.
Note that the above is just one way to state this theorem—another way (given by our text as a corollary) is that every polynomial of degree n has n complex roots.
This is powerful!
Add to that, the fact that every root gives rise to a factor provides us with the last piece of the puzzle: now we can completely factor a polynomial.
If you've got your calculator, then graph the polynomial and find the real zeros. Remove the associated factors until you get the problem down to a quadratic, then use the quadratic formula.
If you don't have your calculator, then start with the possible rational roots. Check them one by one until you find some that work. Remove the factors, and hope that what remains is a quadratic!
[15.] Find all complex zeros of .
If you have a calculator, then start with the graph. Let's pretend that we don't, and begin with the possible rational roots: .
Whew! Start small and work your way up.
is clearly not zero, so 1 is not a root.
is also not zero, so -1 is not a root.
is not zero, so 2 is not a root. Keep trying!
! We found one! Factor out .
That puts us at . There is no need to continue with checking rational roots—we're down to a quadratic, and we have ways of factoring that on our own!
[16.] Find all complex zeros of .
Again, let's pretend that we don't have the calculator. The possible rational roots are .
, so 1 is a root.
Now I've got to find the zeros of …it doesn't look like 1 or -1 will work; I'll move on to two.
, so 2 isn't a root.
, so -2 is a root!
That reduces the problem to …which I don't see how to factor. That leaves the quadratic formula!
The zeros are .
We all know how to FOIL (or, at least, you should…)—however, that only works when multiplying two binomials. What if you need to raise a binomial to some power (other than 2)? You could FOIL in pairs, but then you've got to multiply the resulting trinomials, and then multiply the resulting 4-term polynomials, etc.
There's got to be a better way!
…and there is. It's called the Binomial Theorem.
It looks daunting…but it's just a formula…and we'll see how the calculator can be used to do most of the dirty work for you!
The parts that have two numbers on top of each other are calculated like this: . The exclamation mark is called the factorial, and is calculated like this: . If you've got a calculator, then this will be a lot easier than it looks.
If you don't have a calculator, then you may want to use Pascal's Triangle. It is named for Blaise Pascal (1623-1662), although he is not the first person to have written out numbers in this form.
(There is a famous proposition that says no important discovery will ever be named for the person who actually discovered it…and this saying is not named after the first person who proposed it!)
Pascal's Triangle looks like this:
(the pattern continues infinitely)
Each row's entries are found by adding the entries above—look, you'll see it!
This is how you can calculate without a calculator—go to row n, then count over k entries! Note that the rows begin with zero, and the entries in a row start with zero (math people always start counting at zero). For example, is found on row 2 (starting at the top, go down 4 rows) and entry 2 (starting on the left, go right 2 entries).
[17.] Expand: .
From the formula, , , and . , which expands to:
I can get those coefficients from Pascal's Triangle above! The answer is:
[18.] Expand: .
[19.] Find the term of .
This time, we don't need to do the whole thing—just the one term we need! Note that , , , and (since k is the power of a in the formula). Thus, the answer is .
Page last validated 2010-08-15