You were taught long division of polynomials in Intermediate Algebra. Basically, the procedure is carried out like long division of real numbers. The procedure is explained in the textbook if you're not familiar with it.
One key point about division, and this works for real numbers as well as for polynomial division, needs to be pointed out. When you divide the dividend by the divisor, you get a quotient and a remainder. To check the problem, you multiply the divisor by the quotient and add the remainder to get the dividend. If the remainder is 0, then we say that the divisor divides evenly into the dividend.
Dividend / Divisor = Quotient + Remainder / Divisor
Dividend = Divisor * Quotient + Remainder
Like I said, the same thing can be done with polynomial functions.
f(x) = d(x) * q(x) + r(x)
Where f(x) is the polynomial function being divided into (dividend), d(x) is the polynomial function being divided by (divisor), q(x) is the polynomial function that is the quotient, and r(x) is the polynomial remainder function and will have degree less than the divisor.
If the remainder, r(x), is zero, then f(x) = d(x)*q(x). We have just factored the function f(x) into two factors, d(x) and q(x).
When a polynomial function f is divided by x-k, the remainder r is f(k).
Okay, now in English. If you divide a polynomial by a linear factor, x-k, the remainder is the value you would get if you plugged x=k into the function and evaluated.
Now, tie that into what we just said above. If the remainder is zero, then you have successfully factored the polynomial. If the remainder when dividing by (x-k) is zero, then the function evaluated at x=k is zero and you have found a zero or root of the polynomial. Plus, you now have a factored polynomial (the quotient) which is one less degree than the original polynomial. If the quotient is down to a quadratic or linear factor, then you can solve and find the other solutions.
To divide a polynomial synthetically by x-k, perform the following steps.
Once you have things set up, you can actually start to perform the synthetic division.
You can only use synthetic division as described above to divide by x-k. That is, it must be a linear factor, and the leading coefficient must be a one.
There are similar ways to divide by a quadratic, cubic, etc, but for some reason, they aren't taught anymore (no, they won't die with me, I'm sure someone else knows them, too, but thanks for your concern).
Every polynomial in one variable of degree n, n > 0, has at least one real or complex zero.
Every polynomial in one variable of degree n, n > 0, has exactly n real or complex zeros.
Complex solutions come in pairs. If (a+bi) is a solution, then its complex conjugate (a-bi) is also a solution.
Solutions involving square roots also come in pairs. If a+sqrt(b) is a solution, then its conjugate, a-sqrt(b) will also be a solution. The same is not necessarily true of other roots.
This is not in your text!
Consider: f(x) = 3x^6 + x^5 - x^4 + 3x^3 + 2x^2 - x + 1.
The signs in f(x) are + + - + + - +. There are 4 sign changes (+ to -) or (- to +).
Now, f(-x) = 3x^6 - x^5 - x^4 - 3x^3 + 2x^2 + x + 1.
The signs in f(-x) are + - - - + + +. There are 2 sign changes (+ to -) or (- to +).
Here are the Possible Combinations of Roots
Total | Positive | Negative | Complex |
---|---|---|---|
6 | 4 | 2 | 0 |
6 | 2 | 2 | 2 |
6 | 0 | 2 | 4 |
6 | 4 | 0 | 2 |
6 | 2 | 0 | 4 |
6 | 0 | 0 | 6 |
Notice that the positive and negative values can decrease by two independently of each other.
If a polynomial function has integer coefficients, then every rational zero will have the form p/q where p is a factor of the constant and q is a factor of the leading coefficient.
Example: f(x) = 4x^5 - x^2 + 12
Possible rational zeros will be of the form (factor of 12) over (factor of 4). A division table can help you find all these values
1 | 2 | 3 | 4 | 6 | 12 | |
---|---|---|---|---|---|---|
1 | 1 | 2 | 3 | 4 | 6 | 12 |
2 | 1 / 2 | 1 | 3 / 2 | 2 | 3 | 6 |
4 | 1 / 4 | 1 / 2 | 3 / 4 | 1 | 3 / 2 | 3 |
The division table helps identify the possible rational zeros. You should throw out the duplicates, and list the others in order.
-12, -6, -4, -3, -2, -3 / 2, -1, -3 / 4, -1 / 2, -1 / 4, 1 / 4, 3 / 4, 1, 3 / 2, 2, 3, 4, 6, 12
Now, you perform synthetic division on possible rational zeros until you find one.
Here's where Descartes' Rule of Signs comes in. In this particular problem, there would be a maximum of 2 positive and 1 negative root. That means that you may have 2 or 0 positive roots, but you will always have 1 negative. There is no guarantee that negative is rational, though. Descartes only guaranteed real roots. If there were no negatives, then you would know not to try any.
If you have a polynomial with real coefficients and a positive leading coefficient, then ...
The zero in the bottom row may be considered positive or negative as needed.
Once you have found a zero using synthetic division, use the quotient as a new polynomial for all further divisions. The quotient will be one less degree than the original dividend. Each time you find a root, the quotient becomes one less in degree. Eventually, it will become a quadratic, and then you can factor, extract roots, complete the square, or use the quadratic equation to find the remaining roots.
If you continue to use the original function, you will become very frustrated and waste a lot of time.