A polynomial is the sum of one or more monomials with real coefficients and nonnegative integer exponents, and these functions all contain polynomials of the form:
The degree of the polynomial function is the highest exponent of any variable in the function. Polynomial functions of only one term are called monomials or power functions. Power functinos have the form f(x)=ax^n. One way to tell the degree of a funtion is to count the number of times its graph changes direction. Be careful with this though, because there are other indicators of roots that need to be counted.
The Fundamental Theorum of Algebra: The fundamental theorem of algebra states that every non-constant single-variable polynomial with complex coefficients has at least one complex root. The roots of a function are where is crosses the x-axis. Different degree polynomial functions all have different numbers of roots, and roots look different depending on the degree of the function, and whether they are double roots, triple roots, etc. and whether they are imaginary or not. The degree of the polynomial tells you how many zeroes there are.
End Behavior: If the degree of the function is even, the two ends of the graph will be pointing the same direction; if odd, then in opposite directions. See the section entitled Functions for more information.
Roots: When a polynomial function is completely factored, it can help to identify the roots. Consider the function f(x)=x^2-x+2. It factors to (x-2)(x+1). Because the roots are the zeroes of the function, or where the y-values are zeroes, set each set of parentheses equal to zero. the roots of this function are 2 and -1. For more complicated equations that may not be factorable, there are different methods of finding the roots. The graph of the polynomial function changes direction depending on its roots, and if the graph changes directions without crossing the x-axis, that means those roots are imaginary.
Problem Examples: Find the roots of the following functions.
a. f(x)=X^2-5x+6 Factors to (x-2)(x-3)
Roots are 2 and 3
b. f(x)=x^3+3x^2-4x Factors to x(x+4)(x-1)
Roots are 0, -4, and 1
Problem Examples: Find the roots of the following functions.
a. f(x)=X^2-5x+6 Factors to (x-2)(x-3)
Roots are 2 and 3
b. f(x)=x^3+3x^2-4x Factors to x(x+4)(x-1)
Roots are 0, -4, and 1
Factor Theorem: a polynomial function p(x) has a factor x-c if and only if p(c)=0
Remainder Theorem: if a polynomial p(x) is divided by x-c, then the remainder is r=p(c)
Remainder Theorem: if a polynomial p(x) is divided by x-c, then the remainder is r=p(c)
Polynomial Long Division: This can be used for polynomials of any degree.
Synthetic Division: This can only be used when dividing by a first degree polynomial such as (x+2) or (x-4) with no coefficient on the x.
Roots Again: Some polynomial functions will not be able to factor, have imaginary or irrational roots, etc., and there are different approaches you can take to find their roots. If you are given a non-factorable polynomial function with no roots given, there is a way to tell what some of its possible rational roots are. You take each of the factors of the last term, and put them over each of the factors of the first to get your possible roots. You can then use synthetic division or long division, if you wish, to see which of the numbers does not leave you with a remainder. If the possible root you divide the function by does not have a remainder, that means that it is indeed one of the roots of that function. Similarly, you can just plug the possible roots, or zeroes, in for f(x). If the solution you get is 0, the possible root you used works for that function. After you find one possible root, you divide it out of the original function, and are left with two sets of parentheses, one containing your "root term", or the expression you divided out of your original function that shows your first root, like x-1, or x+4. Then you have the polynomial that was left after you divided out your root. This new polynomial may be factorable, and if so you should factor it and find the rest of your roots that way. If not, you can try your division again using your possible roots. It may be that the polynomial has no more rational roots, and has irrational or imaginary. For example, think of having the polynomial x^2+4 left over. If you set that equal to 0, you get a positive or negative sqrt-4. Because you cannot take the square root of a negative, this function has imaginary roots. Remember that imaginary roots ALWAYS come in pairs; if you have a positive imaginary root, it comes with a negative, and vice versa. It is the same for all complex roots; they come in pairs, and the pairs are always conjugates. So eventually you will end up with a function that consists of your parentheses, all showing your roots. Pay attention to the leading terms in your parentheses, they should all multiply out to be the same leading term as your original function. If there is a 3x^5 for your parentheses, but the original function was 6x^5, there needs to be a 2 added to the beginning of all your parentheses. Looking at the graph of your function, it should cross the x-axis at all your real roots. Again, if the function has imaginary roots, these are shown through a change of direction away from the x-axis. The videos below show some examples of this process. If you are given the zeroes of a function, this can also allow for you to find that particular function; its like the same thing but reversed. There are different tricks for imaginary and complex roots, but it all amounts to getting an expression for each set of roots and multiplying them out.
Problem: Find The zeroes of the polynomial using any method.
a. f(x)= x^5 -5x^4 -5x^3 +25x^2 -6x +30
You could go about this problem in a number of ways, but first we will look at the possible rational roots. These are the factors of 30 over the factors of 1 (not many of those), and can each be positive or negative. We have 1,2,3,5,6,10,15, and 30. You could then either use polynomial long division or synthetic division, or plugging one into the equation to find the first root. Another way would be to graph the function and try to see where the graph crosses the x-axis, and plug in the numbers you think are closest. Lets try plugging in 2. When we try this, 2^5 -5(2)^4 -5(2)^3 +25(2)^2 -6(2) +30 does not give a solution of zero, so 2 is not a root. However, when we plug in 5, we do get zero, so 5 is a root of this function. Again, we could proceed in a number of ways. You could divide the function by (x-5), but we will use synthetic division.
5| 1 -5 -5 25 -6 30
5 0 -25 0 -30
1 0 -5 0 -6 0
Now we have (x-5), and the new equation (x^4-5x^2-6) from the division. 5 is our rational root, and now we need to find the rest. Again, there are different ways to go about this, but we will try just factoring the equation. When you look at what multiplies to -6 and adds to -5, you get a -6 and a 1, but remember that the degree of this equation is 4 with an x^2 term in the middle. Now we have (x^2-6) and (x^2+1). We have factored our original function to (x-5)(x^2-6)(x^2+1). To find the rest of our roots, we can set each set of parentheses equal to 0.
(x^2-6)=0
x^2=6
x= positive or negative sqrt6, and this can also be shown as (x+sqrt6)(x-sqrt6)
(x^2+1)=0
x^2=-1
x= positive or negative sqrt-1 -> uh oh! You can't square root a negative number, so these roots are imaginary. The square root of -1 can also be denoted i, so now we have (x=+i)(x-i)
We have found all our roots. We have 5 roots in all because the degree of the function is 5, and our complex roots come in pairs, so that is one way you can check to see if you're doing it correctly. Our zeroes are
{ 5, sqrt6, -sqrt6, i, -i }
a. f(x)= x^5 -5x^4 -5x^3 +25x^2 -6x +30
You could go about this problem in a number of ways, but first we will look at the possible rational roots. These are the factors of 30 over the factors of 1 (not many of those), and can each be positive or negative. We have 1,2,3,5,6,10,15, and 30. You could then either use polynomial long division or synthetic division, or plugging one into the equation to find the first root. Another way would be to graph the function and try to see where the graph crosses the x-axis, and plug in the numbers you think are closest. Lets try plugging in 2. When we try this, 2^5 -5(2)^4 -5(2)^3 +25(2)^2 -6(2) +30 does not give a solution of zero, so 2 is not a root. However, when we plug in 5, we do get zero, so 5 is a root of this function. Again, we could proceed in a number of ways. You could divide the function by (x-5), but we will use synthetic division.
5| 1 -5 -5 25 -6 30
5 0 -25 0 -30
1 0 -5 0 -6 0
Now we have (x-5), and the new equation (x^4-5x^2-6) from the division. 5 is our rational root, and now we need to find the rest. Again, there are different ways to go about this, but we will try just factoring the equation. When you look at what multiplies to -6 and adds to -5, you get a -6 and a 1, but remember that the degree of this equation is 4 with an x^2 term in the middle. Now we have (x^2-6) and (x^2+1). We have factored our original function to (x-5)(x^2-6)(x^2+1). To find the rest of our roots, we can set each set of parentheses equal to 0.
(x^2-6)=0
x^2=6
x= positive or negative sqrt6, and this can also be shown as (x+sqrt6)(x-sqrt6)
(x^2+1)=0
x^2=-1
x= positive or negative sqrt-1 -> uh oh! You can't square root a negative number, so these roots are imaginary. The square root of -1 can also be denoted i, so now we have (x=+i)(x-i)
We have found all our roots. We have 5 roots in all because the degree of the function is 5, and our complex roots come in pairs, so that is one way you can check to see if you're doing it correctly. Our zeroes are
{ 5, sqrt6, -sqrt6, i, -i }