Back To CourseAlgebra II: High School
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Amy has a master's degree in secondary education and has taught math at a public charter high school.
The rational roots theorem is a very useful theorem. It tells you that given a polynomial function with integer or whole number coefficients, a list of possible solutions can be found by listing the factors of the constant, or last term, over the factors of the coefficient of the leading term. Okay, that's a mouthful. Let me show you what this all means.
If you have a polynomial function such as 2x^2 + 9x + 4, then our possible solutions are -1/1, -2/1, -4/1, -1/2, -2/2, -4/2, 1/1, 2/1, 4/1, 1/2, 2/2 and 4/2. This list actually has duplicates, such as 1/1 and 2/2. Getting rid of these duplicates and simplifying, we get this list of possible solutions: -4, -2, -1, -1/2, 1/2, 1, 2 and 4.
Note that in our list, we took each and every factor of our constant term, our 4, and put it over the factors of our leading coefficient, our 2. All of our numerators are factors of the constant term, and all our denominators are factors of the leading coefficient.
We call this the rational root theorem because all these possible solutions are rational numbers. Having this list is useful because it tells us that our solutions may be in this list. In fact, we can actually check to see that our solutions are part of this list. If we factor our polynomial, we get (2x + 1)(x + 4). Our solutions are thus x = -1/2 and x = -4. As you can see, these solutions are included in our list.
So, if we didn't know where to start to find solutions to our polynomial, we can begin with the rational roots theorem to help us find the solutions. After making our list, we then can plug numbers in one at a time into our polynomial and see which one will give us an answer of 0 and, thus, be a solution.
Notice that I've been using the word 'possible' solutions. This is because the list of numbers that we get from using the rational root theorem is just that. It is a list of possibilities. As you saw from our very first example, most of the numbers will not turn out to be solutions at all. And it is also possible to get a list where none of the numbers are solutions.
When you are using this theorem to get your list of possible solutions, just remember that. You are getting a list of possibilities; you are not getting a list of solutions.
Let's look at this theorem in more detail now with another example. Let's try and find the possible solutions to the polynomial function x^3 - 6x^2 + 11x - 6. First, we locate our constant term, which is our last term. It is -6.
Next, we locate our leading coefficient. It is 1, since there is no number in front of the x^3 term. Now, we need to look for the factors of these two numbers. The factors of -6 are -6, -3, -2, -1, 1, 2, 3 and 6. We have both positive and negative because we can do 1 * -6 as well as -1 * 6. These factors will become our numerators.
Next, what are our factors of the leading coefficient of 1? They are -1 and 1. Remember we can do either 1 * 1 or -1 * -1. These will become our denominators. Now we can go and systematically place our numerators over our denominators one by one. We get -6/1, -3/1, -2/1, -1/1, 1/1, 2/1, 3/1 and 6/1.
We can simplify these fractions to become -6, -3, -2, -1, 1, 2, 3 and 6. If you saw this coming, then you saw that we could have stuck with just the factors of -6 because our denominator is always 1, and anything over 1 is itself.
Now that we have our list, we can now go on and check to see if any of them is a solution. We do this by plugging in the numbers one by one until we find the solutions. Our polynomial is a degree 3 polynomial, so it will have three solutions.
Now, remember our list is a possible list, so it may or may not contain all of the solutions. To begin checking, I am going to pick an easy number. I am going to pick the number 1. Plugging this number into the polynomial, I get 1^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 = 0. It equals 0. So, 1 is a solution. Great, we've found a solution! What happens if we plug in -1? (-1)^3 - 6(-1)^2 + 11(-1) - 6 = -1 - 6 - 11 - 6 = -24. It doesn't equal 0, so -1 is not a solution.
We can keep going through our list to see if there are more solutions. Or we can divide our polynomial by (x - k), where k is a solution to get to a smaller polynomial that hopefully we can factor. However, we will leave that process for another video lesson.
For now, practice with finding the list of possibilities and then checking to see if any of them are solutions. If we keep checking different numbers in this example, we will find that the other two solutions are 2 and 3. So, in this case, all of our answers are rational and therefore are contained in this possible list of solutions.
Let's review what we've learned. We've learned that the rational roots theorem tells us that given a polynomial function with integer or whole number coefficients, a list of possible solutions can be found by listing the factors of the constant, or last term, over the factors of the coefficient of the leading term.
For example, the polynomial 3x^2 + 7x + 2 has a constant term of 2 and a leading coefficient of 3. The factors of 2 are -2, -1, 1 and 2. The factors of 3 are -3, -1, 1 and 3. Placing the factors of the constant term over the factors of the leading coefficient, we get this list of possible solutions: -2/1, -1/1, 1/1, 2/1, -2/3, -1/3, 1/3 and 2/3. Simplifying, we get this list: -2, -1, 1, 2, -2/3, -1/3, 1/3 and 2/3.
To check to see if any of the possible solutions are real solutions, we plug them back into the polynomial to see if it will equal 0. If it equals 0, then it is a solution. If it equals anything other than 0, then it is not a solution.
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Back To CourseAlgebra II: High School
23 chapters | 203 lessons