Rational Root (Zero) Theorem

Rational Root (Zero) Theorem - How it Works - Video

Rational Root (Zero) Theorem

Rational Root (Zero) Theorem:

Here we have the rational root (zero) theorem. We can use this theorem to find possible rational roots of any polynomial. We have to use the leading coefficient, a_n, and the constant, a_o.

Once we find our leading coefficient and our constant, we find the factors of each number. Then we divided the factors of the constant by the factors of the leading coefficient to find the possible rational roots. Don't forget the plus or minus of each one.

Example 1

Example 1:

Here is our first example. We have the function, f(x) = x² + 8x + 5. This parabolic equation cannot be factored. So we already, we don't have any rational roots. One reason is that factors of 5 are 1 and 5. Using the foil method we can the find the last term by multiplying factors of 5. There are two, 1 * 5 and -1 * -5. To find the coefficient of the middle term, we add any combination of ±1 or ±5. Remember we are looking for +8. We cannot reach + 8 with those numbers. That helps us with parabolic equations but what about polynomials with degrees higher than 2.

We start off the same way by using the factors of the leading coefficient and the constant. The factors of the constant, 5, are 1 and 5 and the factor of leading coefficient, 1, is 1. Each number needs to have ± in front the number since we don't know if it is positive or negative.

Now we divide the factors of the constant, 5, by the factor of the leading coefficient, 1. So we have ±1 / ±1 and ±5 / ±1, and that gives us 1, -1, 5, and -5. Those 4 numbers are the numbers that we have to test to figure out if any of the numbers are a root. Let's do test each one.

Now we have the output value of each one: 14 for f(1), -2 for f(-1), 60 for f(5), and -10 for f(-5). Since none of the values are 0, we don't have any rational roots for this polynomial function.

Example 2

Example 2:

Here we have the function, f(x) = 4x⁴ - x³ + 13x² - 4x - 12. This is one a more steps since we don't have a parabolic equation.

There is more than one way to attack the problem. We could brute force this, but that might take a while. You might not get friendly numbers like 4 and 12 in your example. Let's narrow down the possible numbers so we don't have to guess so many times.

Here we used Descartes' rule to narrow the possible positive, negative, and imaginary real zeros that we could have. We found that we only have two possibilities either 3 positive real zeros and 1 negative real zero or 1 positive real zero, 1 negative real zero, and 2 imaginary real zeros.

Here we found the upper bound to narrow down our search even more. We found a few upper bounds until we continued until we found the least upper bound. Luckily for us, the remainder is 0, so we found a root (zero). So 1 is the upper bound and is a root for our function.

How do we know which numbers to guess? Why don't we start the biggest number 12 first? Well, we use a number in the middle so we can eliminate numbers that are higher. In this case, we chose the number, 3. Since all the numbers are positive, we know that any number higher than 3 is going to give us the same result.

Here we found the lower bound to narrow down our search even more. We found a few lower bounds until we continued until we found the greatest lower bound. This time, the remainder is not 0, so we don't have a root, but we have narrow down our search to just a few numbers.

How do we know which numbers to guess? Why don't we start the biggest number -12 first? Well, we use a number in the middle so we can eliminate numbers that are lower. In this case, we chose the number, -3. Since all the numbers alternate, we know that any number lower than -3 is going to give us the same result.

Now we have our interval to guess, [-1, 1].

We have already found one factor, (x - 1). So we can we write our function as f(x) = (4x³ + 3x² + 16x + 12) * (x -1).

Depending on the question, you don't need to break down the function, but it is helpful to work with smaller numbers and you might be able to skip a few steps at the end if you can figure out if the smaller function is reducible or not.

Let's take a look at our interval or real zeros again. We know that the rational zeros are going to be -1 or 1 or in between the two and that our rational zero is going to a fraction since 0 does not give us a rational zero.

Now let's bring back the possible fractions that could be our rational zero. We have ±1/4, ±1/2, ±3/4, and ±3/2. We can eliminate the positive numbers since we already a zero close to 1, in this case it is 1. Most likely our next zero is going to be closer to -1. Now we have -1/4, -1/2, -3/4, and -3/2. We can eliminate -3/2 because -3/2 = -1.5 which is outside our interval. That is why we chose -3/4 as our next factor.

Notice that the top row in our synthetic division has 4 numbers. That is because we want to make our function smaller so we can see the different factors. WAIT A MINUTE. It should have 4 then 3 then 15 then 12. Why do we have 1 then 3/4 then 4 then 3? Our factor is (x + 3/4). It original started out as (4x + 3), so we multiply the factor by 1/4. Since we are breaking it down into a multiplication of factors, when multiply (4x + 3) our divisor, we need to multiply our dividend by 1/4 as well.

So we have broken down our function even more, and now we have f(x) = (x² + 4) * (4x + 3) * (x -1).

Let's focus on x² + 4. This is not reducible. If we solve for it, we get the complex numbers, ±2i. So our rational roots are 1 and -3/4.

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