Finding Actual Roots Using The Rational Root Theorem

Dec 10, 2025 by Andrew McMorgan 53 views

Hey guys! Today, we're diving deep into the world of polynomials and how to nail down their actual roots. We'll be using a super handy tool called the Rational Root Theorem. So, get ready to flex those math muscles because we're tackling a problem from Plastik Magazine that's all about finding the real heroes – the actual roots of the function $f(x) = 2x^2 + 2x - 24$ . We've got a list of potential suspects: -4, -3, 2, 3, and 4. The big question is, which of these potential roots are the actual roots? Let's break it down step-by-step, making sure we understand the why and how behind it all. No confusing jargon here, just straight-up math guidance to help you ace this!

Understanding the Rational Root Theorem

The Rational Root Theorem is your go-to guide when you're trying to find rational roots of a polynomial equation with integer coefficients. A rational root is basically a number that can be expressed as a fraction p/q, where p and q are integers and q is not zero. This theorem gives you a list of all possible rational roots. It doesn't tell you which ones are the actual roots, but it narrows down the field considerably, saving you tons of guesswork.

For a polynomial like $f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0$ , where all the coefficients ( $a_n, a_{n-1}, ..., a_0$ ) are integers, the Rational Root Theorem states that any rational root, when expressed in its simplest form p/q, must have 'p' as a factor of the constant term ( $a_0$ ) and 'q' as a factor of the leading coefficient ( $a_n$ ).

In simpler terms, you take all the numbers that divide evenly into the constant term (those are your 'p' values) and all the numbers that divide evenly into the leading coefficient (those are your 'q' values). Then, you form all possible fractions p/q. These fractions are your potential rational roots. It's like having a suspect list for a crime – you know the culprit is on the list, but you still need to do some investigating to find out who it really is!

For our specific problem, the polynomial is $f(x) = 2x^2 + 2x - 24$ . Here, the constant term ( $a_0$ ) is -24, and the leading coefficient ( $a_n$ ) is 2.

Factors of the constant term (-24): These are the possible values for 'p'. They include ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24.
Factors of the leading coefficient (2): These are the possible values for 'q'. They include ±1, ±2.

Now, we create all possible combinations of p/q:

If q = 1: ±1/1, ±2/1, ±3/1, ±4/1, ±6/1, ±8/1, ±12/1, ±24/1 (which simplifies to ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24)
If q = 2: ±1/2, ±2/2, ±3/2, ±4/2, ±6/2, ±8/2, ±12/2, ±24/2 (which simplifies to ±1/2, ±1, ±3/2, ±2, ±3, ±4, ±6, ±12)

Combining and removing duplicates, the list of all potential rational roots is: ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24, ±1/2, ±3/2.

Now, the problem already gave us a simplified list of potential roots: -4, -3, 2, 3, 4. Notice that all these numbers are indeed within our generated list of potential rational roots. The Rational Root Theorem has done its job by giving us a set of numbers to test. The next step is crucial: we need to plug these potential roots back into the original function to see which ones actually make $f(x) = 0$ . This is where we separate the suspects from the culprits!

Testing the Potential Roots

Alright guys, we've got our list of potential roots from the problem: -4, -3, 2, 3, and 4. The Rational Root Theorem is awesome for giving us these possibilities, but it's up to us to verify which ones are actually roots. A root of a function is a value of x that makes the function equal to zero. So, to check if a number is an actual root, we simply substitute it into the function $f(x)$ and see if the result is 0. Let's go through each potential root one by one.

Our function is $f(x) = 2x^2 + 2x - 24$ . We want to find the values of x for which $f(x) = 0$ .

Testing x = -4: $f(-4) = 2(-4)^2 + 2(-4) - 24$ $f(-4) = 2(16) - 8 - 24$ $f(-4) = 32 - 8 - 24$ $f(-4) = 32 - 32$ $f(-4) = 0$ Success! Since $f(-4) = 0$ , -4 is an actual root of the function.
Testing x = -3: $f(-3) = 2(-3)^2 + 2(-3) - 24$ $f(-3) = 2(9) - 6 - 24$ $f(-3) = 18 - 6 - 24$ $f(-3) = 18 - 30$ $f(-3) = -12$ Nope! Since $f(-3) eq 0$ , -3 is NOT an actual root.
Testing x = 2: $f(2) = 2(2)^2 + 2(2) - 24$ $f(2) = 2(4) + 4 - 24$ $f(2) = 8 + 4 - 24$ $f(2) = 12 - 24$ $f(2) = -12$ Nope! Since $f(2) eq 0$ , 2 is NOT an actual root.
Testing x = 3: $f(3) = 2(3)^2 + 2(3) - 24$ $f(3) = 2(9) + 6 - 24$ $f(3) = 18 + 6 - 24$ $f(3) = 24 - 24$ $f(3) = 0$ Success! Since $f(3) = 0$ , 3 is an actual root of the function.
Testing x = 4: $f(4) = 2(4)^2 + 2(4) - 24$ $f(4) = 2(16) + 8 - 24$ $f(4) = 32 + 8 - 24$ $f(4) = 40 - 24$ $f(4) = 16$ Nope! Since $f(4) eq 0$ , 4 is NOT an actual root.

So, after plugging in all the potential roots, we found that only -4 and 3 resulted in $f(x) = 0$ . These are our actual roots!

Alternative Method: Factoring the Polynomial

For quadratic functions like $f(x) = 2x^2 + 2x - 24$ , there's often a quicker way to find the roots: factoring! While the Rational Root Theorem is a powerful general tool, factoring can be more straightforward for simpler polynomials. Let's see if we can factor our function and verify our results.

First, notice that all the coefficients (2, 2, and -24) are divisible by 2. We can factor out a 2 to simplify the expression:

$f(x) = 2(x^2 + x - 12)$

Now, we need to factor the quadratic expression inside the parentheses: $x^2 + x - 12$ . We're looking for two numbers that multiply to -12 and add up to +1 (the coefficient of the x term).

Let's list pairs of factors for -12:

1 and -12 (sum = -11)
-1 and 12 (sum = 11)
2 and -6 (sum = -4)
-2 and 6 (sum = 4)
3 and -4 (sum = -1)
-3 and 4 (sum = 1)

Bingo! The pair -3 and 4 adds up to +1. So, we can factor $x^2 + x - 12$ as $(x - 3)(x + 4)$ .

Putting it all back together, our factored function is:

$f(x) = 2(x - 3)(x + 4)$

To find the roots, we set $f(x) = 0$ :

$2(x - 3)(x + 4) = 0$

For this equation to be true, one of the factors must be zero. Since 2 is not zero, we look at the other two factors:

$x - 3 = 0

ightarrow x = 3$

$x + 4 = 0

ightarrow x = -4$

And there you have it! The actual roots are indeed 3 and -4. This factoring method confirms our findings from using the Rational Root Theorem and testing each potential root. It's always great when different mathematical approaches lead you to the same correct answer, right?

Conclusion: Identifying the Actual Roots

So, after meticulously applying the Rational Root Theorem to identify potential candidates and then diligently testing each one by substituting them back into the function $f(x) = 2x^2 + 2x - 24$ , we've successfully pinpointed the actual roots. We found that only -4 and 3 satisfy the condition $f(x) = 0$ . This means that when x equals -4 or 3, the function's output is zero. These are the points where the graph of the function would cross the x-axis.

Our investigation involved testing each of the provided potential roots: -4, -3, 2, 3, and 4. The values that resulted in $f(x)=0$ were -4 and 3. Therefore, the correct choice that lists the actual roots is the one that includes only these two numbers.

Looking at the options provided:

A. -4 and 3
B. -4, 2, and 3
C. -3 and 4
D. -3, 2, and 4

Option A, -4 and 3, perfectly matches our findings. We saw that -3, 2, and 4 did not make $f(x)$ equal to zero, so options B, C, and D are incorrect because they include numbers that are not actual roots or exclude numbers that are.

Remember, the Rational Root Theorem is a fantastic starting point, especially for polynomials with higher degrees where factoring might be a nightmare. It gives you a finite list of rational possibilities to check. However, always remember the final step: testing those possibilities by plugging them into the function. It's the crucial step that distinguishes potential roots from actual roots. And as we saw, sometimes factoring a simpler polynomial like a quadratic can be a quicker way to confirm your results. Keep practicing, and you'll be a root-finding pro in no time!