Find Rational Roots Of 4x^3 - 13x^2 + 9x + 2

Hey math enthusiasts! Today, we're diving deep into the fascinating world of polynomial equations, specifically tackling the question of how many rational roots our friend $f(x)=4 x^3-13 x^2+9 x+2$ has. This isn't just about crunching numbers, guys; it's about understanding the fundamental properties of these functions and how we can uncover their secrets using a powerful tool called the Rational Root Theorem. So, grab your calculators, maybe a nice cup of coffee, and let's get ready to unravel this algebraic puzzle together. We're going to break down the process step-by-step, making sure everyone can follow along, no matter where you are on your math journey. This polynomial might look a bit intimidating at first glance with its cubic term and several coefficients, but trust me, by the end of this, you'll be feeling like a root-finding ninja! We'll explore what rational roots are, why they're important, and most importantly, how to systematically find them. So, let's get started on this exciting mathematical adventure!

Understanding Rational Roots and the Rational Root Theorem

Alright, let's kick things off by defining what we mean by rational roots. In simple terms, a rational root of a polynomial is a solution (or a root) to the equation $f(x) = 0$ that can be expressed as a fraction, $p/q$, where both $p$ and $q$ are integers, and $q$ is not zero. Think of numbers like 1/2, -3, 5/4, or even -7/1. These are all rational numbers. Numbers like $\sqrt{2}$ or $\pi$, on the other hand, are irrational and therefore cannot be rational roots. Our main goal today is to figure out how many of these $p/q$ types of numbers satisfy our specific equation, $f(x)=4 x^3-13 x^2+9 x+2=0$. To do this, we lean heavily on a theorem that's a real game-changer in polynomial analysis: the Rational Root Theorem. This theorem provides us with a list of potential rational roots. It states that if a polynomial has integer coefficients (which ours does, thankfully!), then any rational root $p/q$ must have $p$ as a factor of the constant term and $q$ as a factor of the leading coefficient. Our polynomial is $f(x)=4 x^3-13 x^2+9 x+2$. The constant term is $2$, and the leading coefficient (the coefficient of the highest power of $x$, which is $x^3$) is $4$. So, according to the Rational Root Theorem, $p$ must be a factor of $2$, and $q$ must be a factor of $4$. This theorem is super handy because it narrows down the infinite possibilities of numbers to a finite, manageable list. We don't have to guess; we have a strategy! We'll list out all the factors of the constant term and all the factors of the leading coefficient, and then form all possible fractions $p/q$. This list isn't a guarantee that every number on it is a root, but it is a guarantee that all rational roots must be on this list. Pretty neat, right? It's like getting a cheat sheet for the test!

Identifying Potential Rational Roots

Now that we've got the theory down, let's get practical and apply it to our polynomial, $f(x)=4 x^3-13 x^2+9 x+2$. Remember, the Rational Root Theorem tells us that any rational root $p/q$ must have $p$ as a factor of the constant term ($2$) and $q$ as a factor of the leading coefficient ($4$). Let's break this down. First, let's find all the integer factors of the constant term, which is $2$. These are the possible values for $p$. So, $p$ can be $\pm 1$ and $\pm 2$. Easy peasy. Next, we find all the integer factors of the leading coefficient, which is $4$. These are the possible values for $q$. So, $q$ can be $\pm 1$, $\pm 2$, and $\pm 4$. Now comes the fun part: we create all possible combinations of $p/q$. We need to be systematic here to make sure we don't miss anything. We'll take each value of $p$ and divide it by each value of $q$.

• When $p = \pm 1$:

  • $q = \pm 1 \implies \pm 1/1 = \pm 1$
  • $q = \pm 2 \implies \pm 1/2$
  • $q = \pm 4 \implies \pm 1/4$

• When $p = \pm 2$:

  • $q = \pm 1 \implies \pm 2/1 = \pm 2$
  • $q = \pm 2 \implies \pm 2/2 = \pm 1$ (We already have this one, so we don't need to list it again)
  • $q = \pm 4 \implies \pm 2/4 = \pm 1/2$ (Again, already listed)

So, our complete list of potential rational roots is: $\pm 1$, $\pm 2$, $\pm 1/2$, $\pm 1/4$. Let's write this out clearly: $\{-2, -1, -1/2, -1/4, 1/4, 1/2, 1, 2\}$. This list contains 8 possible rational roots. Keep in mind, this is just our candidate list. Not all of these numbers will necessarily be roots of the polynomial. Our next step is to test these candidates to see which ones actually make $f(x)$ equal to zero.

Testing the Potential Rational Roots

Alright, guys, we've got our list of potential rational roots: $\{-2, -1, -1/2, -1/4, 1/4, 1/2, 1, 2\}$. Now, we need to test each one to see if it's a true root. A number is a root if, when you plug it into the polynomial $f(x)=4 x^3-13 x^2+9 x+2$, the result is $0$. We can do this by direct substitution or by using synthetic division, which is often quicker for testing multiple values. Let's start testing!

Test $x = 1$: $f(1) = 4(1)^3 - 13(1)^2 + 9(1) + 2 = 4 - 13 + 9 + 2 = 2$. Since $f(1) \neq 0$, $x=1$ is not a root.

Test $x = -1$: $f(-1) = 4(-1)^3 - 13(-1)^2 + 9(-1) + 2 = 4(-1) - 13(1) - 9 + 2 = -4 - 13 - 9 + 2 = -24$. Since $f(-1) \neq 0$, $x=-1$ is not a root.

Test $x = 2$: $f(2) = 4(2)^3 - 13(2)^2 + 9(2) + 2 = 4(8) - 13(4) + 18 + 2 = 32 - 52 + 18 + 2 = 0$. Success! $x=2$ is a rational root.

Since $x=2$ is a root, we know that $(x-2)$ is a factor of $f(x)$. We can use synthetic division with $2$ to find the remaining quadratic factor:

2 | 4  -13   9   2
  |    8  -10  -2
  ----------------
    4   -5  -1   0 

The result means that $f(x) = (x-2)(4x^2 - 5x - 1)$. Now, we only need to check the remaining potential rational roots for the quadratic equation $4x^2 - 5x - 1 = 0$. Let's continue testing our list with this new polynomial.

Test $x = -2$: $f(-2) = 4(-2)^3 - 13(-2)^2 + 9(-2) + 2 = 4(-8) - 13(4) - 18 + 2 = -32 - 52 - 18 + 2 = -100$. Not a root.

Test $x = 1/2$: $f(1/2) = 4(1/2)^3 - 13(1/2)^2 + 9(1/2) + 2 = 4(1/8) - 13(1/4) + 9/2 + 2 = 1/2 - 13/4 + 18/4 + 8/4 = 2/4 - 13/4 + 18/4 + 8/4 = 15/4$. Not a root.

Test $x = -1/2$: $f(-1/2) = 4(-1/2)^3 - 13(-1/2)^2 + 9(-1/2) + 2 = 4(-1/8) - 13(1/4) - 9/2 + 2 = -1/2 - 13/4 - 18/4 + 8/4 = -2/4 - 13/4 - 18/4 + 8/4 = -25/4$. Not a root.

Test $x = 1/4$: $f(1/4) = 4(1/4)^3 - 13(1/4)^2 + 9(1/4) + 2 = 4(1/64) - 13(1/16) + 9/4 + 2 = 1/16 - 13/16 + 36/16 + 32/16 = 56/16 = 7/2$. Not a root.

Test $x = -1/4$: $f(-1/4) = 4(-1/4)^3 - 13(-1/4)^2 + 9(-1/4) + 2 = 4(-1/64) - 13(1/16) - 9/4 + 2 = -1/16 - 13/16 - 36/16 + 32/16 = -18/16 = -9/8$. Not a root.

So far, only $x=2$ has worked from our original list.

Finding the Remaining Roots

We've successfully identified one rational root, $x=2$. We factored our polynomial as $f(x) = (x-2)(4x^2 - 5x - 1)$. Now, we need to find the roots of the quadratic factor $4x^2 - 5x - 1 = 0$. We can use the quadratic formula for this: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In this equation, $a=4$, $b=-5$, and $c=-1$.

Let's plug in the values: $x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-1)}}{2(4)}$ $x = \frac{5 \pm \sqrt{25 + 16}}{8}$ $x = \frac{5 \pm \sqrt{41}}{8}$

So, the other two roots are $x = \frac{5 + \sqrt{41}}{8}$ and $x = \frac{5 - \sqrt{41}}{8}$.

Now, the crucial question is: are these remaining roots rational? A rational number is a number that can be expressed as a fraction $p/q$ where $p$ and $q$ are integers. The number $\sqrt{41}$ is an irrational number because $41$ is not a perfect square. Since $\sqrt{41}$ is irrational, any number involving it in the form $\frac{5 \pm \sqrt{41}}{8}$ will also be irrational. Therefore, these two roots are irrational.

This means that out of the potential rational roots we tested, only $x=2$ turned out to be a true rational root. The other two roots are irrational.

Conclusion: The Number of Rational Roots

We started this journey by asking a very specific question: How many rational roots does the polynomial $f(x)=4 x^3-13 x^2+9 x+2$ have? We employed the powerful Rational Root Theorem to generate a list of potential rational roots: $\{-2, -1, -1/2, -1/4, 1/4, 1/2, 1, 2\}$. Through careful testing, we discovered that only one of these potential candidates, $x=2$, actually satisfies the equation $f(x)=0$. We then factored the polynomial into $(x-2)(4x^2 - 5x - 1)$ and used the quadratic formula to find the roots of the remaining quadratic factor. These roots, $\frac{5 + \sqrt{41}}{8}$ and $\frac{5 - \sqrt{41}}{8}$, were determined to be irrational. Therefore, after all our hard work and calculations, we can definitively conclude that the polynomial $f(x)=4 x^3-13 x^2+9 x+2$ has exactly one rational root. It's pretty awesome how theorems like the Rational Root Theorem give us such a structured way to solve these kinds of problems. Keep practicing, guys, and you'll be solving even more complex polynomials in no time! The mathematical world is full of these elegant solutions waiting to be discovered.