Solving Polynomials: Find The Zeros

Hey Plastik Magazine readers! Ever stumbled upon a tricky polynomial problem? Today, we're diving deep into the world of polynomials, specifically focusing on how to find their zeros. For all the math enthusiasts out there, this article is for you! We'll break down a specific problem step-by-step, making it super easy to understand. So, grab your notebooks, and let's get started. We'll be solving a cubic polynomial problem, so buckle up. This is going to be fun.

Understanding the Problem: The Remainder Theorem

Let's begin by understanding the problem. We're given a cubic polynomial: $P(x) = Ax^3 + 9x^2 + bx + 15$. We know a few crucial pieces of information. When this polynomial is divided by $(x - 1)$, the remainder is $-12$. Also, when divided by $(x - 2)$, the remainder is $-21$. Our ultimate goal? To find the zeros of the polynomial. Remember, the zeros of a polynomial are the values of x for which $P(x) = 0$. The Remainder Theorem is our best friend here. It states that if you divide a polynomial $P(x)$ by $(x - c)$, the remainder is $P(c)$. It's a lifesaver, really. Using the Remainder Theorem, we can set up two equations based on the given information. When divided by $(x - 1)$, the remainder is $-12$, which means $P(1) = -12$. Similarly, when divided by $(x - 2)$, the remainder is $-21$, so $P(2) = -21$. These two equations will help us find the values of A and b, which are the unknown coefficients in our polynomial. Alright, guys, let’s get into the specifics. Let's apply the Remainder Theorem, and find those equations. It is going to be easy.

Setting Up the Equations

Now, let's turn the information we have into concrete equations. First, let's use the fact that $P(1) = -12$. Substitute $x = 1$ into the polynomial: $P(1) = A(1)^3 + 9(1)^2 + b(1) + 15 = -12$. This simplifies to $A + 9 + b + 15 = -12$. Combining the constants, we get our first equation: $A + b = -36$. Next, let's use the fact that $P(2) = -21$. Substitute $x = 2$ into the polynomial: $P(2) = A(2)^3 + 9(2)^2 + b(2) + 15 = -21$. This simplifies to $8A + 36 + 2b + 15 = -21$. Combining the constants, we get our second equation: $8A + 2b = -72$. We now have a system of two linear equations with two variables, A and b. These are the keys to unlocking the rest of the problem. We're getting closer to finding the zeros! The setup is complete, and we are ready to solve for the unknown coefficients, A and b. Do not worry; it is going to be easy to solve the system of equations.

Solving for the Coefficients A and B

Alright, let’s solve for A and b. We have two equations:

1. $A + b = -36$
2. $8A + 2b = -72$

There are several ways to solve a system of linear equations. We can use the substitution or elimination method. Let's use the substitution method here, because it's usually the easiest for this type of problem. From the first equation, we can express b in terms of A: $b = -36 - A$. Now, substitute this expression for b into the second equation: $8A + 2(-36 - A) = -72$. This simplifies to $8A - 72 - 2A = -72$. Combine the terms with A: $6A - 72 = -72$. Adding 72 to both sides gives us $6A = 0$, which means $A = 0$. Now that we have the value of A, we can find b. Substitute $A = 0$ into the equation $b = -36 - A$: $b = -36 - 0$, so $b = -36$. Bingo! We’ve found the values of A and b. With $A = 0$ and $b = -36$, our polynomial becomes $P(x) = 0x^3 + 9x^2 - 36x + 15$, or simply $P(x) = 9x^2 - 36x + 15$. Notice something interesting? The x^3 term disappeared, and we now have a quadratic polynomial, which makes the next step easier. Now the polynomial is much more manageable.

Simplifying the Polynomial

Before we jump into finding the zeros, let’s simplify our polynomial a bit. We have $P(x) = 9x^2 - 36x + 15$. We can factor out a common factor of 3 from all the terms: $P(x) = 3(3x^2 - 12x + 5)$. Now, we're dealing with a simplified version that makes it easier to find the zeros. This step is not strictly necessary, but it makes the next steps a bit cleaner. It also helps to prevent calculation errors. We always want to be simplifying when possible. The simpler, the better! With our simplified polynomial, we are ready to find the zeros of the polynomial.

Finding the Zeros of the Polynomial

Now comes the fun part: finding the zeros! Since we're dealing with a quadratic polynomial $3x^2 - 12x + 5$, we can use the quadratic formula to find its zeros. Remember, the quadratic formula is a lifesaver for these kinds of problems! The quadratic formula is given by: x = rac{-b ext{ ± } ext{√}(b^2 - 4ac)}{2a}. In our polynomial $3x^2 - 12x + 5$, we have $a = 3$, $b = -12$, and $c = 5$. Substitute these values into the quadratic formula: x = rac{-(-12) ext{ ± } ext{√}((-12)^2 - 4 * 3 * 5)}{2 * 3}. This simplifies to x = rac{12 ext{ ± } ext{√}(144 - 60)}{6}, which further simplifies to x = rac{12 ext{ ± } ext{√}84}{6}. We can simplify the square root of 84: $ ext√}84 = ext{√}(4 * 21) = 2 ext{√}21$. So, our zeros are x = rac{12 ext{ ± } 2 ext{√}21}{6}. Finally, simplify the expression by dividing both terms in the numerator by 2 and then by 6 $x = 2 ext{ ± rac{ ext{√}21}{3}$. Therefore, the zeros of the polynomial are x = 2 + rac{ ext{√}21}{3} and x = 2 - rac{ ext{√}21}{3}. Congratulations, guys! We've found the zeros of the polynomial! We used the Remainder Theorem, solved equations, simplified, and then used the quadratic formula. We did it all! It's a journey, right? Let's take a look at the summary of the solution.

Summary of the Solution

Here’s a quick recap of what we did:

1. Understanding the Remainder Theorem: We used the remainder theorem to set up equations.
2. Setting up and Solving Equations: We formed two equations using the given information and solved for A and b.
3. Simplifying the Polynomial: We substituted the values of A and b back into the original polynomial and simplified.
4. Finding the Zeros: We used the quadratic formula to find the zeros of the simplified polynomial.

This methodical approach helps to break down a complex problem into manageable steps. Remember, guys, practice makes perfect! The more you work through these types of problems, the easier they become. Keep practicing, and you'll become a polynomial pro in no time! So, keep learning, keep exploring, and keep having fun with math! If you have any questions or want to see more examples, leave a comment below. Until next time, Plastik Magazine readers! Keep those math skills sharp!

Disclaimer: This article is for informational purposes only and does not provide financial, legal, or medical advice. Always consult with a qualified professional for any specific needs. The author and publisher are not responsible for any errors or omissions, or for the results obtained from the use of this information.