Polynomial Zeros: Find Other Roots & Multiplicity

Hey guys, let's dive into the fascinating world of polynomials and figure out how to find all their hidden zeros and understand their multiplicities. We've got a specific problem to tackle today: Given that $x=1$ is a zero of $f(x)=x^3-4 x^2+5 x-2$, find the other zeros and their multiplicity. This is a super common type of problem in algebra, and once you get the hang of it, you'll be finding roots like a pro! So, buckle up, grab your virtual calculators, and let's get this mathematical adventure started.

Understanding Polynomials and Zeros

Before we jump into solving our specific problem, let's get on the same page about what we're dealing with. A polynomial is basically an expression consisting of variables and coefficients, that involves only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. Think of it like a mathematical recipe with different ingredients (variables and numbers) mixed in specific ways. The zeros of a polynomial, also known as roots, are the values of the variable (in our case, $x$) that make the polynomial equal to zero. Finding these zeros is like finding the secret keys that unlock the polynomial's behavior. When we say $x=1$ is a zero of $f(x)$, it means that if you plug $x=1$ into the function $f(x)$, the result will be zero. So, $f(1) = (1)^3 - 4(1)^2 + 5(1) - 2 = 1 - 4 + 5 - 2 = 0$. Pretty neat, right?

Now, let's talk about multiplicity. This is a crucial concept when we're talking about zeros. The multiplicity of a zero tells us how many times that particular zero appears as a root of the polynomial. Imagine a zero is like a guest at a party. If a zero has a multiplicity of 1, it means that guest showed up once. If it has a multiplicity of 2, that guest showed up twice, maybe bringing a friend or just really liking the party! In terms of the polynomial's graph, a zero with multiplicity 1 will typically cross the x-axis, while a zero with an even multiplicity (like 2 or 4) will touch the x-axis and bounce back. A zero with an odd multiplicity greater than 1 (like 3 or 5) will cross the x-axis but flatten out momentarily as it does. Understanding multiplicity helps us sketch the graph of a polynomial and understand its behavior more deeply. So, in our problem, we know $x=1$ is a zero, but we don't yet know its multiplicity or what the other zeros are. That's what we're here to find out!

Using the Factor Theorem

Alright guys, so we know that $x=1$ is a zero of our polynomial $f(x)=x^3-4 x^2+5 x-2$. This piece of information is super valuable because it allows us to use a powerful tool called the Factor Theorem. The Factor Theorem is a direct consequence of the Remainder Theorem and essentially states that if $x=c$ is a zero of a polynomial $f(x)$, then $(x-c)$ is a factor of $f(x)$. In simpler terms, if a number makes the polynomial zero, then the expression $(x$ minus that number$)$ will divide evenly into the polynomial. Since we know $x=1$ is a zero, we know that $(x-1)$ must be a factor of $f(x)=x^3-4 x^2+5 x-2$. This is awesome because it means we can divide our polynomial by $(x-1)$ to find out what the remaining factors are.

There are a couple of ways to perform this division: polynomial long division or synthetic division. Synthetic division is often quicker and less prone to errors for linear divisors like $(x-1)$, so let's go with that. To set up synthetic division, we write down the coefficients of our polynomial $f(x)$, which are 1, -4, 5, and -2. We place the known zero, which is 1, to the left. So, we'll have:

1 | 1  -4   5  -2
  |________________
  | 

Now, we bring down the first coefficient (1) below the line. Then, we multiply this number by our zero (1) and write the result (1) under the next coefficient (-4). After that, we add the numbers in the second column (-4 + 1 = -3) and write the sum below the line. We repeat this process: multiply the new number below the line (-3) by the zero (1), write the result (-3) under the next coefficient (5), add them up (5 + -3 = 2), and then multiply again (2 * 1 = 2) and add one last time (-2 + 2 = 0).

Here's how it looks:

1 | 1  -4   5  -2
  |    1  -3   2
  |________________
    1  -3   2 | 0

The last number on the right (0) is the remainder. Since the remainder is 0, this confirms that $(x-1)$ is indeed a factor, as expected. The other numbers below the line (1, -3, 2) are the coefficients of the quotient polynomial. Since we started with a cubic polynomial ($x^3$) and divided by a linear factor ($x-1$), the quotient will be a quadratic polynomial (degree 2). So, our quotient is $1x^2 - 3x + 2$, or simply $x^2 - 3x + 2$. This means we can rewrite our original polynomial as $f(x) = (x-1)(x^2 - 3x + 2)$. Now, our task is to find the zeros of this quadratic factor, $x^2 - 3x + 2$.

Finding the Remaining Zeros

Awesome job, guys! We've successfully factored out $(x-1)$ from our polynomial $f(x)=x^3-4 x^2+5 x-2$, and we're left with a quadratic expression: $x^2 - 3x + 2$. The remaining zeros of $f(x)$ will be the zeros of this quadratic factor. To find these zeros, we simply need to set the quadratic expression equal to zero and solve for $x$:

$x^2 - 3x + 2 = 0$

This is a standard quadratic equation, and we can solve it using a few methods. Factoring is usually the easiest if it's possible. We're looking for two numbers that multiply to 2 (the constant term) and add up to -3 (the coefficient of the $x$ term). Let's think about pairs of factors for 2: (1, 2) and (-1, -2). If we take (-1) and (-2), their product is $(-1) imes (-2) = 2$, and their sum is $(-1) + (-2) = -3$. Perfect! So, we can factor the quadratic as:

$(x - 1)(x - 2) = 0$

Now, to find the zeros, we set each factor equal to zero:

$x - 1 = 0 ightarrow x = 1$

$x - 2 = 0 ightarrow x = 2$

So, the zeros of the quadratic factor are $x=1$ and $x=2$. This means that our original polynomial $f(x)$ can be fully factored as:

$f(x) = (x-1)(x-1)(x-2)$

Or, more compactly, using exponents:

$f(x) = (x-1)^2 (x-2)$

This factorization gives us all the zeros of the polynomial. We found that the zeros are $x=1$ and $x=2$. However, we still need to determine their multiplicities.

Determining Multiplicities

We're in the home stretch, folks! We've successfully found all the zeros of our polynomial $f(x)=x^3-4 x^2+5 x-2$. The zeros are $x=1$ and $x=2$. Now, we need to nail down their multiplicities. Remember, multiplicity tells us how many times a specific factor appears in the polynomial's factorization. Looking at our factored form, $f(x) = (x-1)^2 (x-2)$, we can directly see the multiplicities.

The factor $(x-1)$ is raised to the power of 2. This means that the zero $x=1$ occurs twice. Therefore, the multiplicity of the zero $x=1$ is 2.

The factor $(x-2)$ is raised to the power of 1 (since it's not explicitly written, it's understood to be 1). This means that the zero $x=2$ occurs once. Therefore, the multiplicity of the zero $x=2$ is 1.

So, to summarize, the zeros of $f(x)=x^3-4 x^2+5 x-2$ are $x=1$ with a multiplicity of 2, and $x=2$ with a multiplicity of 1. The sum of the multiplicities (2 + 1 = 3) should equal the degree of the polynomial (which is 3 for $x^3$), and it does! This is a great way to double-check our work. This means that $x=1$ is a repeated root, and $x=2$ is a simple root.

Understanding these multiplicities is key to truly understanding the behavior of the polynomial. For instance, at $x=1$, the graph of $f(x)$ will touch the x-axis and turn around (because the multiplicity is even), whereas at $x=2$, the graph will cross the x-axis (because the multiplicity is odd).

This problem showcases a fundamental technique for analyzing polynomials: using a known zero to reduce the polynomial's degree and then solving the resulting simpler polynomial. Keep practicing this method, and you'll become a polynomial-solving whiz in no time! You guys totally got this!