Find All Roots Of Polynomials Using The Remainder Theorem

Dec 19, 2025 by Alex Johnson 58 views

Unearthing All the Roots: A Polynomial Puzzle Solved with the Remainder Theorem

In the fascinating world of mathematics, polynomials often present us with intriguing challenges, and one common puzzle is to find all of their roots, especially when we're given a head start with one known root. Today, we're diving deep into a specific problem: finding all the roots of the function $f(x) = x^3 + 10x^2 - 25x - 250$ , given that one of its roots is $x = -10$ . We'll be employing a powerful tool in our mathematical arsenal, the Remainder Theorem, to systematically uncover all the solutions. This theorem is not just a theoretical concept; it's a practical key that unlocks the secrets hidden within polynomial equations, allowing us to break down complex problems into more manageable parts. When you're faced with a polynomial of a higher degree, like our cubic function here, knowing even a single root can dramatically simplify the process of finding the others. The Remainder Theorem, in conjunction with polynomial division (or synthetic division, which is a streamlined version of it), allows us to reduce the degree of the polynomial, making it easier to solve. Think of it like peeling back layers of an onion; each layer you remove brings you closer to the core. In this case, knowing that $x = -10$ is a root tells us that $(x + 10)$ is a factor of $f(x)$ . This is a direct consequence of the Factor Theorem, which is a close cousin of the Remainder Theorem. The Factor Theorem states that if $f(c) = 0$ , then $(x - c)$ is a factor of the polynomial $f(x)$ . Since we are given that $x = -10$ is a root, it means that $f(-10) = 0$ . Therefore, $(x - (-10))$ , which simplifies to $(x + 10)$ , must be a factor of $f(x) = x^3 + 10x^2 - 25x - 250$ . Our mission, then, is to divide $f(x)$ by $(x + 10)$ to find the remaining factor, which will be a quadratic polynomial. Solving this quadratic will then reveal the other two roots of our original cubic equation. This methodical approach ensures that we don't miss any solutions and that our findings are mathematically sound. The beauty of these theorems lies in their elegance and their ability to transform a daunting problem into a series of straightforward algebraic manipulations. It's a testament to the structured nature of mathematics, where understanding fundamental principles can unlock solutions to complex problems.

The Power of the Remainder and Factor Theorems

Let's set the stage with the Remainder Theorem. This fundamental principle in algebra states that when a polynomial $f(x)$ is divided by a linear factor $(x - c)$ , the remainder is equal to $f(c)$ . This might seem simple, but its implications are profound. It directly links the value of a polynomial at a specific point to the remainder obtained upon division by a corresponding linear term. For our problem, since we know $x = -10$ is a root, we can confidently say that $f(-10) = 0$ . This is the cornerstone of the Factor Theorem, which is essentially a special case of the Remainder Theorem. The Factor Theorem states that $(x - c)$ is a factor of a polynomial $f(x)$ if and only if $f(c) = 0$ . Given that $x = -10$ is a root of $f(x) = x^3 + 10x^2 - 25x - 250$ , we know that $f(-10) = 0$ . Consequently, $(x - (-10))$ , which is $(x + 10)$ , must be a factor of $f(x)$ . Our next step is to perform polynomial division to find the other factor. We will divide $f(x)$ by $(x + 10)$ . This division will yield a quotient, which will be a polynomial of one degree less than $f(x)$ (in this case, a quadratic), and a remainder of 0, as expected since $(x + 10)$ is a factor. The process of polynomial division can be carried out using either long division or synthetic division. Synthetic division is often preferred for its speed and simplicity, especially when dividing by a linear factor of the form $(x - c)$ . By performing this division, we effectively factorize the original cubic polynomial into the product of a linear factor $(x + 10)$ and a quadratic factor. The roots of the original cubic polynomial are then the root we were given ( $x = -10$ ) along with the roots of the resulting quadratic factor. The Remainder Theorem and its corollary, the Factor Theorem, are incredibly useful for simplifying polynomial equations. They allow us to test potential roots efficiently and, more importantly, to reduce the complexity of the polynomial we need to solve. Instead of tackling a cubic equation directly, we transform it into a linear equation (the known root) and a quadratic equation, which is a much more familiar and solvable form. This systematic approach underscores the power of fundamental algebraic principles in demystifying complex mathematical problems and providing clear pathways to their solutions.

Step-by-Step Solution: Finding All Roots

Let's embark on the process of finding all the roots of $f(x) = x^3 + 10x^2 - 25x - 250$ , knowing that $x = -10$ is one root. As established, because $x = -10$ is a root, $(x + 10)$ must be a factor of $f(x)$ . We will use synthetic division to divide $f(x)$ by $(x + 10)$ . Synthetic division is a quick method for dividing a polynomial by a linear factor of the form $(x - c)$ . The 'c' value in our case is $-10$ . The coefficients of our polynomial $f(x) = x^3 + 10x^2 - 25x - 250$ are 1, 10, -25, and -250.

Here's how the synthetic division works:

Set up the division: Write the root $(-10)$ to the left. Write the coefficients of the polynomial $(1, 10, -25, -250)$ to the right.
```
-10 | 1   10   -25   -250
    |___________________
```
Bring down the first coefficient: Bring down the '1' below the line.
```
-10 | 1   10   -25   -250
    |___________________
      1
```
Multiply and add: Multiply the number you just brought down (1) by the root (-10), and write the result (-10) under the next coefficient (10). Then, add the numbers in that column (10 + -10 = 0).
```
-10 | 1   10   -25   -250
    |    -10
    |___________________
      1    0
```
Repeat the multiply and add process: Multiply the new number below the line (0) by the root (-10), and write the result (0) under the next coefficient (-25). Add the numbers in that column (-25 + 0 = -25).
```
-10 | 1   10   -25   -250
    |    -10     0
    |___________________
      1    0   -25
```
Final multiplication and addition: Multiply the latest number below the line (-25) by the root (-10), and write the result (250) under the final coefficient (-250). Add the numbers in that column (-250 + 250 = 0).
```
-10 | 1   10   -25   -250
    |    -10     0    250
    |___________________
      1    0   -25     0
```

The last number on the right (0) is the remainder. As expected, the remainder is 0, confirming that $(x + 10)$ is indeed a factor. The other numbers on the bottom row $(1, 0, -25)$ are the coefficients of the quotient polynomial. Since we started with a cubic polynomial and divided by a linear factor, the quotient is a quadratic polynomial. The coefficients (1, 0, -25) correspond to $1x^2 + 0x - 25$ , which simplifies to $x^2 - 25$ . So, we have factored $f(x)$ as:

$f(x) = (x + 10)(x^2 - 25)$

Now, to find the remaining roots, we need to solve the quadratic equation $x^2 - 25 = 0$ . This is a simple difference of squares, which can be factored as $(x - 5)(x + 5) = 0$ . Setting each factor to zero gives us:

$x - 5 = 0 ightarrow x = 5$
$x + 5 = 0 ightarrow x = -5$

Therefore, the roots of the polynomial $f(x) = x^3 + 10x^2 - 25x - 250$ are $x = -10$ , $x = 5$ , and $x = -5$ . This systematic approach, powered by the Factor Theorem and synthetic division, allowed us to efficiently find all the roots of the cubic polynomial.

Conclusion: Mastering Polynomial Roots

We have successfully navigated the process of finding all the roots of the polynomial $f(x) = x^3 + 10x^2 - 25x - 250$ , given that $x = -10$ is one of the roots. By leveraging the Factor Theorem, which is a direct application of the Remainder Theorem, we established that $(x + 10)$ is a factor of $f(x)$ . This critical insight enabled us to use synthetic division, a streamlined method for polynomial division, to divide $f(x)$ by $(x + 10)$ . The division yielded a quadratic factor, $x^2 - 25$ , and a remainder of 0, confirming our factorization. The remaining roots were then easily found by solving the quadratic equation $x^2 - 25 = 0$ . This yielded the roots $x = 5$ and $x = -5$ . Thus, the complete set of roots for $f(x) = x^3 + 10x^2 - 25x - 250$ is $\textbf{{-10, 5, and -5}}$ . This problem beautifully illustrates the power and utility of the Remainder and Factor Theorems in simplifying complex polynomial problems. They provide a systematic and efficient pathway to factoring polynomials and finding their roots, transforming what might initially seem like a daunting task into a series of manageable algebraic steps. Mastering these theorems is a significant step in building a strong foundation in algebra and is essential for tackling more advanced mathematical concepts. For further exploration into polynomial functions and their properties, you might find resources from Khan Academy to be incredibly helpful. They offer comprehensive explanations and practice problems that can deepen your understanding of these fundamental mathematical principles.