Divide X 2 10x 30 X 5

Deconstructing and Solving: Divide x² + 10x + 30 by x + 5

This article delves into the intricacies of polynomial division, specifically focusing on how to divide the polynomial x² + 10x + 30 by the binomial x + 5. We'll explore multiple methods, explaining the underlying principles and providing a step-by-step guide for each. Understanding polynomial division is crucial for various mathematical applications, from solving equations to simplifying complex expressions. We'll also touch upon the broader context of polynomial division within algebra and its real-world applications.

Understanding Polynomial Division

Before diving into the specifics of dividing x² + 10x + 30 by x + 5, let's establish a foundational understanding of polynomial division. Polynomial division is the process of dividing one polynomial by another polynomial of lower or equal degree. The result consists of a quotient and a remainder. This is analogous to long division with numbers.

The general form of polynomial division can be represented as:

(Dividend) = (Divisor) * (Quotient) + (Remainder)

Where:

Dividend: The polynomial being divided. In our case, it's x² + 10x + 30.
Divisor: The polynomial doing the dividing. In our case, it's x + 5.
Quotient: The result of the division. This is what we aim to find.
Remainder: The amount left over after the division. This may be zero or another polynomial of a degree lower than the divisor.

Method 1: Long Division

Long division is a classic and widely understood method for polynomial division. Let's apply it to our problem:

(x² + 10x + 30) ÷ (x + 5)

Set up the long division:

             _______
x + 5 | x² + 10x + 30

Divide the first term of the dividend (x²) by the first term of the divisor (x):

x² / x = x

Write this 'x' above the line as the first term of the quotient:

             x      
             _______
x + 5 | x² + 10x + 30

Multiply the divisor (x + 5) by the first term of the quotient (x):

x * (x + 5) = x² + 5x

Write this result below the dividend:

             x      
             _______
x + 5 | x² + 10x + 30
       x² + 5x

Subtract the result from the dividend:

(x² + 10x) - (x² + 5x) = 5x

             x      
             _______
x + 5 | x² + 10x + 30
       x² + 5x   
       -------
             5x + 30

Bring down the next term of the dividend (+30):

             x      
             _______
x + 5 | x² + 10x + 30
       x² + 5x   
       -------
             5x + 30

Divide the first term of the new dividend (5x) by the first term of the divisor (x):

5x / x = 5

Write this '5' above the line as the next term of the quotient:

             x + 5  
             _______
x + 5 | x² + 10x + 30
       x² + 5x   
       -------
             5x + 30

Multiply the divisor (x + 5) by the new term of the quotient (5):

5 * (x + 5) = 5x + 25

Write this result below the new dividend:

             x + 5  
             _______
x + 5 | x² + 10x + 30
       x² + 5x   
       -------
             5x + 30
             5x + 25

Subtract the result from the new dividend:

(5x + 30) - (5x + 25) = 5

This is the remainder.

             x + 5  
             _______
x + 5 | x² + 10x + 30
       x² + 5x   
       -------
             5x + 30
             5x + 25
             -------
                   5

Therefore, the quotient is x + 5, and the remainder is 5. We can express this as:

x² + 10x + 30 = (x + 5)(x + 5) + 5

Method 2: Synthetic Division

Synthetic division offers a more concise method for dividing polynomials, especially when the divisor is of the form (x - c). Let's use synthetic division for (x² + 10x + 30) ÷ (x + 5):

Identify the divisor: The divisor is x + 5, which means c = -5 (since x + 5 = x - (-5)).
Set up the synthetic division:

-5 | 1   10   30
   |
   |---------

Bring down the first coefficient (1):

-5 | 1   10   30
   |
   |---------
   | 1

Multiply the last entry (1) by the divisor (-5) and add it to the next coefficient (10):

1 * (-5) + 10 = 5

-5 | 1   10   30
   |     -5  
   |---------
   | 1    5

Repeat the process for the next coefficient (30):

5 * (-5) + 30 = 5

-5 | 1   10   30
   |     -5  -25
   |---------
   | 1    5    5

The last entry (5) represents the remainder. The other entries (1 and 5) represent the coefficients of the quotient. Therefore, the quotient is x + 5, and the remainder is 5, confirming the result from long division.

Real-World Applications of Polynomial Division

Polynomial division isn't just a theoretical exercise; it has practical applications across various fields:

Engineering: Used extensively in control systems, signal processing, and circuit analysis.
Computer Science: Essential in algorithm design and numerical analysis.
Physics: Used in modeling physical phenomena, such as projectile motion and wave propagation.
Economics: Used in mathematical modeling of economic systems.
Financial Modeling: Crucial for calculating present and future values of investments.

Further Exploration: Remainder Theorem and Factor Theorem

Two important theorems are closely related to polynomial division:

The Remainder Theorem: States that when a polynomial f(x) is divided by (x - c), the remainder is f(c). In our case, f(x) = x² + 10x + 30, and c = -5. f(-5) = (-5)² + 10(-5) + 30 = 25 - 50 + 30 = 5, which confirms our remainder.
The Factor Theorem: States that (x - c) is a factor of f(x) if and only if f(c) = 0. Since our remainder is 5 (not 0), (x + 5) is not a factor of x² + 10x + 30.

Conclusion

This detailed exploration of dividing x² + 10x + 30 by x + 5 highlights the importance of understanding polynomial division. Both long division and synthetic division provide effective methods for solving such problems. The Remainder Theorem and Factor Theorem further solidify the theoretical underpinnings of this crucial algebraic process, demonstrating its practical applications beyond abstract mathematical concepts and into the real world. Mastering polynomial division is a fundamental step towards more advanced mathematical concepts and problem-solving. Remember to practice consistently to build your proficiency and confidence in tackling polynomial division problems.