Maths Problem Solved: Factorization, Remainders & Equations

Hey guys! Let's dive into some cool algebra problems. We're gonna tackle a few questions that involve factors, equations, factorization, and remainders. This is a classic type of problem that tests your understanding of polynomials. Don't worry, it's not as scary as it sounds. We'll break it down step by step and make sure you understand every concept. So, grab your pencils and let's get started!

(a) Finding the Equation for k and Proving k = 19

Alright, first things first. We're given that $(x + 2)$ is a factor of the polynomial $12x^3 + kx^2 - 13x - 6$. This is super important because it gives us a massive clue. Remember the Factor Theorem? It states that if $(x - a)$ is a factor of a polynomial, then the polynomial equals zero when $x = a$.

In our case, since $(x + 2)$ is a factor, that means $x = -2$ makes the polynomial equal to zero. Let's substitute $x = -2$ into the polynomial and set it equal to zero to find the value of k. We have:

$12(-2)^3 + k(-2)^2 - 13(-2) - 6 = 0$

Let's simplify that bad boy! $(-2)^3$ is $-8$, and $(-2)^2$ is $4$. So we get:

$12(-8) + 4k + 26 - 6 = 0$

$-96 + 4k + 26 - 6 = 0$

Now, combine those constants:

$-76 + 4k = 0$

Finally, isolate k:

$4k = 76$

$k = 19$

BOOM! We've found the equation satisfied by k, which is $-76 + 4k = 0$, and we've shown that $k = 19$. Pretty neat, huh? This whole process utilizes the Factor Theorem effectively. It's a fundamental concept for understanding the relationship between factors and the roots of a polynomial. The key takeaway here is the direct link between a factor and the value of x that makes the polynomial zero. Understanding this connection unlocks the ability to solve many polynomial problems.

Now we've got the value of k to proceed to the next parts of this problem. Remember, these types of problems often build upon each other, so the answer we got from the first part, in this case the value of k, is used in the second one.

Practical Application of the Factor Theorem

The ability to use the Factor Theorem is crucial for a number of real-world and mathematical situations. Firstly, in engineering and physics, polynomial equations are often used to model various phenomena. The factor theorem helps in breaking these complex equations into simpler parts, which makes the analysis and interpretation much easier. For example, if you are designing a bridge, you need to understand the forces acting on each of the bridge's components. Polynomial equations are used to model these forces, and the Factor Theorem can help in finding critical points where the stress is maximum, hence helping engineers to make informed decisions about the material and structure of the bridge.

Secondly, the Factor Theorem finds application in computer science. Algorithms for root finding in polynomial equations are essential in many areas, including computer graphics, simulations, and data analysis. If a computer program needs to find a root of a polynomial, then the Factor Theorem can guide the algorithm, leading to faster and more efficient root-finding methods. Also, in the field of cryptography, factorization of large numbers into prime factors is at the heart of many encryption algorithms. Understanding factors can provide insight into how secure an encryption method is. So, grasping this theorem is not just about passing exams, it helps build a foundation in understanding areas that shape our technological world.

(b) Factorising $12x^3 + 19x^2 - 13x - 6$

Okay, now that we know $k = 19$, our polynomial becomes $12x^3 + 19x^2 - 13x - 6$. We already know that $(x + 2)$ is a factor. This means we can use polynomial long division or synthetic division to divide the polynomial by $(x + 2)$ and get a quadratic expression. Let's use polynomial long division for this part. It’s a bit more involved, but it's a solid method.

Divide $12x^3 + 19x^2 - 13x - 6$ by $(x + 2)$.

          12x^2 - 5x - 3
      ______________________
x + 2 | 12x^3 + 19x^2 - 13x - 6
         -(12x^3 + 24x^2)
         ______________________
                -5x^2 - 13x
                -(-5x^2 - 10x)
                ______________________
                        -3x - 6
                        -(-3x - 6)
                        ______________________
                                0

So, the result of the division is $12x^2 - 5x - 3$. Therefore:

$12x^3 + 19x^2 - 13x - 6 = (x + 2)(12x^2 - 5x - 3)$

Now, we need to factorise the quadratic $12x^2 - 5x - 3$. We're looking for two numbers that multiply to $(12 imes -3) = -36$ and add up to $-5$. Those numbers are $-9$ and $4$. So, let's rewrite the quadratic:

$12x^2 - 5x - 3 = 12x^2 - 9x + 4x - 3$

Now, factor by grouping:

$12x^2 - 9x + 4x - 3 = 3x(4x - 3) + 1(4x - 3)$

$ = (3x + 1)(4x - 3)$

Putting it all together, the fully factorized form of the original cubic polynomial is:

$12x^3 + 19x^2 - 13x - 6 = (x + 2)(3x + 1)(4x - 3)$

We successfully factorized the cubic polynomial. This is the goal here, using the information from the first part, where we calculated the value of k to simplify our work. Factoring helps simplify the expression, which makes finding roots (the values of x that make the equation equal to zero) easier. Also, remember that the factored form reveals the roots of the equation directly, in this case, the roots are -2, -1/3 and 3/4. The skills developed here are essential for further mathematical topics. This includes more complex equations, graphing, and calculus, forming the foundation of problem-solving. It involves a systematic approach to breaking down the problem into more manageable parts, a key problem-solving skill in all disciplines.

Benefits of Factorization

Factorization offers multiple benefits, the first being simplifying expressions. A factorized expression is simpler than the original form, often allowing you to solve the equations faster. Reducing complexity makes solving much easier. Another benefit is simplification of complex equations. Factoring is useful for simplifying complex expressions and it helps in finding the roots of polynomial equations, which are the values that make the expression equal to zero. These roots are critical for understanding the behavior of the polynomial. In various fields of science and engineering, the understanding of roots is crucial for modeling real-world phenomena. In a financial context, factorization is used in portfolio optimization and risk assessment. Also, in computer science, factoring is a fundamental part of the algorithm design in various contexts, from cryptography to data compression. So, in summary, it is a versatile tool with both theoretical and practical applications.

(c) Finding the Remainder When Dividing by $2x - 1$

Alright, let's wrap this up by finding the remainder when we divide $12x^3 + 19x^2 - 13x - 6$ by $(2x - 1)$. We can use the Remainder Theorem, which is a cousin of the Factor Theorem. The Remainder Theorem states that if you divide a polynomial $f(x)$ by $(x - a)$, the remainder is $f(a)$.

In our case, we're dividing by $(2x - 1)$. To use the Remainder Theorem directly, we need to rewrite this in the form $(x - a)$. We can do this by factoring out a 2:

2x - 1 = 2(x - rac{1}{2})

Now, let's call our polynomial $f(x) = 12x^3 + 19x^2 - 13x - 6$. We need to find f(rac{1}{2}):

f(rac{1}{2}) = 12(rac{1}{2})^3 + 19(rac{1}{2})^2 - 13(rac{1}{2}) - 6

f(rac{1}{2}) = 12(rac{1}{8}) + 19(rac{1}{4}) - rac{13}{2} - 6

f(rac{1}{2}) = rac{3}{2} + rac{19}{4} - rac{26}{4} - rac{24}{4}

f(rac{1}{2}) = rac{6}{4} + rac{19}{4} - rac{26}{4} - rac{24}{4}

f(rac{1}{2}) = rac{6 + 19 - 26 - 24}{4}

f(rac{1}{2}) = rac{-25}{4}

So, the remainder when $12x^3 + 19x^2 - 13x - 6$ is divided by $2x - 1$ is -rac{25}{4}. That wraps it up! We successfully utilized the Remainder Theorem to find our final answer. Understanding the Remainder Theorem and its application simplifies complex polynomial divisions, thus saving time and effort. This theorem is also valuable because it can be used to check your work when you have factored a polynomial or when you have performed polynomial long division. This way, the process of finding the remainder helps in validating our calculations, ensuring accuracy in our mathematical solutions.

Application of Remainder Theorem

The Remainder Theorem also has several practical applications. It is used in computer science for designing efficient algorithms to evaluate polynomial expressions. Because finding the remainder can be performed much faster than polynomial division, this theorem is used to optimize code. Similarly, in signal processing, it is used for analyzing and manipulating signals, allowing engineers to design filters and processing techniques, for instance, in the field of digital image processing and audio processing.

In the field of cryptography, the Remainder Theorem provides a foundation for more advanced mathematical methods. It plays an important role in the construction of cryptographic systems. More specifically, the Remainder Theorem is used to ensure the integrity of the encryption and decryption processes.

In various scientific and engineering disciplines, polynomial equations are employed to model diverse phenomena. Applying the Remainder Theorem allows for simpler evaluation of these models by checking specific points. This is particularly useful in scenarios where computational efficiency is important or when quick calculations are necessary.

I hope this explanation was helpful, guys! Keep practicing, and you'll become a pro at these types of problems. Remember the key takeaways: the Factor Theorem, the Remainder Theorem, and the importance of breaking down problems into manageable steps. Keep practicing, and keep having fun with math! Bye!