Finding 'a' When (x - 1) Is A Factor Of A²x³ - 4ax + 4a - 1
Introduction
In this article, we delve into the fascinating world of polynomial factorization. Polynomials, the unsung heroes of algebra, are expressions consisting of variables and coefficients, combined using addition, subtraction, and multiplication, with non-negative integer exponents. Factoring polynomials, the art of breaking them down into simpler expressions, is a fundamental skill in mathematics. It unlocks a treasure trove of problem-solving techniques and paves the way for understanding more advanced mathematical concepts. In this particular problem, we embark on a journey to find the specific value of 'a' that makes (x - 1) a factor of the polynomial a²x³ - 4ax + 4a - 1. This exploration will not only enhance your understanding of the Factor Theorem but also demonstrate how it can be applied to solve intricate algebraic problems.
Understanding the Factor Theorem
Before we dive into the solution, let's brush up on the Factor Theorem. This theorem is our guiding star, illuminating the path to solving this problem. The Factor Theorem states that for a polynomial p(x), (x - c) is a factor of p(x) if and only if p(c) = 0. In simpler terms, if we substitute 'c' into the polynomial and the result is zero, then (x - c) is indeed a factor of that polynomial. Conversely, if (x - c) is a factor, then substituting 'c' into the polynomial will always yield zero. This elegant relationship between factors and roots is the cornerstone of our solution.
Applying the Factor Theorem
In our case, we are given that (x - 1) is a factor of the polynomial a²x³ - 4ax + 4a - 1. According to the Factor Theorem, this implies that if we substitute x = 1 into the polynomial, the result should be zero. This is our key insight. By setting p(1) = 0, we can create an equation that involves 'a'. Solving this equation will reveal the value(s) of 'a' that satisfy the given condition. This step-by-step process exemplifies the power of the Factor Theorem in solving polynomial problems.
Step-by-Step Solution
1. Apply the Factor Theorem
Given that (x - 1) is a factor of the polynomial p(x) = a²x³ - 4ax + 4a - 1, we can use the Factor Theorem. Substitute x = 1 into the polynomial:
p(1) = a²(1)³ - 4a(1) + 4a - 1
2. Simplify the Expression
Simplify the expression obtained in the previous step:
p(1) = a² - 4a + 4a - 1 p(1) = a² - 1
3. Set p(1) to Zero
According to the Factor Theorem, if (x - 1) is a factor, then p(1) must be equal to zero:
a² - 1 = 0
4. Solve for 'a'
Now we have a simple quadratic equation in terms of 'a'. Solve for 'a':
a² = 1
Taking the square root of both sides, we get:
a = ±1
5. Verify the Solutions
It's crucial to verify our solutions by plugging the values of 'a' back into the original polynomial and checking if (x - 1) is indeed a factor. This step ensures that we haven't made any errors in our calculations.
Case 1: a = 1
If a = 1, the polynomial becomes:
p(x) = (1)²x³ - 4(1)x + 4(1) - 1 p(x) = x³ - 4x + 4 - 1 p(x) = x³ - 4x + 3
To check if (x - 1) is a factor, we can use synthetic division or substitute x = 1:
p(1) = (1)³ - 4(1) + 3 = 1 - 4 + 3 = 0
Since p(1) = 0, (x - 1) is a factor when a = 1.
Case 2: a = -1
If a = -1, the polynomial becomes:
p(x) = (-1)²x³ - 4(-1)x + 4(-1) - 1 p(x) = x³ + 4x - 4 - 1 p(x) = x³ + 4x - 5
To check if (x - 1) is a factor, substitute x = 1:
p(1) = (1)³ + 4(1) - 5 = 1 + 4 - 5 = 0
Since p(1) = 0, (x - 1) is a factor when a = -1.
Solutions
Therefore, the values of 'a' for which (x - 1) is a factor of the polynomial a²x³ - 4ax + 4a - 1 are a = 1 and a = -1. These values satisfy the condition set by the Factor Theorem, making (x - 1) a genuine factor of the polynomial.
Alternative Methods
While the Factor Theorem provides a direct and efficient method for solving this problem, it's worth noting that other techniques can also be employed. These alternative approaches not only offer a different perspective but also enhance your problem-solving toolkit.
1. Polynomial Long Division
Polynomial long division is a powerful technique for dividing one polynomial by another. In this case, we could divide a²x³ - 4ax + 4a - 1 by (x - 1) and set the remainder to zero. This would yield an equation in terms of 'a', which we could then solve. While this method is more computationally intensive than using the Factor Theorem, it provides a deeper understanding of polynomial division.
To perform polynomial long division, we would set up the division as follows:
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x - 1 | a²x³ + 0x² - 4ax + 4a - 1
Note that we've added a 0x² term to maintain the proper place values. The process involves iteratively dividing the leading term of the dividend by the leading term of the divisor, multiplying the result by the divisor, and subtracting it from the dividend. This process is repeated until the degree of the remainder is less than the degree of the divisor.
In this case, after performing the long division, we would obtain a quotient and a remainder. The condition for (x - 1) to be a factor is that the remainder must be zero. Setting the remainder to zero would give us an equation in terms of 'a', which we could then solve to find the possible values of 'a'. This method, while more involved, reinforces the concept of polynomial division and its relationship to factorization.
2. Synthetic Division
Synthetic division is a streamlined method for dividing a polynomial by a linear factor of the form (x - c). It's a more efficient alternative to polynomial long division, especially when dealing with linear divisors. In our case, we could use synthetic division to divide a²x³ - 4ax + 4a - 1 by (x - 1). The remainder obtained from synthetic division must be zero for (x - 1) to be a factor.
To perform synthetic division, we would set up a table with the coefficients of the polynomial and the value 'c' from the divisor (x - c). In this case, c = 1. The process involves bringing down the first coefficient, multiplying it by 'c', adding the result to the next coefficient, and repeating the process. The last number in the bottom row represents the remainder.
Setting the remainder to zero would give us an equation in terms of 'a', which we could then solve. Like polynomial long division, synthetic division provides a visual and efficient way to determine the remainder when dividing by a linear factor. This method is particularly useful for quick calculations and can be a valuable tool in your problem-solving arsenal.
By exploring these alternative methods, we gain a more comprehensive understanding of polynomial factorization and the various techniques available to tackle such problems. Each method offers a unique perspective and reinforces the fundamental principles of algebra.
Conclusion
In conclusion, by skillfully applying the Factor Theorem, we successfully determined that the values of 'a' for which (x - 1) is a factor of the polynomial a²x³ - 4ax + 4a - 1 are a = 1 and a = -1. This problem highlights the importance of understanding fundamental theorems and their applications in solving algebraic problems. Factoring polynomials is a critical skill in mathematics, and mastering techniques like the Factor Theorem opens doors to more advanced mathematical concepts and problem-solving strategies. Remember, the journey of mathematical discovery is paved with practice and a deep understanding of the underlying principles.
