Finding Coefficients In Polynomials Without Division A Factorial Approach

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Introduction

In the realm of polynomial equations, understanding the relationship between factors and coefficients is paramount. This article delves into a fascinating problem where we're given that the quadratic expression $x^2 + 5x + 6$ is a factor of the quartic expression $x^4 + rx^2 + s$. Our mission is to determine the values of the unknown coefficients, r and s, without resorting to long or synthetic division. This endeavor will showcase the power of factoring, algebraic manipulation, and the application of key theorems. Let's embark on this mathematical journey together, unraveling the intricacies of polynomial factorization and coefficient determination.

Factoring the Quadratic Expression

Our initial step involves factoring the given quadratic expression, $x^2 + 5x + 6$. This factorization will serve as the cornerstone of our solution, allowing us to establish a crucial link between the factors of the quadratic and the quartic polynomial. Factoring a quadratic expression entails identifying two binomials that, when multiplied together, yield the original quadratic. We seek two numbers that add up to the coefficient of the x term (which is 5) and multiply to the constant term (which is 6). These numbers are 2 and 3, as 2 + 3 = 5 and 2 * 3 = 6. Therefore, we can express the quadratic expression as a product of two binomials:

$x^2 + 5x + 6 = (x + 2)(x + 3)$

This factorization reveals that $(x + 2)$ and $(x + 3)$ are factors of the quadratic expression. This understanding is pivotal because it directly implies that these binomials are also factors of the quartic expression, $x^4 + rx^2 + s$, since the quadratic is given to be a factor of the quartic. This connection forms the basis for our subsequent steps in determining the values of r and s. Knowing that $(x + 2)$ and $(x + 3)$ are factors means that if we substitute $x = -2$ and $x = -3$ into the quartic expression, the result must be zero, according to the Factor Theorem. This gives us a powerful tool to create equations involving r and s, which we can then solve to find their values. In essence, the ability to factor the quadratic expression has unlocked a pathway to unravel the coefficients of the quartic expression, showcasing the interconnectedness of factors and coefficients in polynomial algebra.

Leveraging the Factor Theorem

The Factor Theorem is a cornerstone of polynomial algebra, providing a direct link between the roots of a polynomial and its factors. In essence, the Factor Theorem states that for a polynomial P(x), if P(a) = 0 for some value a, then (x - a) is a factor of P(x). Conversely, if (x - a) is a factor of P(x), then P(a) = 0. This theorem is instrumental in our problem because it allows us to translate the fact that $(x + 2)$ and $(x + 3)$ are factors of $x^4 + rx^2 + s$ into concrete equations.

Since $(x + 2)$ is a factor, we know that when $x = -2$, the quartic expression must equal zero. Substituting $x = -2$ into $x^4 + rx^2 + s$, we get:

$(-2)^4 + r(-2)^2 + s = 0$

Simplifying this equation yields:

$16 + 4r + s = 0$ (Equation 1)

Similarly, since $(x + 3)$ is a factor, substituting $x = -3$ into the quartic expression must also result in zero:

$(-3)^4 + r(-3)^2 + s = 0$

Simplifying this equation, we obtain:

$81 + 9r + s = 0$ (Equation 2)

Now, we have a system of two linear equations with two unknowns, r and s. This system of equations is a direct consequence of applying the Factor Theorem, which has transformed the problem of finding factors into a problem of solving a system of linear equations. The power of the Factor Theorem lies in its ability to convert abstract concepts about factors and roots into tangible algebraic relationships, allowing us to manipulate and solve for unknown coefficients. The next step involves employing algebraic techniques to solve this system of equations, thereby revealing the values of r and s.

Solving the System of Equations

With the application of the Factor Theorem, we have successfully established a system of two linear equations in two unknowns:

1. $16 + 4r + s = 0$
2. $81 + 9r + s = 0$

Our next objective is to solve this system to determine the values of r and s. There are several methods to solve such systems, including substitution, elimination, and matrix methods. For this particular system, the elimination method presents a straightforward approach. We can eliminate the variable s by subtracting Equation 1 from Equation 2:

$(81 + 9r + s) - (16 + 4r + s) = 0 - 0$

Simplifying this subtraction yields:

$65 + 5r = 0$

Now, we can solve for r:

$5r = -65$

$r = -13$

Having found the value of r, we can substitute it back into either Equation 1 or Equation 2 to solve for s. Let's substitute r = -13 into Equation 1:

$16 + 4(-13) + s = 0$

$16 - 52 + s = 0$

$-36 + s = 0$

$s = 36$

Therefore, we have determined the values of the coefficients r and s to be r = -13 and s = 36. This solution demonstrates the effectiveness of the elimination method in solving systems of linear equations, a technique that is widely applicable in various mathematical contexts. By systematically eliminating variables, we can reduce the complexity of the system and isolate the unknowns, ultimately leading to the desired solution. The values of r and s are crucial as they fully define the quartic expression, allowing us to understand its behavior and properties.

Verifying the Solution

After obtaining the values of r and s, it is prudent to verify our solution to ensure its accuracy. This verification step involves substituting the calculated values back into the original problem statement and confirming that the given conditions are satisfied. In our case, we found that r = -13 and s = 36. Substituting these values into the quartic expression, we get:

$x^4 - 13x^2 + 36$

Now, we need to confirm that $x^2 + 5x + 6$ is indeed a factor of this quartic expression. Recall that we factored $x^2 + 5x + 6$ as $(x + 2)(x + 3)$. This means that if we can express the quartic expression as a product of $(x + 2)(x + 3)$ and another quadratic expression, our solution is verified.

Let's assume that the other quadratic factor is of the form $x^2 + ax + b$. Then, we should have:

$x^4 - 13x^2 + 36 = (x + 2)(x + 3)(x^2 + ax + b)$

Expanding $(x + 2)(x + 3)$, we get $x^2 + 5x + 6$. So, we have:

$x^4 - 13x^2 + 36 = (x^2 + 5x + 6)(x^2 + ax + b)$

Expanding the right side, we get:

$x^4 + ax^3 + bx^2 + 5x^3 + 5ax^2 + 5bx + 6x^2 + 6ax + 6b$

Combining like terms, we have:

$x^4 + (a + 5)x^3 + (b + 5a + 6)x^2 + (5b + 6a)x + 6b$

Now, we can equate the coefficients of the corresponding terms on both sides of the equation:

• Coefficient of $x^4$: 1 = 1 (This is already satisfied)
• Coefficient of $x^3$: $a + 5 = 0$ => $a = -5$
• Coefficient of $x^2$: $b + 5a + 6 = -13$
• Coefficient of $x$: $5b + 6a = 0$
• Constant term: $6b = 36$ => $b = 6$

Substituting a = -5 and b = 6 into the equation for the coefficient of $x^2$, we get:

$6 + 5(-5) + 6 = -13$

$12 - 25 = -13$

$-13 = -13$ (This is satisfied)

Substituting a = -5 and b = 6 into the equation for the coefficient of x, we get:

$5(6) + 6(-5) = 0$

$30 - 30 = 0$

$0 = 0$ (This is satisfied)

Since all the coefficients match, we have successfully factored the quartic expression as:

$x^4 - 13x^2 + 36 = (x^2 + 5x + 6)(x^2 - 5x + 6)$

This verification confirms that our solution, r = -13 and s = 36, is correct. The process of verification is a crucial step in problem-solving, as it ensures that the obtained solution not only satisfies the derived equations but also aligns with the original problem statement. It reinforces our confidence in the accuracy of the solution and provides a deeper understanding of the underlying mathematical relationships.

Alternative Approach: Vieta's Formulas

While we have successfully determined the coefficients r and s using the Factor Theorem and solving a system of equations, an alternative approach leveraging Vieta's formulas offers a different perspective. Vieta's formulas provide a direct relationship between the coefficients of a polynomial and the sums and products of its roots. This approach can be particularly insightful and efficient in certain polynomial problems.

Recall that we factored $x^2 + 5x + 6$ as $(x + 2)(x + 3)$. This means the roots of the quadratic factor are $x = -2$ and $x = -3$. Since $x^2 + 5x + 6$ is a factor of $x^4 + rx^2 + s$, the roots of the quadratic factor are also roots of the quartic polynomial. Let's denote the roots of the quartic polynomial as $x_1, x_2, x_3,$ and $x_4$. We know that two of these roots are -2 and -3. Let's assume $x_1 = -2$ and $x_2 = -3$.

Since the quartic polynomial only has even powers of x, it implies that if a is a root, then -a is also a root. This is because substituting -a into the polynomial will yield the same result as substituting a due to the even powers. Therefore, if -2 and -3 are roots, then 2 and 3 must also be roots. So, we can say $x_3 = 2$ and $x_4 = 3$.

Now, we can use Vieta's formulas for a quartic polynomial of the form $ax^4 + bx^3 + cx^2 + dx + e = 0$. In our case, the polynomial is $x^4 + 0x^3 + rx^2 + 0x + s = 0$, so a = 1, b = 0, c = r, d = 0, and e = s. Vieta's formulas state:

• Sum of roots: $x_1 + x_2 + x_3 + x_4 = -b/a$
• Sum of pairwise products: $x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 = c/a$
• Sum of triple products: $x_1x_2x_3 + x_1x_2x_4 + x_1x_3x_4 + x_2x_3x_4 = -d/a$
• Product of roots: $x_1x_2x_3x_4 = e/a$

Applying these formulas to our roots -2, -3, 2, and 3, we get:

• Sum of roots: $-2 + (-3) + 2 + 3 = 0 = -0/1$ (This is consistent)
• Sum of pairwise products: $(-2)(-3) + (-2)(2) + (-2)(3) + (-3)(2) + (-3)(3) + (2)(3) = 6 - 4 - 6 - 6 - 9 + 6 = -13 = r/1$ => $r = -13$
• Sum of triple products: $(-2)(-3)(2) + (-2)(-3)(3) + (-2)(2)(3) + (-3)(2)(3) = 12 + 18 - 12 - 18 = 0 = -0/1$ (This is consistent)
• Product of roots: $(-2)(-3)(2)(3) = 36 = s/1$ => $s = 36$

Thus, Vieta's formulas provide an alternative pathway to arrive at the same solution, r = -13 and s = 36. This approach highlights the power of understanding the relationships between roots and coefficients, offering a different perspective on polynomial problems. The use of Vieta's formulas not only confirms our previous solution but also enriches our understanding of polynomial algebra.

Conclusion

In this article, we embarked on a journey to determine the coefficients r and s of the quartic polynomial $x^4 + rx^2 + s$, given that $x^2 + 5x + 6$ is a factor. We successfully navigated this problem without resorting to long or synthetic division, instead employing the powerful tools of factoring, the Factor Theorem, and solving a system of equations. Furthermore, we explored an alternative approach using Vieta's formulas, which provided a different lens through which to view the relationship between roots and coefficients.

The key takeaways from this exploration are the importance of factoring in polynomial algebra, the direct connection between factors and roots as embodied in the Factor Theorem, and the utility of Vieta's formulas in relating coefficients to roots. These concepts are fundamental to understanding and manipulating polynomial expressions. The ability to solve problems like this enhances our mathematical toolkit and deepens our appreciation for the elegance and interconnectedness of algebraic principles. The journey of solving this problem has not only yielded a solution but also illuminated the rich landscape of polynomial algebra, encouraging further exploration and discovery.

Polynomial equations, Factor Theorem, Vieta's formulas, Coefficients, Roots, Factoring, Algebraic manipulation, System of equations, Quadratic expression, Quartic expression