Problem 26 - Algebra for College Students

Question

For Problems \(21-26\), verify that the equations do not have any rational number solutions. $$ x^{5}-2 x^{4}+3 x^{3}+4 x^{2}+7 x-1=0 $$

Step-by-Step Solution

Answer

The equation has no rational number solutions.

Step 1: Apply the Rational Root Theorem

The Rational Root Theorem states that any rational solution, expressed in its lowest terms \( \frac{p}{q} \), is such that \( p \) divides the constant term \(-1\) and \( q \) divides the leading coefficient \(1\). In this equation, \( p \) can be \( \pm 1 \) and \( q \) can also be \( \pm 1 \). Therefore, the possible rational roots are \( \pm 1 \).

Step 2: Test Possible Rational Roots

We test each possible root \( x = 1 \) and \( x = -1 \) in the polynomial: 1. If \( x = 1 \), the polynomial becomes: \(1^5 - 2(1)^4 + 3(1)^3 + 4(1)^2 + 7(1) - 1 = 1 - 2 + 3 + 4 + 7 - 1 = 12 \). Since 12 is not zero, \( x = 1 \) is not a root.2. If \( x = -1 \), the polynomial becomes: \((-1)^5 - 2(-1)^4 + 3(-1)^3 + 4(-1)^2 + 7(-1) - 1 = -1 - 2 - 3 + 4 - 7 - 1 = -10 \). Since -10 is not zero, \( x = -1 \) is not a root.

Step 3: Conclusion

Since neither \( x = 1 \) nor \( x = -1 \) satisfy the equation \( x^5 - 2x^4 + 3x^3 + 4x^2 + 7x - 1 = 0 \), the polynomial has no rational roots according to the Rational Root Theorem.

Key Concepts

Polynomial Equations

A polynomial equation is a mathematical expression that equates a polynomial to zero. It consists of variables and coefficients, combined using only addition, subtraction, multiplication, and non-negative integer exponents of variables. The polynomial in this exercise is given by:\[ x^{5} - 2x^{4} + 3x^{3} + 4x^{2} + 7x - 1 = 0 \]This equation is set equal to zero, which is typical when looking for the roots or solutions of a polynomial. Each unique power of the variable \(x\) in this expression is associated with a coefficient that dictates its contribution to the overall polynomial expression. Polynomial equations are fundamental in algebra, and understanding them is crucial for solving complex mathematical problems.

Rational Roots

Rational roots are potential solutions to a polynomial equation that can be expressed as a fraction of integers, \( \frac{p}{q} \), where \(p\) is a factor of the constant term and \(q\) is a factor of the leading coefficient. In our exercise, the Rational Root Theorem helps determine the possible rational roots for testing.Given our polynomial:\[ x^{5} - 2x^{4} + 3x^{3} + 4x^{2} + 7x - 1 = 0 \]Here:- The constant term is -1- The leading coefficient is 1According to the Rational Root Theorem:- Possible values for \(p\) are ±1 (factors of -1)- Possible values for \(q\) are ±1 (factors of 1)Hence, the only possible rational roots are ±1. This theorem is a powerful tool, allowing us to quickly narrow down the potential rational solutions to only a few candidates.

Root Verification

Root verification involves checking if a possible rational root indeed satisfies the polynomial equation. It's like testing a candidate to see if they fit the job requirements. For our polynomial:\[ x^{5} - 2x^{4} + 3x^{3} + 4x^{2} + 7x - 1 = 0 \]We need to test the possible roots \(x = 1\) and \(x = -1\) obtained from the Rational Root Theorem.1. **Checking \(x = 1\):** - Substituting \(x = 1\) gives: \(1^5 - 2(1)^4 + 3(1)^3 + 4(1)^2 + 7(1) - 1 = 12\) - Since 12 is not zero, \(x = 1\) is not a root.2. **Checking \(x = -1\):** - Substituting \(x = -1\) gives: \((-1)^5 - 2(-1)^4 + 3(-1)^3 + 4(-1)^2 + 7(-1) - 1 = -10\) - Since -10 is not zero, \(x = -1\) is not a root.Therefore, neither candidate is a root, confirming there are no rational roots.

Degree of Polynomial

The degree of a polynomial is the highest power of the variable in the equation. It plays a significant role in determining the polynomial's behavior and the number of roots it might have.For the polynomial given in our exercise:\[ x^{5} - 2x^{4} + 3x^{3} + 4x^{2} + 7x - 1 = 0 \]- The highest power of \(x\) is 5, which makes this a fifth-degree polynomial.Some key characteristics of polynomial degree:- A polynomial of degree \(n\) can have up to \(n\) roots.- The degree gives insights into the potential complexity and the nature of the equation's solutions.In our case, even though the polynomial is fifth-degree, due to the verification process, we've determined there are no rational roots—a reminder that the degree indicates possibilities but not certainties.