Problem 21
Question
In Exercises \(17-24\) a. List all possible rational roots. b. Use synthetic division to test the possible rational roots and find an actual root. c. Use the quotient from part (b) to find the remaining roots and solve the equation. $$ 6 x^{3}+25 x^{2}-24 x+5=0 $$
Step-by-Step Solution
Verified Answer
The roots of the cubic polynomial equation \(6x^3 + 25x^2 - 24x + 5 = 0\) are -5, -\(\frac{1}{6}\), and \(\frac{1}{3}\).
1Step 1: Listing Possible Rational Roots
The Rational Root Theorem states that all possible rational roots of an equation can be determined by taking all the factors of constant term and dividing them by all the factors of the coefficient of the highest power term. In this case, the constant term is 5 and the coefficient of the highest power term (\(x^3\)) is 6. Factors of 5 are ±1, ±5 and factors of 6 are ±1, ±2, ±3, ±6. So, all possible roots will be ±\(\frac{1}{1}\), ±\(\frac{1}{2}\), ±\(\frac{1}{3}\), ±\(\frac{1}{6}\), ±(5), ±\(\frac{5}{2}\), ±\(\frac{5}{3}\), ±\(\frac{5}{6}\)
2Step 2: Testing Possible Roots Using Synthetic Division
From the list of possible roots, synthetic division will be used to test them one by one and to find an actual root. A possible root will be an actual root if the remainder from the synthetic division test is zero. The result of the synthetic division will be coefficients for a decreased order polynomial (from cubic to quadratic here). After testing, it is found that root -5 yields a remainder of 0. Thus, -5 is an actual root of the equation.
3Step 3: Finding Remaining Roots
The quotient resulting from synthetic division is 6x^2 + x - 1. This quadratic equation can be solved to give the remaining roots of the initial equation. By using the quadratic formula, remaining roots are found to be -\(\frac{1}{6}\) and \(\frac{1}{3}\).
Key Concepts
Rational Root TheoremPolynomial RootsQuadratic Formula
Rational Root Theorem
The Rational Root Theorem is an essential tool for solving polynomial equations, particularly when searching for rational solutions. This theorem offers a systematic way to list all possible rational roots of a polynomial equation with integer coefficients. According to the theorem, if a polynomial has a rational root \frac{p}{q}, where p and q are integers with no common factors other than 1, and q is not zero, then p must be a factor of the constant term and q must be a factor of the leading coefficient.
When faced with a polynomial equation such as \(6x^3 + 25x^2 - 24x + 5 = 0\), where the constant term is 5 and the leading coefficient is 6, the Rational Root Theorem enables us to list potential rational roots. These roots are formed by dividing the factors of the constant term (±1, ±5) by the factors of the leading coefficient (±1, ±2, ±3, ±6). Following this approach provides a finite list of possible roots including ±1, ±\frac{1}{2}, ±\frac{1}{3}, ±\frac{1}{6}, and ±5, ±\frac{5}{2}, ±\frac{5}{3}, ±\frac{5}{6}.
By methodically testing these possible roots through processes such as synthetic division or substitution, we can pinpoint which, if any, are actual roots of the polynomial equation.
When faced with a polynomial equation such as \(6x^3 + 25x^2 - 24x + 5 = 0\), where the constant term is 5 and the leading coefficient is 6, the Rational Root Theorem enables us to list potential rational roots. These roots are formed by dividing the factors of the constant term (±1, ±5) by the factors of the leading coefficient (±1, ±2, ±3, ±6). Following this approach provides a finite list of possible roots including ±1, ±\frac{1}{2}, ±\frac{1}{3}, ±\frac{1}{6}, and ±5, ±\frac{5}{2}, ±\frac{5}{3}, ±\frac{5}{6}.
By methodically testing these possible roots through processes such as synthetic division or substitution, we can pinpoint which, if any, are actual roots of the polynomial equation.
Polynomial Roots
Polynomial roots, also known as zeros or x-intercepts, are the values for which the polynomial is equal to zero. Finding these roots is crucial for fully understanding the behavior of polynomial functions and solving polynomial equations. Roots can be real or complex, and their multiplicity can vary, which influences the shape and the graph of the polynomial function.
In the context of the exercise \(6x^3 + 25x^2 - 24x + 5 = 0\), we aim to determine the roots of this cubic polynomial. After applying the Rational Root Theorem, synthetic division is used to test possible rational roots. Once we identify one rational root, the polynomial can be factored or simplified into a lower-degree polynomial. In this case, after finding that -5 is a root, we are left with a quadratic polynomial \(6x^2 + x - 1\) to further analyze and find the remaining roots. Knowing all the roots helps us to fully solve the equation and provides insight into the polynomial's graph and x-intercepts.
In the context of the exercise \(6x^3 + 25x^2 - 24x + 5 = 0\), we aim to determine the roots of this cubic polynomial. After applying the Rational Root Theorem, synthetic division is used to test possible rational roots. Once we identify one rational root, the polynomial can be factored or simplified into a lower-degree polynomial. In this case, after finding that -5 is a root, we are left with a quadratic polynomial \(6x^2 + x - 1\) to further analyze and find the remaining roots. Knowing all the roots helps us to fully solve the equation and provides insight into the polynomial's graph and x-intercepts.
Quadratic Formula
The quadratic formula is a powerful algebraic tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It states that the solutions to the equation can be found using the formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). This process involves calculating the discriminant \(b^2-4ac\), which determines the number and nature of the roots.
For instance, in our exercise, after discovering the root -5 of the cubic polynomial, we obtain the quadratic equation \(6x^2 + x - 1 = 0\). By applying the quadratic formula with values a = 6, b = 1, and c = -1, we are left with the calculation of \(x = \frac{-1 \pm \sqrt{1^2-4(6)(-1)}}{2(6)}\), which simplifies to \(x = -\frac{1}{6}\) and \(x = \frac{1}{3}\). These two solutions are the remaining roots of the original cubic equation. Understanding the quadratic formula is crucial as it is a reliable method that applies to any quadratic equation and is the foundation for tackling more complex algebraic problems.
For instance, in our exercise, after discovering the root -5 of the cubic polynomial, we obtain the quadratic equation \(6x^2 + x - 1 = 0\). By applying the quadratic formula with values a = 6, b = 1, and c = -1, we are left with the calculation of \(x = \frac{-1 \pm \sqrt{1^2-4(6)(-1)}}{2(6)}\), which simplifies to \(x = -\frac{1}{6}\) and \(x = \frac{1}{3}\). These two solutions are the remaining roots of the original cubic equation. Understanding the quadratic formula is crucial as it is a reliable method that applies to any quadratic equation and is the foundation for tackling more complex algebraic problems.
Other exercises in this chapter
Problem 21
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