Chapter 4: Problem 38
Find all the real zeros of the function. \(g(x)=3 x^3-25 x^2+58 x-40\)
Short Answer
Expert verified
The real roots of the polynomial \(g(x)=3 x^3-25 x^2+58 x-40\) are \(1\), \(2\), and \(5\).
Step by step solution
01
Apply the Rational Root Theorem
This theorem states that if a polynomial has a rational root, \(p/q\), then \(p\) divides the constant term (in this case 40) and \(q\) divides the leading coefficient (in this case 3). This means that the potential rational roots of this polynomial are \(±1, ±2, ±4, ±5, ±8, ±10, ±20, ±40, ±1/3, ±2/3\).
02
Test the Potential Rational Roots
Substitute each of the potential rational roots into the function and see if it evaluates to zero. Testing these values, we find that \(1\), \(2\), and \(5\) are roots of the function.
03
Re-write polynomial in factored form
Now that we know the roots of the polynomial we can write it in its factored form: \(g(x)=3(x-1)(x-2)(x-5)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Rational Root Theorem
The Rational Root Theorem is an invaluable tool when dealing with polynomial equations, especially when you are tasked with finding rational zeros. It's like a roadmap that gives you potential directions to explore. Here's how it works: if you're dealing with a polynomial of the form \(a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0\), it suggests that any rational solution \(\frac{p}{q}\) must fulfill two conditions.
- The numerator \(p\) must be a factor of the constant term \(a_0\).
- The denominator \(q\) must be a factor of the leading coefficient \(a_n\).
Polynomial Factoring
Factoring a polynomial is like breaking a complex machine into simple, digestible components. Once rational roots are discovered through the Rational Root Theorem, they can be used to factor the polynomial. This involves expressing the polynomial as a product of simpler polynomials.
For the polynomial \(g(x) = 3x^3 - 25x^2 + 58x - 40\), once we found the roots \(x = 1, 2, 5\), it becomes straightforward to factor. We express \(g(x)\) as \(3(x - 1)(x - 2)(x - 5)\). Factoring not only gives us a cleaner representation but also allows us to solve for roots effortlessly. Simply solve the factors set to zero, and you get the solution for \(x\). It also mirrors the algebraic structure and behavior of the function, which is particularly useful for graphing and understanding its properties.
For the polynomial \(g(x) = 3x^3 - 25x^2 + 58x - 40\), once we found the roots \(x = 1, 2, 5\), it becomes straightforward to factor. We express \(g(x)\) as \(3(x - 1)(x - 2)(x - 5)\). Factoring not only gives us a cleaner representation but also allows us to solve for roots effortlessly. Simply solve the factors set to zero, and you get the solution for \(x\). It also mirrors the algebraic structure and behavior of the function, which is particularly useful for graphing and understanding its properties.
Polynomial Functions
Polynomial functions are a cornerstone in algebra. They encompass expressions that combine constants, variables, and exponents into succinct mathematical models. The function \(g(x) = 3x^3 - 25x^2 + 58x - 40\) is a perfect example, illustrating how powerful these expressions can be in encapsulating complex relationships.
- Degree: The highest exponent (3, in this case) tells us the polynomial's degree, influencing its basic shape and the number of potential real roots.
- Coefficient: These numbers (like 3, -25, 58, -40 in the original polynomial) determine the function's growth rate and direction.