/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 Find a polynomial function \(f(x... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a polynomial function \(f(x)\) of degree 3 with real coefficients that satisfies the given conditions. See Example 4. Zeros of \(-3,1,\) and \(4 ; f(2)=30\)

Short Answer

Expert verified
\[ f(x) = -3(x + 3)(x - 1)(x - 4) \]

Step by step solution

01

Understand the Structure of a Cubic Polynomial

Start by recognizing the general form of a cubic polynomial with given roots. A polynomial of degree 3 with real coefficients and zeros at oindent oindent oindent: -3, 1, and 4 can be expressed as: oindent \[ f(x) = a(x + 3)(x - 1)(x - 4) \]. oindent oindent oindent.\ oindent The constant 'a' is unknown and will be determined. oindent.
02

Apply Given Function Value

Use the condition oindent oindent \f(2) = 30 \ to find the value of 'a'. Substitute 2 into the polynomial: oindent \[ f(2) = a(2 + 3)(2 - 1)(2 - 4) \]. oindent This yields: oindent \[ 30 = a(5)(1)(-2) \].
03

Solve for 'a'

Solve for 'a' by dividing both sides by -10: oindent \[ 30 = a(-10) \] oindent oindent oindent \[ a = \frac{30}{-10} \] \[ a = -3 \].
04

Form the Final Polynomial

Substitute the value of 'a' back into the polynomial to get the function: oindent \[ f(x) = -3(x + 3)(x - 1)(x - 4) \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Zeros of Polynomial
To find the zeros of a cubic polynomial function, you need to understand that the 'zeros' (or 'roots') are the values of x which make the polynomial equal to zero. In other words, if you have a polynomial function denoted as \(f(x)\), the zeros are the x-values for which \(f(x) = 0\). For a polynomial of degree 3, you generally expect it to have three zeros. In our given problem, the polynomial is specified to have zeros at -3, 1, and 4. This means the polynomial touches or intersects the x-axis at these points. Mathematically, you can represent a cubic polynomial with these zeros as \[f(x) = a(x + 3)(x - 1)(x - 4)\]. Here, 'a' is a constant that affects the vertical stretch or compression of the graph but does not shift the location of the zeros.
Polynomial Coefficients
The coefficients of a polynomial are the numerical factors in terms of the polynomial. In a cubic polynomial, these coefficients will be associated with different powers of x. When given the zeros of the polynomial, we can write it in a factored form. For example, with zeros at -3, 1, and 4, the polynomial is expressed as \(f(x) = a(x + 3)(x - 1)(x - 4)\). To determine the specific coefficients for each power of x, we need to expand this equation. However, without expanding, we focus on finding the appropriate value for the constant 'a' first. This constant is crucial because it gives us the final form of the polynomial with its accurate coefficients.
Solving for Constant a
To fully determine the polynomial function, we need to solve for the constant 'a.' This is achieved using another condition provided in the problem: \(f(2) = 30\). Start by substituting x = 2 into the polynomial: \[f(2) = a(2 + 3)(2 - 1)(2 - 4)\] Simplify the expression inside the parentheses: \[f(2) = a (5) (1) (-2)\] Since \(f(2) = 30\), you have: \[30 = a (5) (1) (-2)\] Now, solve for 'a' by isolating it on one side of the equation: \[30 = a(-10)\] Divide both sides by -10 to find: \[a = \frac{30}{-10} = -3\] Finally, substitute \(a = -3\) back into the polynomial function to get: \[f(x) = -3(x + 3)(x - 1)(x - 4)\] This is the complete polynomial function with the constant 'a' known, satisfying all given conditions.

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Most popular questions from this chapter

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