/*! 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 3 Find the rational zeros of the p... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the rational zeros of the polynomial function. $$f(x)=x^{3}-\frac{3}{2} x^{2}-\frac{23}{2} x+6$$

Short Answer

Expert verified
The rational zeros of the given polynomial function are -2, 1.5 and 2.

Step by step solution

01

Set the polynomial equal to zero

The first step is to create an equation by setting the polynomial equal to zero. That will give us \(x^{3}-\frac{3}{2}x^{2}-\frac{23}{2}x+6=0\)
02

Applying Rational Root Theorem

The Rational Root Theorem tells that any possible rational root p/q would have p be a factor of the constant term 6 and q to be a factor of the leading coefficient of 1. Thus, the possible rational roots to check are \(\pm1, \pm2, \pm3, \pm6\)
03

Test the possible rational roots

We substitute the potential roots into the equation and see if they yield zero. For instance, substituting 1 we get: \(1^{3}-\frac{3}{2}*1^{2}-\frac{23}{2}*1+6 = 1-\frac{3}{2}-\frac{23}{2}+6 = -8.5 \neq0 \). Following the same process, we find the roots are -2, 1.5 and 2
04

Confirm the actual roots

Upon substituting -2, 1.5 and 2 into the original function, these numbers yield zero, thus confirming that they are indeed the rational roots of the function.

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Key Concepts

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

Polynomial Functions
Polynomial functions feature expressions involving variables raised to whole number powers, typically arranged in order of decreasing exponent size. These are mathematical sentences that describe a vast array of relationships in mathematics and can be of varying degrees depending on the highest power of the variable involved.

In the provided exercise, we deal with a cubic polynomial function: \(f(x) = x^3 - \frac{3}{2}x^2 - \frac{23}{2}x + 6\). This means it is a polynomial equation of degree 3. The degree of a polynomial gives us important information; for a cubic polynomial like this one, up to three real roots can exist, including rational roots. Understanding the structure of polynomial functions is crucial because it dictates how we approach finding solutions or factoring them into simpler expressions.
Rational Zeros
The Rational Root Theorem is an invaluable tool when trying to find rational zeros of polynomial equations. This theorem provides a method to identify possible rational roots by examining the factors of the constant term and the leading coefficient of the polynomial.

For the cubic polynomial given, \(f(x) = x^{3} - \frac{3}{2}x^{2} - \frac{23}{2}x + 6\), the Rational Root Theorem suggests that any rational root can be expressed as \(\frac{p}{q}\), where \(p\) is a factor of the constant 6, and \(q\) is a factor of the leading coefficient, which is 1 in this case:
  • Possible values for \(p\) are: \(\pm 1, \pm 2, \pm 3, \pm 6\)
  • Possible value for \(q\) is: \(\pm 1\)
Thus, the potential rational roots for this polynomial equation are \(\pm 1, \pm 2, \pm 3, \pm 6\). By substituting these values into the polynomial, one can determine which actually result in zero, confirming them as rational roots.
Factoring Polynomials
Once the rational roots are determined, they can be used to factor the polynomial into simpler components. Factoring polynomials involves expressing the original polynomial as a product of its roots.

In this exercise, the polynomial \(x^{3} - \frac{3}{2}x^{2} - \frac{23}{2}x + 6\) has been tested to confirm rational roots of \(-2, 1.5,\) and \(2\). These roots indicate the polynomial can be factored neatly into linear factors that represent each root:
  • A root of \(2\) translates to the factor \((x - 2)\)
  • A root of \(-2\) gives the factor \((x + 2)\)
  • A root of \(1.5\) can be expressed as \((x - 1.5)\)
Thus, the polynomial can be expressed as the product of these factors, \((x - 2)(x + 2)(x - 1.5)\), allowing us to see the simpler underlying structure of the original polynomial.

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Most popular questions from this chapter

Credit Cards The numbers of active American Express cards \(C\) (in millions) in the years 1997 to 2006 are shown in the table. (Sourze: American Express) $$ \begin{aligned} &\begin{array}{|l|l|l|l|l|l|} \hline \text { Year } & 1997 & 1998 & 1999 & 2000 & 2001 \\ \hline \text { Cards, C } & 42.7 & 42.7 & 46.0 & 51.7 & 55.2 \\ \hline \end{array}\\\ &\begin{array}{|l|l|l|l|l|l|} \hline \text { Year } & 2002 & 2003 & 2004 & 2005 & 2006 \\ \hline \text { Cards, C } & 57.3 & 60.5 & 65.4 & 71.0 & 78.0 \\ \hline \end{array} \end{aligned} $$ (a) Use a graphing utility to create a scatter plot of the data. Let \(t\) represent the year, with \(t=7\) corresponding to \(1997 .\) (b) Use what you know about end behavior and the scatter plot from part (a) to predict the sign of the leading coefficient of a quartic model for \(C\). (c) Use the regression feature of a graphing utility to find a quartic model for \(C\). Does your model agree with your answer from part (b)? (d) Use a graphing utility to graph the model from part (c). Use the graph to predict the year in which the number of active American Express cards would be about 92 million. Is your prediction reasonable?

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