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Chapter 11: Problem 6

Fill in each blank with the appropriate response. Consider the following function. $$ \begin{array}{l} f(x)=2 x^{4}+6 x^{3}-5 x^{2}+3 x+8 \\ f(x)=(x-2)\left(2 x^{3}+10 x^{2}+15 x+33\right)+74 \end{array} $$ By inspection, we can state that \(f(2)=\) ____.

Short Answer

Expert verified
74

Step by step solution

01

Identify Given Information

We are given the function \( f(x) \) and its factored form.
02

Understand the Factored Form

The factored form of \( f(x) \) is \( f(x) = (x-2)(2x^3 + 10x^2 + 15x + 33) + 74 \).
03

Evaluate at x=2

Substitute \( x = 2 \) into the factored form: \( f(2) = (2-2)(2(2)^3 + 10(2)^2 + 15(2) + 33) + 74 \).
04

Simplify the Expression

Notice that \( (2-2) \) is zero. Hence, \( f(2) = 0 + 74 \).
05

Final Value

Therefore, \( f(2) = 74 \).

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Key Concepts

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

Factored Form of Polynomials
The factored form of a polynomial can make evaluating functions easier. When a polynomial is factored, it is expressed as a product of its simpler components, often multiplied by a constant or added to a constant term. This form helps in breaking down complex expressions.
For example, consider the polynomial function \( f(x) \) given:
\[ f(x)=2x^{4}+6x^{3}-5x^{2}+3x+8 \]
Its factored form is:
\[ f(x)=(x-2)(2x^3+10x^2+15x+33)+74 \]
In this example, expressing \( f(x) \) in factored form allows us to see the roots (like \( x=2 \)) directly influence the value of the function.
Evaluating Functions
Evaluating functions means finding the value of the function for a given value of its variable. This involves substituting the value into the function and simplifying.
Let's use the given exercise as an example. We need to find \( f(2) \) for the function:
\[ f(x)=(x-2)(2x^3+10x^2+15x+33)+74 \]
To evaluate, we substitute \( x=2 \) into the function:
\[ f(2) = (2-2)(2(2)^3 + 10(2)^2 + 15(2) + 33) + 74 \]
This approach can quickly show how each part of the expression behaves.
Substitution in Algebra
Substitution involves replacing a variable with a given numerical value. This is a key step in evaluating polynomials. It simplifies the expression to a form that can be easily calculated.
In our exercise, we substituted \( x=2 \) into the factored form:
\[ f(2) = (2-2)(2(2)^3 + 10(2)^2 + 15(2) + 33) + 74 \]
Observing the term \( (2-2)=0 \), we see that the entire product becomes zero:
\[ (2-2)(2(2)^3 + 10(2)^2 + 15(2) + 33) = 0 \]
Hence, only the constant term remains:
\[ f(2)=0 + 74=74 \]
Substitution simplifies complex algebra by breaking it down into manageable parts.

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Most popular questions from this chapter

Show that the real zeros of each polynomial function satisfy the given conditions. \(f(x)=3 x^{4}+2 x^{3}-4 x^{2}+x-1 ;\) no real zero greater than 1

The table contains incidence ratios by age for deaths due to coronary heart disease (CHD) and lung cancer (LC) when comparing smokers ( \(21-39\) cigarettes per day) to nonsmokers. $$\begin{array}{|c|c|c|}\hline \text { Age } & \text { CHD } & \text { LC } \\\\\hline 55-64 & 1.9 & 10 \\\\\hline 65-74 & 1.7 & 9 \\\\\hline\end{array}$$ The incidence ratio of 10 means that smokers are 10 times more likely than nonsmokers to die of lung cancer between the ages of 55 and \(64 .\) If the incidence ratio is \(x,\) then the percent \(P\) (in decimal form) of deaths caused by smoking can be calculated using the rational function$$P(x)=\frac{x-1}{x}$$ (Data from Walker, A., Observation and Inference: An Introduction to the Methods of Epidemiology, Epidemiology 91影视 Inc.) (a) As \(x\) increases, what value does \(P(x)\) approach? (b) Why might the incidence ratios be slightly less for ages \(65-74\) than for ages \(55-64 ?\)

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