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Chapter 11: Problem 50

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

Short Answer

Expert verified
The polynomial function is \( f(x) = -3x^3 - 6x^2 - 3x - 6 \).

Step by step solution

01

Identify the Zeros and Their Conjugates

The zeros given are -2, i, and -i. Since polynomial functions with real coefficients have complex roots that occur in conjugate pairs, we know that -i is the conjugate of i. So the polynomial must have the form: \( f(x) = k(x + 2)(x - i)(x + i) \) where k is a constant coefficient.
02

Simplify the Polynomial

Next, simplify the polynomial function by multiplying out the factors involving the imaginary roots: \( (x-i)(x+i) = x^2 + 1 \) Thus, the polynomial becomes: \( f(x) = k(x + 2)(x^2 + 1) \)
03

Expand the Polynomial

Now, expand the product to form a single polynomial expression: \( f(x) = k[(x + 2)(x^2 + 1)] \) \( = k[x^3 + x + 2x^2 + 2] \) \( = k(x^3 + 2x^2 + x + 2) \)
04

Determine the Constant k Using the Given Condition

Given that \( f(-3) = 30 \), substitute -3 into the polynomial and solve for k: \( f(-3) = k((-3)^3 + 2(-3)^2 + (-3) + 2) \) \( = k(-27 + 18 - 3 + 2) \) \( = k(-10) \) \( 30 = -10k \) \( k = -3 \)
05

Write the Final Polynomial

Substitute k back into the polynomial: \( f(x) = -3(x^3 + 2x^2 + x + 2) \) Expand to get: \( f(x) = -3x^3 - 6x^2 - 3x - 6 \)

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Key Concepts

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

Zeros of a Polynomial
A polynomial's zeros (also called roots) are the values of the variable that make the polynomial equal to zero. For example, if we have a polynomial function like \(P(x) = 0\), the solutions to this equation represent the zeros. In our given exercise, the zeros are -2, i, and -i. For polynomials with real coefficients, remember that complex zeros must come in conjugate pairs. This is why we have both i and -i as zeros.
Complex Conjugate Roots
Complex roots are solutions to polynomial equations that are not real numbers, and they appear in the form of \(a + bi\) where \(i\) is the imaginary unit. Its conjugate would be \(a - bi\). Since our polynomial has real coefficients, the complex roots appear as pairs; if \(i\) is a root, so is its conjugate, \(-i\). This property helps us construct the polynomial function accurately. In this exercise, since \(i\) is given as a zero, its conjugate \(-i\) must also be a zero.
Polynomial Expansion
Once we determine the zeros and understand that complex roots must appear in conjugate pairs, we can start building the polynomial. Starting from the factored form given the zeros, we have: \( f(x) = k(x + 2)(x - i)(x + i) \). To simplify this expression, we first deal with the complex conjugates: \( (x - i)(x + i) = x^2 + 1 \). Thus, the polynomial function simplifies to \(f(x) = k(x + 2)(x^2 + 1)\). Now, expand the polynomial by distributing: \( f(x) = k(x^3 + x + 2x^2 + 2) = k(x^3 + 2x^2 + x + 2) \).
Determining Constants
After expanding the polynomial, we need to find the constant \(k\). This is done using the additional condition provided in the exercise. We know \(f(-3) = 30\), so substitute \(-3\) into our expanded polynomial to solve for \(k\): \( f(-3) = k((-3)^3 + 2(-3)^2 + (-3) + 2) = k(-27 + 18 - 3 + 2) = k(-10) \). Given \(30 = -10k \), we find \(k = -3 \). Finally, substitute \(k\) back into the polynomial to get the final function: \( f(x) = -3(x^3 + 2x^2 + x + 2) = -3x^3 - 6x^2 - 3x - 6 \). This is our polynomial function with the given conditions.

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Most popular questions from this chapter

For each of the following, (a) show that the polynomial function has a zero between the two given integers and (b) approximate all real zeros to the nearest thousandth. \(f(x)=x^{4}+x^{3}-6 x^{2}-20 x-16 ;\) between -2 and -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 \(55-64 ?\)

Approximate all real zeros of each function to the nearest hundredth. \(f(x)=\sqrt{7} x^{3}+\sqrt{5} x^{2}+\sqrt{17}\)

We have seen the close connection between polynomial division and writing a quotient of polynomials in lowest terms after factoring the numerator. We can also show a connection between dividing one polynomial by another and factoring the first polynomial. letting $$ f(x)=2 x^{2}+5 x-12 $$ Evaluate \(f(-4)\)

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
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