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Answer :
We are given the polynomial
[tex]$$
f(x)=4x^3+19x^2-41x+9.
$$[/tex]
We will solve the problem step-by-step.
–––––––
Step 1. Maximum Number of Zeros
A polynomial of degree [tex]$n$[/tex] has at most [tex]$n$[/tex] real zeros. Since the given polynomial has degree 3, it can have at most
[tex]$$
3 \text{ zeros}.
$$[/tex]
–––––––
Step 2. Possible Rational Zeros
The Rational Zero Theorem tells us that any rational zero, expressed in simplest form as [tex]$\frac{p}{q}$[/tex], must have a numerator [tex]$p$[/tex] that is a factor of the constant term (9) and a denominator [tex]$q$[/tex] that is a factor of the leading coefficient (4).
The factors of 9 are:
[tex]$$
\pm1,\ \pm3,\ \pm9.
$$[/tex]
The factors of 4 are:
[tex]$$
\pm1,\ \pm2,\ \pm4.
$$[/tex]
Thus, the possible rational zeros are all numbers of the form
[tex]$$
\pm \frac{p}{q} \quad\text{where } p \in \{1,3,9\} \text{ and } q \in \{1,2,4\}.
$$[/tex]
Listing these, we have:
[tex]\[
\begin{aligned}
\pm 1, &\quad\pm 3, \quad\pm 9,\\[1mm]
\pm \frac{1}{2}, &\quad\pm \frac{3}{2}, \quad\pm \frac{9}{2},\\[1mm]
\pm \frac{1}{4}, &\quad\pm \frac{3}{4}, \quad\pm \frac{9}{4}.
\end{aligned}
\][/tex]
–––––––
Step 3. Leading Term Test
The leading term of our polynomial is [tex]$4x^3$[/tex]. To understand the end behavior of the polynomial, we look at this term.
- Since the degree is odd ([tex]$3$[/tex]) and the leading coefficient is positive ([tex]$4$[/tex]), the polynomial behaves as follows:
- As [tex]$x \to +\infty$[/tex], the term [tex]$4x^3$[/tex] (and hence [tex]$f(x)$[/tex]) grows toward [tex]$+\infty$[/tex].
- As [tex]$x \to -\infty$[/tex], the term [tex]$4x^3$[/tex] (and hence [tex]$f(x)$[/tex]) goes toward [tex]$-\infty$[/tex].
Thus, by the Leading Term Test, the graph of the polynomial will fall to negative infinity on the left and rise to positive infinity on the right.
–––––––
Summary of Answers
1. The polynomial has at most [tex]$$3$$[/tex] zeros.
2. The possible rational zeros, by the Rational Zero Theorem, are:
[tex]$$
\pm 1,\ \pm 3,\ \pm 9,\ \pm \frac{1}{2},\ \pm \frac{3}{2},\ \pm \frac{9}{2},\ \pm \frac{1}{4},\ \pm \frac{3}{4},\ \pm \frac{9}{4}.
$$[/tex]
3. By the Leading Term Test, because the polynomial is of odd degree with a positive leading coefficient, as [tex]$$x \to -\infty,\ f(x) \to -\infty$$[/tex] and as [tex]$$x \to +\infty,\ f(x) \to +\infty.$$[/tex]
[tex]$$
f(x)=4x^3+19x^2-41x+9.
$$[/tex]
We will solve the problem step-by-step.
–––––––
Step 1. Maximum Number of Zeros
A polynomial of degree [tex]$n$[/tex] has at most [tex]$n$[/tex] real zeros. Since the given polynomial has degree 3, it can have at most
[tex]$$
3 \text{ zeros}.
$$[/tex]
–––––––
Step 2. Possible Rational Zeros
The Rational Zero Theorem tells us that any rational zero, expressed in simplest form as [tex]$\frac{p}{q}$[/tex], must have a numerator [tex]$p$[/tex] that is a factor of the constant term (9) and a denominator [tex]$q$[/tex] that is a factor of the leading coefficient (4).
The factors of 9 are:
[tex]$$
\pm1,\ \pm3,\ \pm9.
$$[/tex]
The factors of 4 are:
[tex]$$
\pm1,\ \pm2,\ \pm4.
$$[/tex]
Thus, the possible rational zeros are all numbers of the form
[tex]$$
\pm \frac{p}{q} \quad\text{where } p \in \{1,3,9\} \text{ and } q \in \{1,2,4\}.
$$[/tex]
Listing these, we have:
[tex]\[
\begin{aligned}
\pm 1, &\quad\pm 3, \quad\pm 9,\\[1mm]
\pm \frac{1}{2}, &\quad\pm \frac{3}{2}, \quad\pm \frac{9}{2},\\[1mm]
\pm \frac{1}{4}, &\quad\pm \frac{3}{4}, \quad\pm \frac{9}{4}.
\end{aligned}
\][/tex]
–––––––
Step 3. Leading Term Test
The leading term of our polynomial is [tex]$4x^3$[/tex]. To understand the end behavior of the polynomial, we look at this term.
- Since the degree is odd ([tex]$3$[/tex]) and the leading coefficient is positive ([tex]$4$[/tex]), the polynomial behaves as follows:
- As [tex]$x \to +\infty$[/tex], the term [tex]$4x^3$[/tex] (and hence [tex]$f(x)$[/tex]) grows toward [tex]$+\infty$[/tex].
- As [tex]$x \to -\infty$[/tex], the term [tex]$4x^3$[/tex] (and hence [tex]$f(x)$[/tex]) goes toward [tex]$-\infty$[/tex].
Thus, by the Leading Term Test, the graph of the polynomial will fall to negative infinity on the left and rise to positive infinity on the right.
–––––––
Summary of Answers
1. The polynomial has at most [tex]$$3$$[/tex] zeros.
2. The possible rational zeros, by the Rational Zero Theorem, are:
[tex]$$
\pm 1,\ \pm 3,\ \pm 9,\ \pm \frac{1}{2},\ \pm \frac{3}{2},\ \pm \frac{9}{2},\ \pm \frac{1}{4},\ \pm \frac{3}{4},\ \pm \frac{9}{4}.
$$[/tex]
3. By the Leading Term Test, because the polynomial is of odd degree with a positive leading coefficient, as [tex]$$x \to -\infty,\ f(x) \to -\infty$$[/tex] and as [tex]$$x \to +\infty,\ f(x) \to +\infty.$$[/tex]
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