Answer :
We can use the Rational Root Theorem. This theorem tells us that any rational zero of a polynomial
[tex]$$
a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0
$$[/tex]
must be of the form
[tex]$$
\frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] is a factor of the constant term [tex]$a_0$[/tex] and [tex]$q$[/tex] is a factor of the leading coefficient [tex]$a_n$[/tex]. We now apply this to each function.
–––––––––––––––––––––––––––––––––––
1. For
[tex]$$
f(x) = 3x^4 - x^3 + 4:
$$[/tex]
- The constant term is [tex]$4$[/tex]. Its positive factors are [tex]$1$[/tex], [tex]$2$[/tex], and [tex]$4$[/tex], so the possible values of [tex]$p$[/tex] are
[tex]$$
\pm 1,\, \pm 2,\, \pm 4.
$$[/tex]
- The leading coefficient is [tex]$3$[/tex]. Its positive factors are [tex]$1$[/tex] and [tex]$3$[/tex], so the possible values of [tex]$q$[/tex] are
[tex]$$
\pm 1,\, \pm 3.
$$[/tex]
Thus, the possible rational zeros are obtained by taking each factor of [tex]$4$[/tex] divided by each factor of [tex]$3$[/tex]:
[tex]$$
\pm 1,\ \pm 2,\ \pm 4,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3}.
$$[/tex]
Writing them in increasing order (when expressed as decimals):
[tex]$$
-4,\ -2,\ -\frac{4}{3},\ -1,\ -\frac{2}{3},\ -\frac{1}{3},\ \frac{1}{3},\ \frac{2}{3},\ 1,\ \frac{4}{3},\ 2,\ 4.
$$[/tex]
–––––––––––––––––––––––––––––––––––
2. For
[tex]$$
f(x) = x^4 + 7x^3 - 15:
$$[/tex]
- The constant term is [tex]$-15$[/tex]. Its factors are [tex]$1$[/tex], [tex]$3$[/tex], [tex]$5$[/tex], and [tex]$15$[/tex], so the possible values of [tex]$p$[/tex] are
[tex]$$
\pm 1,\, \pm 3,\, \pm 5,\, \pm 15.
$$[/tex]
- The leading coefficient is [tex]$1$[/tex]. Its only factors are [tex]$1$[/tex] (or [tex]$-1$[/tex]), so the possible values of [tex]$q$[/tex] are
[tex]$$
\pm 1.
$$[/tex]
Thus, the possible rational zeros are
[tex]$$
\pm 1,\ \pm 3,\ \pm 5,\ \pm 15.
$$[/tex]
Sorted from smallest to largest:
[tex]$$
-15,\ -5,\ -3,\ -1,\ 1,\ 3,\ 5,\ 15.
$$[/tex]
–––––––––––––––––––––––––––––––––––
3. For
[tex]$$
f(x)=x^3-6x^2-8x+24:
$$[/tex]
- The constant term is [tex]$24$[/tex]. Its factors are [tex]$1$[/tex], [tex]$2$[/tex], [tex]$3$[/tex], [tex]$4$[/tex], [tex]$6$[/tex], [tex]$8$[/tex], [tex]$12$[/tex], and [tex]$24$[/tex], so the possible values of [tex]$p$[/tex] are
[tex]$$
\pm 1,\, \pm 2,\, \pm 3,\, \pm 4,\, \pm 6,\, \pm 8,\, \pm 12,\, \pm 24.
$$[/tex]
- The leading coefficient is [tex]$1$[/tex], whose only factors are [tex]$1$[/tex] (and [tex]$-1$[/tex]), so the possible value of [tex]$q$[/tex] is
[tex]$$
\pm 1.
$$[/tex]
Thus, the possible rational zeros are
[tex]$$
\pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 8,\ \pm 12,\ \pm 24.
$$[/tex]
When arranged in order:
[tex]$$
-24,\ -12,\ -8,\ -6,\ -4,\ -3,\ -2,\ -1,\ 1,\ 2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24.
$$[/tex]
–––––––––––––––––––––––––––––––––––
Summary of the answers:
[tex]\[
\begin{array}{ll}
\textbf{23. } f(x)=3x^4-x^3+4: & \pm 1,\ \pm 2,\ \pm 4,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3} \\
\textbf{24. } f(x)=x^4+7x^3-15: & \pm 1,\ \pm 3,\ \pm 5,\ \pm 15 \\
\textbf{25. } f(x)=x^3-6x^2-8x+24: & \pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 8,\ \pm 12,\ \pm 24
\end{array}
\][/tex]
These are all the possible rational zeros for the given functions.
[tex]$$
a_n x^n + a_{n-1} x^{n-1} + \cdots + a_0
$$[/tex]
must be of the form
[tex]$$
\frac{p}{q},
$$[/tex]
where [tex]$p$[/tex] is a factor of the constant term [tex]$a_0$[/tex] and [tex]$q$[/tex] is a factor of the leading coefficient [tex]$a_n$[/tex]. We now apply this to each function.
–––––––––––––––––––––––––––––––––––
1. For
[tex]$$
f(x) = 3x^4 - x^3 + 4:
$$[/tex]
- The constant term is [tex]$4$[/tex]. Its positive factors are [tex]$1$[/tex], [tex]$2$[/tex], and [tex]$4$[/tex], so the possible values of [tex]$p$[/tex] are
[tex]$$
\pm 1,\, \pm 2,\, \pm 4.
$$[/tex]
- The leading coefficient is [tex]$3$[/tex]. Its positive factors are [tex]$1$[/tex] and [tex]$3$[/tex], so the possible values of [tex]$q$[/tex] are
[tex]$$
\pm 1,\, \pm 3.
$$[/tex]
Thus, the possible rational zeros are obtained by taking each factor of [tex]$4$[/tex] divided by each factor of [tex]$3$[/tex]:
[tex]$$
\pm 1,\ \pm 2,\ \pm 4,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3}.
$$[/tex]
Writing them in increasing order (when expressed as decimals):
[tex]$$
-4,\ -2,\ -\frac{4}{3},\ -1,\ -\frac{2}{3},\ -\frac{1}{3},\ \frac{1}{3},\ \frac{2}{3},\ 1,\ \frac{4}{3},\ 2,\ 4.
$$[/tex]
–––––––––––––––––––––––––––––––––––
2. For
[tex]$$
f(x) = x^4 + 7x^3 - 15:
$$[/tex]
- The constant term is [tex]$-15$[/tex]. Its factors are [tex]$1$[/tex], [tex]$3$[/tex], [tex]$5$[/tex], and [tex]$15$[/tex], so the possible values of [tex]$p$[/tex] are
[tex]$$
\pm 1,\, \pm 3,\, \pm 5,\, \pm 15.
$$[/tex]
- The leading coefficient is [tex]$1$[/tex]. Its only factors are [tex]$1$[/tex] (or [tex]$-1$[/tex]), so the possible values of [tex]$q$[/tex] are
[tex]$$
\pm 1.
$$[/tex]
Thus, the possible rational zeros are
[tex]$$
\pm 1,\ \pm 3,\ \pm 5,\ \pm 15.
$$[/tex]
Sorted from smallest to largest:
[tex]$$
-15,\ -5,\ -3,\ -1,\ 1,\ 3,\ 5,\ 15.
$$[/tex]
–––––––––––––––––––––––––––––––––––
3. For
[tex]$$
f(x)=x^3-6x^2-8x+24:
$$[/tex]
- The constant term is [tex]$24$[/tex]. Its factors are [tex]$1$[/tex], [tex]$2$[/tex], [tex]$3$[/tex], [tex]$4$[/tex], [tex]$6$[/tex], [tex]$8$[/tex], [tex]$12$[/tex], and [tex]$24$[/tex], so the possible values of [tex]$p$[/tex] are
[tex]$$
\pm 1,\, \pm 2,\, \pm 3,\, \pm 4,\, \pm 6,\, \pm 8,\, \pm 12,\, \pm 24.
$$[/tex]
- The leading coefficient is [tex]$1$[/tex], whose only factors are [tex]$1$[/tex] (and [tex]$-1$[/tex]), so the possible value of [tex]$q$[/tex] is
[tex]$$
\pm 1.
$$[/tex]
Thus, the possible rational zeros are
[tex]$$
\pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 8,\ \pm 12,\ \pm 24.
$$[/tex]
When arranged in order:
[tex]$$
-24,\ -12,\ -8,\ -6,\ -4,\ -3,\ -2,\ -1,\ 1,\ 2,\ 3,\ 4,\ 6,\ 8,\ 12,\ 24.
$$[/tex]
–––––––––––––––––––––––––––––––––––
Summary of the answers:
[tex]\[
\begin{array}{ll}
\textbf{23. } f(x)=3x^4-x^3+4: & \pm 1,\ \pm 2,\ \pm 4,\ \pm \frac{1}{3},\ \pm \frac{2}{3},\ \pm \frac{4}{3} \\
\textbf{24. } f(x)=x^4+7x^3-15: & \pm 1,\ \pm 3,\ \pm 5,\ \pm 15 \\
\textbf{25. } f(x)=x^3-6x^2-8x+24: & \pm 1,\ \pm 2,\ \pm 3,\ \pm 4,\ \pm 6,\ \pm 8,\ \pm 12,\ \pm 24
\end{array}
\][/tex]
These are all the possible rational zeros for the given functions.
