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Answer :
To find the possible rational roots of
[tex]$$3x^5 - 10x^4 + 12x^3 - 15x^2 + 25x - 35 = 0,$$[/tex]
we can use the Rational Root Theorem. This theorem tells us that any rational solution (root) of the polynomial equation can be written in the form
[tex]$$x = \pm \frac{p}{q},$$[/tex]
where
- [tex]$p$[/tex] is a factor of the constant term, and
- [tex]$q$[/tex] is a factor of the leading coefficient.
Let’s work through this step by step.
1. Identify the Constant and Leading Coefficient
- The constant term is [tex]$-35$[/tex]. Its factors (ignoring the sign) are:
[tex]$$1,\, 5,\, 7,\, 35.$$[/tex]
- The leading coefficient is [tex]$3$[/tex]. Its factors (ignoring the sign) are:
[tex]$$1,\, 3.$$[/tex]
2. List the Possible Values for [tex]$x$[/tex]
According to the theorem, the candidate rational roots are given by
[tex]$$x = \pm \frac{p}{q},$$[/tex]
where [tex]$p \in \{1, 5, 7, 35\}$[/tex] and [tex]$q \in \{1, 3\}$[/tex]. This gives the following list of potential rational roots:
[tex]$$\begin{array}{cccc}
\pm \frac{1}{1}, & \pm \frac{5}{1}, & \pm \frac{7}{1}, & \pm \frac{35}{1},\\[1mm]
\pm \frac{1}{3}, & \pm \frac{5}{3}, & \pm \frac{7}{3}, & \pm \frac{35}{3}.
\end{array}$$[/tex]
3. Write the Final List in a Clear Format
Arranging the factors, the potential rational solutions are:
[tex]$$\pm 1, \quad \pm 5, \quad \pm 7, \quad \pm 35, \quad \pm \frac{1}{3}, \quad \pm \frac{5}{3}, \quad \pm \frac{7}{3}, \quad \pm \frac{35}{3}.$$[/tex]
4. Sorting by Absolute Value (Optional)
Sometimes, it is helpful to list the roots in order of increasing absolute value. When sorted by the absolute value, the list becomes:
[tex]$$\pm \frac{1}{3}, \quad \pm 1, \quad \pm \frac{5}{3}, \quad \pm 5, \quad \pm \frac{7}{3}, \quad \pm 7, \quad \pm \frac{35}{3}, \quad \pm 35.$$[/tex]
This is the complete set of potential rational roots determined by the Rational Root Theorem.
Thus, the list of all potential rational solutions to the equation
[tex]$$3x^5 - 10x^4 + 12x^3 - 15x^2 + 25x - 35 = 0$$[/tex]
is
[tex]$$\boxed{\pm \frac{1}{3}, \; \pm 1, \; \pm \frac{5}{3}, \; \pm 5, \; \pm \frac{7}{3}, \; \pm 7, \; \pm \frac{35}{3}, \; \pm 35.}$$[/tex]
[tex]$$3x^5 - 10x^4 + 12x^3 - 15x^2 + 25x - 35 = 0,$$[/tex]
we can use the Rational Root Theorem. This theorem tells us that any rational solution (root) of the polynomial equation can be written in the form
[tex]$$x = \pm \frac{p}{q},$$[/tex]
where
- [tex]$p$[/tex] is a factor of the constant term, and
- [tex]$q$[/tex] is a factor of the leading coefficient.
Let’s work through this step by step.
1. Identify the Constant and Leading Coefficient
- The constant term is [tex]$-35$[/tex]. Its factors (ignoring the sign) are:
[tex]$$1,\, 5,\, 7,\, 35.$$[/tex]
- The leading coefficient is [tex]$3$[/tex]. Its factors (ignoring the sign) are:
[tex]$$1,\, 3.$$[/tex]
2. List the Possible Values for [tex]$x$[/tex]
According to the theorem, the candidate rational roots are given by
[tex]$$x = \pm \frac{p}{q},$$[/tex]
where [tex]$p \in \{1, 5, 7, 35\}$[/tex] and [tex]$q \in \{1, 3\}$[/tex]. This gives the following list of potential rational roots:
[tex]$$\begin{array}{cccc}
\pm \frac{1}{1}, & \pm \frac{5}{1}, & \pm \frac{7}{1}, & \pm \frac{35}{1},\\[1mm]
\pm \frac{1}{3}, & \pm \frac{5}{3}, & \pm \frac{7}{3}, & \pm \frac{35}{3}.
\end{array}$$[/tex]
3. Write the Final List in a Clear Format
Arranging the factors, the potential rational solutions are:
[tex]$$\pm 1, \quad \pm 5, \quad \pm 7, \quad \pm 35, \quad \pm \frac{1}{3}, \quad \pm \frac{5}{3}, \quad \pm \frac{7}{3}, \quad \pm \frac{35}{3}.$$[/tex]
4. Sorting by Absolute Value (Optional)
Sometimes, it is helpful to list the roots in order of increasing absolute value. When sorted by the absolute value, the list becomes:
[tex]$$\pm \frac{1}{3}, \quad \pm 1, \quad \pm \frac{5}{3}, \quad \pm 5, \quad \pm \frac{7}{3}, \quad \pm 7, \quad \pm \frac{35}{3}, \quad \pm 35.$$[/tex]
This is the complete set of potential rational roots determined by the Rational Root Theorem.
Thus, the list of all potential rational solutions to the equation
[tex]$$3x^5 - 10x^4 + 12x^3 - 15x^2 + 25x - 35 = 0$$[/tex]
is
[tex]$$\boxed{\pm \frac{1}{3}, \; \pm 1, \; \pm \frac{5}{3}, \; \pm 5, \; \pm \frac{7}{3}, \; \pm 7, \; \pm \frac{35}{3}, \; \pm 35.}$$[/tex]
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