Get the answers you've been looking for with the help of IDNLearn.com's expert community. Join our knowledgeable community and get detailed, reliable answers to all your questions.
Sagot :
To answer this question, we will use the Rational Zero Theorem. This theorem states that any rational solution of the polynomial equation [tex]\( P(x) = 0 \)[/tex] is a fraction [tex]\(\frac{p}{q}\)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term and [tex]\( q \)[/tex] is a factor of the leading coefficient.
Given the polynomial equation:
[tex]\[ 8x^3 + 50x^2 - 41x + 7 = 0 \][/tex]
1. Identify the constant term and its factors:
The constant term is [tex]\( 7 \)[/tex]. The factors of [tex]\( 7 \)[/tex] are:
[tex]\[ \pm 1, \pm 7 \][/tex]
2. Identify the leading coefficient and its factors:
The leading coefficient is [tex]\( 8 \)[/tex]. The factors of [tex]\( 8 \)[/tex] are:
[tex]\[ \pm 1, \pm 2, \pm 4, \pm 8 \][/tex]
3. Formulate all possible rational roots:
According to the Rational Zero Theorem, the possible rational roots are given by all possible values of [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term ([tex]\( \pm 1, \pm 7 \)[/tex]) and [tex]\( q \)[/tex] is a factor of the leading coefficient ([tex]\( \pm 1, \pm 2, \pm 4, \pm 8 \)[/tex]).
Combining these, we get the possible rational roots:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{4}, \pm \frac{7}{8} \][/tex]
Thus, listing all the possible rational roots, we have:
[tex]\[ \{\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}\} \][/tex]
Given the multiple-choice options:
A. [tex]\(\pm 1, \pm 7\)[/tex]
B. [tex]\(\pm 1, \pm 2, \pm 4, \pm 8\)[/tex]
C. [tex]\(\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{7}, \pm \frac{2}{7}, \pm \frac{4}{7}, \pm \frac{8}{7}\)[/tex]
D. [tex]\(\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}\)[/tex]
The correct answer is:
D. [tex]\(\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}\)[/tex]
Given the polynomial equation:
[tex]\[ 8x^3 + 50x^2 - 41x + 7 = 0 \][/tex]
1. Identify the constant term and its factors:
The constant term is [tex]\( 7 \)[/tex]. The factors of [tex]\( 7 \)[/tex] are:
[tex]\[ \pm 1, \pm 7 \][/tex]
2. Identify the leading coefficient and its factors:
The leading coefficient is [tex]\( 8 \)[/tex]. The factors of [tex]\( 8 \)[/tex] are:
[tex]\[ \pm 1, \pm 2, \pm 4, \pm 8 \][/tex]
3. Formulate all possible rational roots:
According to the Rational Zero Theorem, the possible rational roots are given by all possible values of [tex]\( \frac{p}{q} \)[/tex], where [tex]\( p \)[/tex] is a factor of the constant term ([tex]\( \pm 1, \pm 7 \)[/tex]) and [tex]\( q \)[/tex] is a factor of the leading coefficient ([tex]\( \pm 1, \pm 2, \pm 4, \pm 8 \)[/tex]).
Combining these, we get the possible rational roots:
[tex]\[ \pm 1, \pm \frac{1}{2}, \pm \frac{1}{4}, \pm \frac{1}{8}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{4}, \pm \frac{7}{8} \][/tex]
Thus, listing all the possible rational roots, we have:
[tex]\[ \{\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}\} \][/tex]
Given the multiple-choice options:
A. [tex]\(\pm 1, \pm 7\)[/tex]
B. [tex]\(\pm 1, \pm 2, \pm 4, \pm 8\)[/tex]
C. [tex]\(\pm 1, \pm 2, \pm 4, \pm 8, \pm \frac{1}{7}, \pm \frac{2}{7}, \pm \frac{4}{7}, \pm \frac{8}{7}\)[/tex]
D. [tex]\(\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}\)[/tex]
The correct answer is:
D. [tex]\(\pm 1, \pm 7, \pm \frac{1}{2}, \pm \frac{7}{2}, \pm \frac{1}{4}, \pm \frac{7}{4}, \pm \frac{1}{8}, \pm \frac{7}{8}\)[/tex]
We value your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for visiting IDNLearn.com. We’re here to provide accurate and reliable answers, so visit us again soon.
