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To determine the possible rational roots of the given polynomial function [tex]\( f(x)=10 x^6+7 x-7 \)[/tex], we will use the Rational Root Theorem, which states that any rational root of a polynomial equation [tex]\( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \)[/tex] is a fraction [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].
1. Identify the constant term and its factors:
- The constant term [tex]\( a_0 \)[/tex] in [tex]\( f(x) = 10x^6 + 7x - 7 \)[/tex] 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 [tex]\( a_n \)[/tex] in [tex]\( f(x) = 10x^6 + 7x - 7 \)[/tex] is [tex]\( 10 \)[/tex].
- The factors of [tex]\( 10 \)[/tex] are [tex]\( \pm 1, \pm 2, \pm 5, \pm 10 \)[/tex].
3. Form all possible rational roots [tex]\( \frac{p}{q} \)[/tex]:
- Possible combinations of [tex]\( \frac{p}{q} \)[/tex] are formed by dividing the factors of [tex]\( \pm 7 \)[/tex] by the factors of [tex]\( \pm 10 \)[/tex]:
- [tex]\( \frac{\pm 1}{\pm 1} = \pm 1 \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 2} = \pm \frac{1}{2} \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 5} = \pm \frac{1}{5} \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 10} = \pm \frac{1}{10} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 1} = \pm 7 \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 2} = \pm \frac{7}{2} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 5} = \pm \frac{7}{5} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 10} = \pm \frac{7}{10} \)[/tex]
4. List all possible rational roots:
- [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{5}, \pm \frac{7}{10} \)[/tex]
5. Compare the given options with possible rational roots:
- The given options are [tex]\( \frac{2}{7}, \frac{10}{7}, \frac{5}{7}, \frac{7}{10} \)[/tex].
- From the possible roots list above, [tex]\( \frac{2}{7}, \frac{10}{7}, \frac{5}{7} \)[/tex] are not possible.
- Only [tex]\( \frac{7}{10} \)[/tex] matches one of the possible rational roots.
Therefore, the possible rational root at point [tex]\( P \)[/tex] is [tex]\( \frac{7}{10} \)[/tex].
1. Identify the constant term and its factors:
- The constant term [tex]\( a_0 \)[/tex] in [tex]\( f(x) = 10x^6 + 7x - 7 \)[/tex] 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 [tex]\( a_n \)[/tex] in [tex]\( f(x) = 10x^6 + 7x - 7 \)[/tex] is [tex]\( 10 \)[/tex].
- The factors of [tex]\( 10 \)[/tex] are [tex]\( \pm 1, \pm 2, \pm 5, \pm 10 \)[/tex].
3. Form all possible rational roots [tex]\( \frac{p}{q} \)[/tex]:
- Possible combinations of [tex]\( \frac{p}{q} \)[/tex] are formed by dividing the factors of [tex]\( \pm 7 \)[/tex] by the factors of [tex]\( \pm 10 \)[/tex]:
- [tex]\( \frac{\pm 1}{\pm 1} = \pm 1 \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 2} = \pm \frac{1}{2} \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 5} = \pm \frac{1}{5} \)[/tex]
- [tex]\( \frac{\pm 1}{\pm 10} = \pm \frac{1}{10} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 1} = \pm 7 \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 2} = \pm \frac{7}{2} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 5} = \pm \frac{7}{5} \)[/tex]
- [tex]\( \frac{\pm 7}{\pm 10} = \pm \frac{7}{10} \)[/tex]
4. List all possible rational roots:
- [tex]\( \pm 1, \pm \frac{1}{2}, \pm \frac{1}{5}, \pm \frac{1}{10}, \pm 7, \pm \frac{7}{2}, \pm \frac{7}{5}, \pm \frac{7}{10} \)[/tex]
5. Compare the given options with possible rational roots:
- The given options are [tex]\( \frac{2}{7}, \frac{10}{7}, \frac{5}{7}, \frac{7}{10} \)[/tex].
- From the possible roots list above, [tex]\( \frac{2}{7}, \frac{10}{7}, \frac{5}{7} \)[/tex] are not possible.
- Only [tex]\( \frac{7}{10} \)[/tex] matches one of the possible rational roots.
Therefore, the possible rational root at point [tex]\( P \)[/tex] is [tex]\( \frac{7}{10} \)[/tex].
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